Advanced Integration 9781532394089

Citation preview

Advanced Integration

Manual Title: Advanced Integration Author: Dan Hamilton Editor: John Hamilton Cover design by: John Hamilton Copyright  2002 All rights reserved. Printed in the United States of America. No part of this manual may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the author. Request for permission or further information should be addressed to Hamilton Education Guides via [email protected]. First published in 2002 Library of Congress Catalog Card Number 20022-81883 Library of Congress Cataloging-in-Publication Data ISBN 978-1-5323-9408-9

Hamilton Education Guides Book Series

____________________________________________________________________________________

eBook and paperback versions available in the Amazon Kindle Store

Hamilton Education Guides Manual Series

____________________________________________________________________________________

eManual versions available in the Amazon Kindle Store

Hamilton Education Guides Manual Series

____________________________________________________________________________________

eManual versions available in the Amazon Kindle Store

Contents Advanced Integration

Quick Reference to Problems 1 1.1

Integration by Parts ................................................................................................... 2

1.2

Integration Using Trigonometric Substitution........................................................ 23

1.3

Integration by Partial Fractions ............................................................................... 35 Case I - The Denominator Has Distinct Linear Factors 35 Case II - The Denominator Has Repeated Linear Factors 42 Case III - The Denominator Has Distinct Quadratic Factors 49 Case IV - The Denominator Has Repeated Quadratic Factors 59

1.4

Integration of Hyperbolic Functions ........................................................................ 65

Appendix – Exercise Solutions.............................................................................................. 94 Section 1.1 94 Section 1.4 113

Hamilton Education Guides

Section 1.2 102

Section 1.3 106

i

Acknowledgments The primary motivating factor in writing the Hamilton Education Guides manual series is to provide students with specific subjects on mathematics. I am grateful to John Hamilton for his editorial comments, cover design, and suggestions on easier presentation of the topics. I would also like to acknowledge the original contributors of the Hamilton Education Guides books for their editorial reviews. Finally, I would like to thank my family for their understanding and patience in allowing me to prepare this manual.

Hamilton Education Guides

ii

Introduction and Overview It is my belief that the key to learning mathematics is through positive motivation. Students can be greatly motivated if subjects are presented concisely and the problems are solved in a detailed step by step approach. This keeps students motivated and provides a great deal of encouragement in wanting to learn the next subject or to solve the next problem. During my teaching career, I found this method to be an effective way of teaching. I hope by presenting equations in this format, more students will become interested in the subject of mathematics. This manual is a chapter from my Calculus 1 – Differentiation and Integration book with the primary focus on the subject of advanced integration. The scope of this manual is intended for educational levels ranging from the 12th grade to adult. The manual can also be used by students in home study programs, parents, teachers, special education programs, preparatory schools, and adult educational programs including colleges and universities as a supplementary manual. A fundamental understanding of basic mathematical operations such as addition, subtraction, multiplication, and division is required. This manual addresses advanced integrals and how they are simplified and mathematically operated. Students learn topics that include the integration of functions using the Integration by Parts, Trigonometric Substitution, and Partial Fractions methods; and how to find the integral of hyperbolic functions. Detailed solutions to the exercises are provided in the Appendix. Students are encouraged to solve each problem in the same detail and step by step format as shown in the text. It is my hope that all Hamilton Education Guides books and manuals stand apart in their understandable treatment of the presented subjects and for their clarity and special attention to detail. I hope readers of this manual will find it useful. With best wishes, Dan Hamilton

Hamilton Education Guides

iii

Advanced Integration

Quick Reference to Problems

Advanced Integration Quick Reference to Problems 1.1

Integration by Parts ................................................................................................... 2

∫e 1.2

2x

∫e

x

∫ x cos 3x dx

sin x dx = ;

=

Integration Using Trigonometric Substitution........................................................ 23

∫ 1.3

cos 2 x dx = ;

x 2 dx 36 − x

=;

2

dx

∫ ( 9 + x 2 )2

dx

∫ x4

=;

x 2 −1

=

Integration by Partial Fractions ............................................................................... 35 Case I - The Denominator Has Distinct Linear Factors 35

∫

x +1 dx x (x − 2 )(x + 3)

=;

1

∫ (x + 1)(x + 2)

∫

=;

dx

x 2 +1 dx = x(x − 1)(x + 1)

Case II - The Denominator Has Repeated Linear Factors 42 1

x+3

∫ x (x − 1)2 dx = ; ∫ x 2 (x − 1) dx

=;

5

∫ x (x − 1)2 dx =

Case III - The Denominator Has Distinct Quadratic Factors 49 x2 − x + 3

∫ x (x 2 + 1)

dx

=;

1

∫ x (x 2 + 25)

dx = ;

1

∫ x 2 (x 2 + 16)

dx =

Case IV - The Denominator Has Repeated Quadratic Factors 59 x2

∫ (x 2 + 1)2 1.4

dx

=;

x 2 +1

∫ (x 2 + 4)2

dx

=;

x3

∫ (x 2 + 2)2 dx

=

Integration of Hyperbolic Functions ........................................................................ 65 1

∫ cosh 5 x dx

Hamilton Education Guides

=;

∫ (sinh 4 x + cosh 2 x ) dx

=;

∫x

2

csc h 2 x 3 dx =

1

Advanced Integration The objective of this manual is to improve the student’s ability to solve additional problems involving integration. A method used to integrate functions, known as Integration by Parts, is addressed in Section 1.1. Integration of functions using the Trigonometric Substitution method is discussed in Section 1.2. Integration of functions using the Partial Fractions technique is addressed in Section 1.3. Four different cases, depending on the denominator having distinct linear factors, repeated linear factors, distinct quadratic factors, or repeated quadratic factors, are addressed in this section. Finally, integration of hyperbolic functions is discussed in Section 1.4. Each section is concluded by solving examples with practice problems to further enhance the student’s ability.

1.1

Integration by Parts

Integration by parts is a technique for replacing hard to integrate integrals by ones that are easier to integrate. This technique applies mainly to integrals that are in the form of ∫ f (x ) g (x ) dx where in most cases f (x ) can be differentiated several times to become zero and g (x ) can be integrated several times without difficulty.

∫x

For example, given the integral

2 3x

e dx the

function f (x ) = x 2 can be differentiated three times to become zero and the function g (x ) = e 3 x can be integrated several times easily. On the other hand, integrals such as

∫e

−3 x

∫e

2x

cos 2 x dx and

sin 3 x dx do not fall under the category described above. In this section we will learn how to

apply Integration by Parts method in solving various integrals. The formula for integration by parts comes from the product rule, i.e., where u and v are differentiable we obtain

d ( uv ) = u dv + v du dx dx dx functions of x , multiplying

both sides of the equation by dx

d ( uv ) = u dv + v du

rearranging the terms we then have

u dv = d ( uv ) − v du

integrating both sides of the equation we obtain

∫ u dv = uv − ∫ v du

The above formula is referred to as the Integration by Parts Formula. Note that in using the above equality we must first select dv such that it is easily integrable and second ensure that ∫ u dv is easier to evaluate than ∫ u dv . In the following examples we will solve problems using

the Integration by Parts method.

Example 1.1-1: Evaluate the following indefinite integrals: a.

∫xe

d.

∫ ln x dx

x

dx =

=

Hamilton Education Guides

b.

∫x

e dx =

c.

∫xe

e.

∫ x ln x dx =

f.

∫x

2 3x

3

−2 x

dx =

ln x dx =

2

Advanced Integration

g.

∫e

2x

1.1 Integration by Parts

cos 2 x dx =

∫e

h.

x

sin x dx =

i.

∫ x cos 3x dx =

Solutions: a. Given

∫ xe

x

dx let u = x and dv = e x dx then du = dx and

∫ dv = ∫ e

x

dx which implies v = e x .

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ xe

x

∫

dx = xe x − e x dx = xe x − e x + c

= e x ( x − 1) + c

Check: Let y = e x (x − 1) + c , then y ′ = e x ⋅ (x − 1) + 1 ⋅ e x + 0 = xe x − e x + e x = xe x − e x + e x = xe x b. Given

∫x

2 3x

e dx let u = x 2 and dv = e 3 x dx then du = 2 x dx and

∫ dv = ∫ e

3x

dx which implies v =

1 3x e . 3

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫

x 2 e 3x 2 1 3x 1 x 2 e 3 x dx = x 2 ⋅ e 3 x − − x e 3 x dx e ⋅ 2 x dx = 3 3 3 3

To integrate v=

1 3x e . 3

∫xe

3x

(1 )

∫

∫

dx let u = x and dv = e 3 x dx then du = dx and

Therefore,

∫

∫ dv = ∫ e

3x

dx which implies

xe 3 x e 3 x 1 1 3x x e 3 x dx = x ⋅ e 3 x − − +c e dx = 3 9 3 3

(2)

∫

Combining equations ( 1 ) and ( 2 ) together we obtain:

∫

x 2 e 3 x dx =

x 2 e 3x 2 − 3 3

∫

1 3

x e 3 x dx

=

 1 2 3x 2 3x 2 3x x 2 e 3 x 2  xe 3 x e 3 x x e − xe + e +c − − + c =   9 27 3 3 3 3 9 

(

+

c. Given v=−

∫xe

2 ⋅ 3e 3 x + 0 27

∫xe

−2 x

1 −2 x . e 2

−2 x

=

) (

1 2 2 2 3x 2 x ⋅ e 3 x + 3e 3 x ⋅ x 2 − 1 ⋅ e 3 x + 3e 3 x ⋅ x e + c , then y ′ = 3 9 9 27 2 2 2 3x 2 xe + x 2 e 3 x − e 3 x − xe 3 x + e 3 x = x 2 e 3 x 9 3 3 9

Check: Let y = x 2 e 3 x − xe 3 x +

dx let u = x and dv = e −2 x dx then du = dx and

∫ dv = ∫ e

−2 x

)

dx which implies

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

dx = x ⋅ −

1 −2 x 1 e + e − 2 x dx 2 2

∫

1 2

1 4

1 2

= − x e −2 x +

1 e − 2 x dx 2

Check: Let y = − x e −2 x − e −2 x + c , then y ′ = − xe − 2 x +

1 −2 x e 2

Hamilton Education Guides

∫

(

1 2

1 4

= − x e−2 x − e−2 x + c

)

1 1 1 ⋅ e − 2 x − 2e − 2 x ⋅ x − ⋅ −2e − 2 x + 0 2 4

1 2

= − e −2 x

= xe −2 x

3

Advanced Integration

1.1 Integration by Parts 1 x

d. Given ∫ ln x dx let u = ln x and dv = dx then du = dx and ∫ dv = ∫ dx which implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ ln x dx

1 = ln x ⋅ x − ∫ x ⋅ dx = x ln x − ∫ dx = x ln x − x + c x

1 x

Check: Let y = x ln x − x + c , then y ′ =  1 ⋅ ln x + ⋅ x  − 1 + 0 = ln x + 1 − 1 = ln x 

e. Given



1

1

∫ x ln x dx let u = ln x and dv = x dx then du = x dx and ∫ dv = ∫ x dx which implies v = 2 x

2

.

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain 1

1

1

∫ x ln x dx = ln x ⋅ 2 x 2 − ∫ 2 x 2 ⋅ x dx = 1 2

1 2 1 x ln x − 2 2

1 4

Check: Let y = x 2 ln x − x 2 + c , then y ′ = f. Given

∫x

3

∫ x dx

1 2 1 x ln x − x 2 + c 2 4

=

1 1 2 1  2 x ⋅ ln x + ⋅ x  − ⋅ 2 x + 0 x 2  4 1 dx x

ln x dx let u = ln x and dv = x 3 dx then du =

1 2

1 2

= x ln x + x − x = x ln x 1 4

and ∫ dv = ∫ x 3 dx which implies v = x 4 .

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain 1

1

1

∫ x ln x dx = ln x ⋅ 4 x 4 − ∫ 4 x 4 ⋅ x dx = 1 4

Check: Let y = x 4 ln x − −

1 3 x 4

1 4 x +c, 16

1 4 1 x ln x − 4 4

∫x

3

dx

=

1 4 1 4 x ln x − x +c 4 16

1 1 1 1 then y ′ =  4 x 3 ⋅ ln x + ⋅ x 4  − ⋅ 4 x 3 + 0 = x 3 ln x + x 3 4

4

 16

x

= x 3 ln x

g. Given ∫ e 2 x cos 2 x dx let u = e 2 x and dv = cos 2 x dx then du = 2e 2 x dx and ∫ dv = ∫ cos 2 x dx which 1 2

implies v = sin 2 x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we 1 2

obtain ∫ e 2 x cos 2 x dx = e 2 x ⋅ sin 2 x −

1 sin 2 x ⋅ 2e 2 x dx 2

∫

=

e 2 x sin 2 x − e 2 x sin 2 x dx 2

(1 )

∫

To integrate ∫ e 2 x sin 2 x dx let u = e 2 x and dv = sin 2 x dx then du = 2e 2 x dx and ∫ dv = ∫ sin 2 x dx which 1 2

1 2

implies v = − cos 2 x . Thus, ∫ e 2 x sin 2 x dx = e 2 x ⋅ − cos 2 x +

∫

+ e 2 x cos 2 x dx

1 cos 2 x ⋅ 2e 2 x dx 2

∫

= −

e 2 x cos 2 x 2

( 2 ) . Combining equations ( 1 ) and ( 2 ) together we obtain:

Hamilton Education Guides

4

Advanced Integration

1.1 Integration by Parts

e 2 x sin 2 x − e 2 x sin 2 x dx 2

∫

e 2 x cos 2 x dx =

+

e 2 x cos 2 x − e 2 x cos 2 x dx . 2

∫

∫

 e 2 x sin 2 x  e 2 x cos 2 x − − + e 2 x cos 2 x dx    2 2  

∫

=

∫

Check: Let y = −

e 2 x sin 2 x e 2 x cos 2 x + 2 2

(

and

)

1 2x e sin 2 x + e 2 x cos 2 x + c , 4

2 sin 2 x ⋅ e 2 x + 0 4

=

e 2 x sin 2 x 2

Taking the ∫ e 2 x cos 2 x dx from the right hand side of the equation

to the left hand side we obtain ∫ e 2 x cos 2 x dx + ∫ e 2 x cos 2 x dx = 2 e 2 x cos 2 x dx =

=

∫e

2x

cos 2 x dx =

then y ′ =

e 2 x sin 2 x e 2 x cos 2 x + 2 2

which implies

(

)

1 2x e sin 2 x + e 2 x cos 2 x + c 4

2 2x 2 2 e ⋅ sin 2 x + cos 2 x ⋅ e 2 x + e 2 x ⋅ cos 2 x 4 4 4

2 2x 4 2 e sin 2 x + e 2 x cos 2 x − e 2 x sin 2 x 4 4 4

=

4 2x e cos 2 x 4

= e 2 x cos 2 x

h. Given ∫ e x sin x dx let u = e x and dv = sin x dx then du = e x dx and ∫ dv = ∫ sin x dx which implies v = − cos x .

∫e

x

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫

(1 )

∫

sin x dx = e x ⋅ − cos x − − cos x ⋅ e x dx = − e x cos x + e x cos x dx

To integrate ∫ e x cos x dx let u = e x and dv = cos x dx then du = e x dx and ∫ dv = ∫ cos x dx which

(2)

implies v = sin x . Thus, ∫ e x cos x dx = e x ⋅ sin x − ∫ sin x ⋅ e x dx = e x sin x − ∫ e x sin x dx Combining equations ( 1 ) and ( 2 ) together we obtain:

∫e

x

∫

∫

sin x dx = − e x cos x + e x cos x dx = − e x cos x + e x sin x − e x sin x dx

Taking the ∫ e x sin x dx from the right hand side of the equation to the left hand side we obtain

∫e

x

and

∫

∫

sin x dx + e x sin x dx = − e x cos x + e x sin x which implies 2 e x sin x dx = − e x cos x + e x sin x

∫e

x

sin x dx = −

1 x 1 e cos x + e x sin x + c 2 2 1 2

1 2

1 2

1 2

1 2

1 2

Check: Let y = − e x cos x + e x sin x + c , then y ′ = − e x ⋅ cos x + sin x ⋅ e x + e x ⋅ sin x + cos x ⋅ e x 1 2

1 2

1 2

1 2

= − e x cos x + e x sin x + e x sin x + e x cos x = i. Given

∫ x cos 3x dx let u = x

Hamilton Education Guides

1 x 1 e sin x + e x sin x 2 2

= e x sin x

and dv = cos 3x dx then du = dx and ∫ dv = ∫ cos 3x dx which implies

5

Advanced Integration

v=

1 sin 3 x . 3

∫ x cos 3x dx

1.1 Integration by Parts

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain 1 3

= x ⋅ sin 3x −

1 sin 3 x dx 3

1 3

∫

=

1 1 x sin 3 x + cos 3 x + c 9 3

1 9

Check: Let y = x sin 3x + cos 3x + c , then y ′ = +

1 3 x cos 3 x − sin 3 x 3 3

1 ( 1⋅ sin 3x + cos 3x ⋅ 3 ⋅ x ) − 1 ⋅ sin 3x ⋅ 3 + 0 3 9

=

1 sin 3 x 3

= x cos 3x

Example 1.1-2: Evaluate the following indefinite integrals: a.

∫ x sec

d.

∫ arc sin 6 x dx

g.

∫

x dx 2

2

b.

∫ x sec

=

e.

=

h.

5 x dx =

arc cos

2

(x +1) dx =

x

c.

∫ 3 sin 2 x dx =

∫ 5 arc sin y dy =

1

f.

∫ arc cos x dx =

∫

arc tan 10 x dx =

i.

x ex

∫ (1 + x )2 dx =

Solutions:

∫ x sec

a. Given v=

1 tan 5 x . 5

∫ x sec

2

2

5 x dx let u = x and dv = sec 2 5 x dx then du = dx and

2

5 x dx which implies

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

1 1 1 1 x tan 5 x − ln sec 5 x + c 5 x dx = x ⋅ tan 5 x − tan 5 x dx = 25 5 5 5

∫

1 5

Check: Let y = x tan 5 x − = b. Given

∫ dv = ∫ sec

1 ln sec 5 x + c , 25

tan 5 x 5 x sec 2 5 x 5 sec 5 x tan 5 x + − 5 5 25 sec 5 x

∫ x sec

2

(x +1) dx let

u=x

then y ′ = =

(

)

sec 5 x tan 5 x 1 ⋅5 + 0 1 ⋅ tan 5 x + sec 2 5 x ⋅ 5 ⋅ x − 5 25 sec 5 x

tan 5 x tan 5 x + x sec 2 5 x − 5 5

= x sec 2 5 x

and dv = sec 2 (x + 1) dx then du = dx and ∫ dv = ∫ sec 2 (x + 1) dx

which implies v = tan (x + 1) . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ x sec

2

(x +1) dx =

x ⋅ tan (x + 1) − tan (x + 1) dx = x tan ( x + 1) − ln sec ( x + 1) + c

∫

Check: Let y = x tan (x + 1) − ln sec (x + 1) + c , then y ′ = 1 ⋅ tan (x + 1) + sec 2 (x + 1) ⋅ x −

sec (x + 1) tan (x + 1) +0 sec (x + 1)

= tan (x + 1) + x sec 2 (x + 1) − tan (x + 1) = x sec 2 (x + 1) c. Given

x

x

∫ 3 sin 2 x dx let u = 3

Hamilton Education Guides

and dv = sin 2 x dx then du =

dx 3

and ∫ dv = ∫ sin 2 x dx which implies

6

Advanced Integration

v=−

1 cos 2 x . 2

x

∫ 3 sin 2 x dx

1.1 Integration by Parts

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

=

1 6

x 1 dx 1 ⋅ − cos 2 x + cos 2 x 3 2 2 3

∫

= − x cos 2 x +

1 sin 2 x + c 12

1 1 1 sin 2 x + c , then y ′ = − ( 1 ⋅ cos 2 x − sin 2 x ⋅ 2 ⋅ x ) + cos 2 x ⋅ 2 + 0 12 12 6 1 1 1 x − cos 2 x + x sin 2 x + cos 2 x = sin 2 x 6 3 6 3 1 6

Check: Let y = − x cos 2 x + =

6 dx

d. Given ∫ arc sin 6 x dx let u = arc sin 6 x and dv = dx then du =

and ∫ dv = ∫ dx which

1 − (6 x )2

implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ arc sin 6 x dx To integrate dx = −

= −

dw 72 x

6 dx

= arc sin 6 x ⋅ x − ∫ x ⋅

∫

x dx 1 − 36 x

Therefore,

2 −1 1 1 ⋅ w 2 72 2−1

2

= −

2

∫

1 − 36 x

x dx

= x arc sin 6 x − 6∫

2

1 − 36 x

(1 )

2

use the substitution method by letting w = 1 − 36 x 2 then x dx 1 − 36 x 2

=

x

dw − w 72 x

∫

= −

1 72

(

1 1 1 2 12 1 ⋅ w = − w2 = − 1 − 36 x 2 36 72 1 36

)

∫

dw

and

−1 1− 1 1 1 1 w 2 dw = − ⋅ w 2 72 72 1 − 1 2

∫

= −

w

dw = −72 x dx

1 2

(2)

Combining equations ( 1 ) and ( 2 ) together we obtain

∫

arc sin 6 x dx = x arc sin 6 x − 6

Check: Let y = x arc sin 6 x + = arc sin 6 x + e. Given v= y. 1

x dx

∫

1 − 36 x 2

(

1 1 − 36 x 2 6

6x 1 − 36 x 2

= x arc sin 6 x +

−

)

1 2

+c ,

6x

1

)

1 2

+ c = x arc sin 6 x +

6x

then y ′ = arc sin 6 x +

1 − 36 x 2

−

1 12

(

1 1 − 36 x 2 6

72 x 1 − 36 x 2

+0

= arc sin 6 x

1 − 36 x 2

∫ 5 arc sin y dy let u = arc sin y

(

6 1 − 36 x 2 36

and dv = dy then du =

dy 1− y 2

and ∫ dv = ∫ dy which implies

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ 5 arc sin y dy =

1 1 arc sin y ⋅ y − 5 5

Hamilton Education Guides

∫ y⋅

dy 1− y

2

=

1 1 y arc sin y − 5 5

∫

y dy 1− y 2

(1 )

7

)

1 2

+c

Advanced Integration

To integrate dy = − 1 2

dw 2y

1.1 Integration by Parts

y dy

∫

1− y 2

Therefore,

1

= − ⋅ 2−1 w

2 −1 2

∫

1 2 2 1

y dy 1− y 2

y

dw − w 2y

∫

=

(

1

1

= −

= − ⋅ w 2 = − w 2 = − 1− y 2

2

dw = −2 y dy

use the substitution method by letting w = 1 − y 2 then

)

1 2

∫

dw

= −

w

−1 1 w 2 dw 2

and

1− 1 1 1 w 2 2 1− 1 2

∫

= − ⋅

1 2

(2)

Combining equations ( 1 ) and ( 2 ) together we obtain:

∫

1

1 arc sin y dy 5

1 5

Check: Let w = y arc sin y + = f. Given v=x.

1

y dy

(

)

1 1 arc sin y + 5 5

1 1− y 2 5

y 1− y 2

1

(

+c ,

1− y 2

then w′ =

)

1 1 arc sin y + 5 5

1 2

+c

y 1 − y2

1 1 ⋅ ⋅ 5 2

−

2y 1 − y2

+0

1 arc sin y 5

=

− dx

and dv = dx then du =

1− x 2

and ∫ dv = ∫ dx which implies

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

To integrate

1 2

1 2

y

1 5

−

∫ arc cos x dx let u = arc cos x

∫ arc cos x dx = arc cos x ⋅ x − ∫ x ⋅

dx = −

1

= 5 y arc sin y − 5 ∫ = y arc sin y + 1 − y 2 2 5 5 1− y

dw 2x

x dx

∫

1

2 −1 2

1− x 2

1− x 2

∫

1 2 2 1

x dx 1− x 2

=

∫

x

dw w − 2x

(

1

1

= − ⋅ w 2 = − w 2 = − 1− x 2

2

x dx

= x arc cos x + ∫

(1 )

1− x 2

use the substitution method by letting w = 1 − x 2 then

Therefore,

= − ⋅ 2−1 w

−dx

= −

)

1 2

dw

∫

w

= −

−1 1 w 2 dw 2

∫

dw = −2 x dx

and

1− 1 1 1 w 2 2 1− 1 2

= − ⋅

1 2

(2)

Combining equations ( 1 ) and ( 2 ) together we obtain:

∫

arc cos x dx = x arc cos x +

∫

(

x dx 1− x 2

Check: Let y = x arc cos x − 1 − x 2 −

x 1− x

2

+

Hamilton Education Guides

x 1− x 2

)

1 2

(

= x arc cos x − 1 − x 2 +c,

)

1 2

then y ′ = arc cos x −

+c

x 1− x

2

−

1 2

−2 x 1− x

2

+ 0 = arc cos x

= arc cos x

8

Advanced Integration

g. Given

1.1 Integration by Parts

x

x

∫ arc cos 2 dx let u = arc cos 2

implies v = x . x

x

∫

x dx 1−

2 −1 2

2

−dx 2 1−

x 2

= x arc cos +

()

x 2 2

1 2

we obtain

x dx

∫

1−

(1 )

()

x 2 2

dw 2x 1 x =− ⋅ =− dx 2 2 2

2

(2x )2

2 x

1

2 1−

use the substitution method by letting w = 1 − (2x ) then

and dx = − dw Therefore,

= − 2 ⋅ 2−1 w

and ∫ dv = ∫ dx which

(2x )2 Using the integration by parts formula ∫ u dv = u v − ∫ v du

∫ arc cos 2 dx = arc cos 2 ⋅ x − ∫ x ⋅ To integrate

− dx

and dv = dx then du =

x dx

∫

1−

=

()

x 2 2

x −2dw w x

∫

= − 2∫

−1

dw

= − 2∫ w 2 dw = − 2 ⋅

w

1 1 − 12

1− 1

w

1

1 2 2 2 1 = − 2 ⋅ w 2 = − 4w 2 = − 41 − (2x )  1  

2

(2)

Combining equations ( 1 ) and ( 2 ) together we obtain:

∫

arc cos

x x 1 dx = x arc cos + 2 2 2

Check: Let

x dx

∫

1−

x y = x arc cos − 2 1 − 2  x 2

= arc cos −

x

()

2 2 1 − 2x

+

()

()

x 2 2 1

x 22 2 

x

4

x

= x arc cos − 1 − 2 2 +c,

(2x )2 

1 2

+ c = x arc cos

x

x then y ′ = arc cos − 2

= arc cos

()

2 2 1 − 2x

2 1−

(2x )2

−

x  − 2 1 − 2 

x 1−

(2x )2

⋅−

1 2

( 2x ) 2 

1 +0 2

x 2

h. Given ∫ arc tan 10 x dx let u = arc tan 10 x and dv = dx then du =

10 dx 1 + (10 x )2

and ∫ dv = ∫ dx which

implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ arc tan 10 x dx To integrate And dx =

= arc tan 10 x ⋅ x − ∫ x ⋅ x dx

∫ 1 + (10 x )2

dw 200 x

Thus,

10 dx 1 + (10 x )

2

= x arc tan 10 x − 10∫

x dx

(1 )

1 + (10 x )2

use the substitution method by letting w = 1 + (10 x )2 then x dx

∫ 1 + (10 x )2

=

x dw

∫ w 200 x

=

1 200

∫

dw w

=

1 ln w 200

=

dw = 200 x dx

1 ln 1 + (10 x )2 200

(2)

Combining equations ( 1 ) and ( 2 ) together we obtain:

Hamilton Education Guides

9

+c

Advanced Integration

∫ arc tan 10 x dx

1.1 Integration by Parts

= x arc tan 10 x − 10∫

Check: Let y = x arc tan 10 x − x

arc tan 10 x + x ex

∫ (1 + x )2 dx let u = xe

i. Given

which implies v = − x ex

∫ (1 + x )2 =

1 + ( 10 x )

dx = xe x ⋅

1 . 1+ x

−1 − 1+ x

− xe x + e x + xe x +c 1+ x

Check: Let y =

=

2

x

x dx 1 + (10 x )

2

= x arc tan 10 x −

1 ln 1 + (10 x )2 + c , 20

−

x

1 ln 1 + (10 x ) 2 + c 20

then y ′ = arc tan 10 x +

x

1 + ( 10 x )

−

2

1 20 x +0 20 1 + ( 10 x )2

= arc tan 10 x

1 + ( 10 x )2

and dv = ( 1 + x )−2 dx then du = e x ( 1 + x ) dx and ∫ dv = ∫ ( 1 + x )−2 dx

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫

−1 ⋅ e x ( 1 + x ) dx ( 1 + x)

=

− xe x + e x dx 1+ x

∫

=

− xe x +ex +c 1+ x

=

− xe x + e x ( 1 + x ) +c 1+ x

ex +c 1+ x

ex +c , 1+ x

then y ′ =

e x ( 1 + x) − e x

( 1 + x )2

+0 =

e x + xe x − e x

=

( 1 + x )2

xe x

( 1 + x )2

Example 1.1-3: Evaluate the following indefinite integrals: b.

∫ sin

2

x dx =

c.

∫ arctan x dx =

∫ sin ( ln x ) dx =

e.

∫x

2

e x dx =

f.

∫x

∫x

h.

∫e

−x

i.

∫e

a.

∫ sin

d. g.

3

2

x dx =

cos 3 x dx =

cos x dx =

3

sin x dx =

−3 x

sin 3 x dx =

Solutions: a. Given ∫ sin 3 x dx = ∫ sin 2 x ⋅ sin x dx let u = sin 2 x and dv = sin x dx then du = 2 sin x cos x dx and

∫ dv = ∫ sin x dx which implies v = − cos x .

Using the integration by parts formula ∫ u dv = u v − ∫ v du

we obtain ∫ sin 3 x dx = sin 2 x ⋅ − cos x + ∫ cos x ⋅ 2 sin x cos x dx = − sin 2 x cos x + 2∫ cos 2 x sin x dx

(1 )

To integrate ∫ cos 2 x sin x dx use the integration by parts method again, i.e., let u = cos 2 x and dv = sin x dx then du = −2 sin x cos x dx and

∫ cos

2

∫

∫ dv = ∫ sin x dx which implies v = − cos x .

Therefore,

∫

x sin x dx = cos 2 x ⋅ − cos x − cos x ⋅ 2 sin x cos x dx = − cos 3 x − 2 cos 2 x sin x dx . Taking the

integral − 2∫ cos 2 x sin x dx from the right hand side of the equation to the left side we obtain Hamilton Education Guides

10

Advanced Integration

∫ cos

1.1 Integration by Parts

∫

2

x sin x dx + 2 cos 2 x sin x dx = − cos 3 x . Therefore,

∫ cos

1 x sin x dx = − cos 3 x 3

2

(2)

Combining equations ( 1 ) and ( 2 ) together we have

∫ sin

3

2 1 x dx = − sin 2 x cos x + 2 cos 2 x sin x dx = − sin 2 x cos x + 2 ⋅ − cos 3 x + c = − sin 2 x cos x − cos 3 x + c 3 3

∫

(

)

− 1 − cos 2 x cos x −

2 2  cos 3 x + c = − cos x +  cos 3 x − cos 3 3 3 

1  x  + c = cos 3 x − cos x + c 3 

Note that another method of solving the above problem (as was shown in Section 4.3) is in the following way:

∫ sin

3

∫ sin

x dx =

Therefore,

(

∫ sin

3

2

x ⋅ sin x dx =

x dx =

)

= − ∫ 1 − u 2 du =

∫u

2

∫ sin

2

2 ∫ (1 − cos x )⋅ sin x dx

x ⋅ sin x dx =

−1 du =

1 3

1 3 u −u +c 3

2 ∫ (1 − cos x )⋅ sin x dx

=

Check: Let y = cos 3 x − cos x + c , then y ′ =

(

let u = cos x , then

)

=

du = − sin x dx

and dx = −

du sin x

.

du 2 ∫ (1 − u )⋅ sin x ⋅ − sin x

1 cos 3 x − cos x + c 3 1 ⋅ 3 cos 2 x ⋅ − sin x + sin x + 0 = − cos 2 x sin x + sin x 3

= sin x 1 − cos 2 x = sin x sin 2 x = sin 3 x b. Given ∫ sin 2 x dx = ∫ sin x ⋅ sin x dx let u = sin x and dv = sin x dx then du = cos x dx and ∫ dv = ∫ sin x dx which implies v = − cos x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ sin

2

∫

∫

x dx = sin x ⋅ − cos x + cos x ⋅ cos x dx = − sin x cos x + cos 2 x dx = − sin x cos x +

2 ∫ (1 − sin x ) dx

= − sin x cos x + x − ∫ sin 2 x dx . Taking the integral ∫ sin 2 x dx from the right hand side of the equation to the left side we have ∫ sin 2 x dx + ∫ sin 2 x dx = − sin x cos x + x . Therefore,

∫

2 sin 2 x dx = − sin x cos x + x and

∫ sin

2

x dx = −

1 x 1 x sin x cos x + + c = − sin 2 x + + c 4 2 2 2

or, we can solve the given integral in the following way:

∫ sin

2

x dx =

1

∫ 2 ( 1 − cos 2 x ) dx 1 4

=

1 1 dx − cos 2 x dx 2 2

∫

∫

1 4

x 2

=

x 1 sin 2 x − ⋅ +c 2 2 2

1 2

1 2

Check: Let y = − sin 2 x + + c , then y ′ = − cos 2 x ⋅ 2 + + 0 = − cos 2 x + 1 2

x 2

1 2

1 2

1 2

1 4

x 2

= − sin 2 x + + c 1 1 = ( 1 − cos 2 x ) = sin 2 x or, 2 2 1 2

1 2

Let y = − sin x cos x + + c , then y ′ = − cos x cos x + sin x sin x + + 0 = − cos 2 x + sin 2 x +

Hamilton Education Guides

1 2

11

Advanced Integration

= −

1.1 Integration by Parts

(

)

1 1 1 1 1 1 1 1 − sin 2 x + sin 2 x + = − + sin 2 x + sin 2 x + 2 2 2 2 2 2 2

c. Given ∫ arctan x dx let u = arc tan x and dv = dx then du = v=x.

dx 1+ x 2

=

1 1 sin 2 x + sin 2 x 2 2

= sin 2 x

and ∫ dv = ∫ dx which implies

Substituting the integral with its equivalent value ∫ u dv = u v − ∫ v du we obtain

∫ x arctan x dx

= arc tan x ⋅ x − ∫ x ⋅

To integrate

∫ 1+ x 2

Therefore,

x dx

x dx

∫ 1+ x 2

dx 1+ x

2

= x arc tan x − ∫

x dx

(1 )

1+ x 2

use the substitution method by letting w = 1 + x 2 then

=

x dw

∫ w 2x

=

1 2

∫

dw w

=

1 ln w 2

=

dw = 2x dx

and dx =

1 ln 1 + x 2 2

dw 2x

(2)

Combining equations ( 1 ) and ( 2 ) together we obtain:

∫ x arctan x dx

= x arc tan x −

x dx

∫ 1+ x 2

1 2

= x arc tan x − ln 1 + x 2 + c

1 Check: Let y = x arc tan x − ln 1 + x 2 + c , then y ′ = arc tan x + 2

+

x 1+ x

2

−

x 1+ x 2

x 1+ x

2

−

1 2x +0 2 1+ x 2

= arc tan x

= arc tan x

d. Given ∫ sin ( ln x ) dx let u = sin ( ln x ) and dv = dx then du =

cos ( ln x ) dx x

and ∫ dv = ∫ dx which

implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ sin ( ln x ) dx = sin ( ln x )⋅ x − ∫ x ⋅

cos ( ln x ) dx x

= x sin ( ln x ) − ∫ cos ( ln x ) dx

(1 )

To integrate ∫ cos ( ln x ) dx use the integration by parts formula again, i.e., let u = cos ( ln x ) and dv = dx

then du =

− sin ( ln x ) dx x

= cos ( ln x ) ⋅ x + ∫ x ⋅

sin ( ln x ) dx x

and

∫ dv = ∫ dx

which implies v = x . Therefore, ∫ cos ( ln x ) dx

= x cos ( ln x ) + ∫ sin ( ln x ) dx

(2)

Combining equations ( 1 ) and ( 2 ) together we have

∫ sin ( ln x ) dx = x sin ( ln x ) − ∫ cos ( ln x ) dx

= x sin ( ln x ) − x cos ( ln x ) − ∫ sin ( ln x ) dx

Taking the integral − ∫ sin ( ln x ) dx from the right hand side of the equation to the left hand side we obtain ∫ sin ( ln x ) dx + ∫ sin ( ln x ) dx = x sin ( ln x ) − x cos ( ln x ) + c Therefore,

Hamilton Education Guides

12

Advanced Integration

1.1 Integration by Parts

2 sin ( ln x ) dx = x sin ( ln x ) − x cos ( ln x ) + c and thus

∫

Check: Let y =

∫x

=

x x sin ( ln x ) − cos ( ln x ) + c 2 2

sin ( ln x ) x cos ( ln x ) cos ( ln x ) x sin ( ln x ) x x sin ( ln x ) − cos ( ln x ) + c , then y ′ = + − + +0 2 2 2 2x 2 2x

sin ( ln x ) cos ( ln x ) cos ( ln x ) sin ( ln x ) sin ( ln x ) sin ( ln x ) = = sin ( ln x ) + − + + 2 2 2 2 2 2

= e. Given

∫ sin ( ln x ) dx

2

e x dx let u = x 2 and dv = e x dx then du = 2 x dx and

∫ dv = ∫ e

x

dx which implies v = e x .

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫x

2

∫

To integrate

∫ xe

Then du = dx and

∫ xe

(1 )

∫

e x dx = x 2 ⋅ e x − e x ⋅ 2 x dx = x 2 e x − 2 x e x dx

x

x

dx use the integration by parts formula again, i.e., let u = x and dv = e x dx

∫ dv = ∫ dx ∫

which implies v = e x . Therefore,

(2)

∫

dx = x ⋅ e x − e x ⋅ dx = xe x − e x dx = xe x − e x

Combining equations ( 1 ) and ( 2 ) together we have

∫x

2

(

∫

)

e x dx = x 2 e x − 2 x e x dx = x 2 e x − 2 xe x − e x + c

= x 2 e x − 2 xe x + 2e x + c

Check: Let y = x 2 e x − 2 xe x + 2e x + c , then y ′ = 2 xe x + x 2 e x − 2e x − 2 xe x + 2e x + 0 = x 2 e x

f. Given

∫x

3

v = − cos x .

∫x

3

sin x dx let u = x 3 and dv = sin x dx then du = 3 x 2 dx and

∫ dv = ∫ sin x dx which implies

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

(1 )

∫

∫

sin x dx = x 3 ⋅ − cos x + cos x ⋅ 3 x 2 dx = − x 3 cos x + 3 x 2 cos x dx

To integrate

∫x

2

cos x dx use the integration by parts formula again, i.e., let u = x 2 and dv = cos x dx

then du = 2 x dx and

∫ dv = ∫ cos x dx which implies v = sin x .

Using the integration by parts

formula ∫ u dv = u v − ∫ v du we obtain

∫x

2

∫

(2)

∫

cos x dx = x 2 ⋅ sin x − sin x ⋅ 2 x dx = x 2 sin x − 2 x sin x dx

To integrate

∫ x sin x dx

then du = dx and

use the integration by parts formula again, i.e., let u = x and dv = sin x dx

∫ dv = ∫ sin x dx which implies v = − cos x .

Hamilton Education Guides

Using the integration by parts

13

Advanced Integration

1.1 Integration by Parts

formula ∫ u dv = u v − ∫ v du we obtain

(3 )

∫ x sin x dx = x ⋅ − cos x + ∫ cos x ⋅ dx = − x cos x + sin x Combining equations ( 1 ) , ( 2 ) and ( 3 ) together we have

∫x

3

sin x dx = − x 3 cos x + 3 x 2 cos x dx = − x 3 cos x + 3 x 2 sin x − 6 x sin x dx = − x 3 cos x + 3 x 2 sin x

∫

∫

+ 6 x cos x − 6 sin x + c

(

Check: Let y = − x 3 cos x + 3x 2 sin x + 6 x cos x − 6 sin x + c , then y ′ = − 3x 2 cos x + x 3 sin x

(

)

+ 6 x sin x + 3 x 2 cos x + (6 cos x − 6 x sin x ) − 6 cos x + 0

g. Given

∫x

2

implies v =

∫x

2

= x 3 sin x

cos 3 x dx let u = x 2 and dv = cos 3 x dx then du = 2 x dx and sin 3 x 3

∫ dv = ∫ cos 3x dx which

. Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

cos 3 x dx = x 2 ⋅

To integrate

sin 3 x sin 3 x − ⋅ 2 x dx 3 3

∫

=

2 1 2 x sin 3 x − 3 3

(1 )

∫ x sin 3x dx

∫ x sin 3x dx use the integration by parts formula again, i.e., let u = x

then du = dx and

)

∫ dv = ∫ sin 3x dx which implies v =

− cos 3 x . 3

and dv = sin 3x dx

Using the integration by parts

formula ∫ u dv = u v − ∫ v du we obtain

∫ x sin 3x dx = x ⋅

− cos 3 x cos 3 x ⋅ dx + 3 3

∫

1 3

= − x cos 3x +

1 cos 3 x dx 3

(2)

∫

Combining equations ( 1 ) and ( 2 ) together we have

∫x =

2

cos 3 x dx =

1 2 2 x sin 3 x − 3 3

∫ x sin 3x dx

=

1 2 2 1 2 1 x sin 3 x − ⋅ − x cos 3 x − ⋅ cos 3 x dx 3 3 3 3 3

∫

1 2 2 2 x sin 3 x + x cos 3 x − sin 3 x + c 3 9 27 1 3

2 9

(

)

2 1 sin 3 x + c , then y ′ = 2 x sin 3 x + 3 x 2 cos 3 x 27 3 6 2 2 2 2 2 3 + ( cos 3 x − 3 x sin 3 x ) − cos 3 x + c = x sin 3 x + x 2 cos 3 x + cos 3 x − x sin 3 x − cos 3 x 9 27 3 9 3 3 9

Check: Let y = x 2 sin 3x + x cos 3x −

=

3 2 x cos 3 x 3

= x 2 cos 3x

h. Given ∫ e − x cos x dx let u = cos x and dv = e − x dx then du = − sin x dx and ∫ dv = ∫ e − x dx which Hamilton Education Guides

14

Advanced Integration

1.1 Integration by Parts

implies v = −e − x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫e

−x

∫

(1 )

∫

cos x dx = cos x ⋅ −e − x − e − x ⋅ sin x dx = − e − x cos x − e − x sin x dx

To integrate ∫ e − x sin x dx use the integration by parts formula again, i.e., let u = sin x and then du = cos x dx and

dv = e − x dx

∫e

−x

∫ dv = ∫ e

−x

dx which implies v = −e − x . Therefore,

∫

(2)

∫

sin x dx = sin x ⋅ −e − x + e − x ⋅ cos x dx = − e − x sin x + e − x cos x dx

Combining equations ( 1 ) and ( 2 ) together we have

∫e

−x

∫

∫

cos x dx = − e − x cos x − e − x sin x dx = − e − x cos x + e − x sin x − e − x cos x dx

Taking the integral

∫e

−x

cos x dx from the right hand side of the equation to the left hand side

we obtain ∫ e − x cos x dx + ∫ e − x cos x dx = − e − x cos x + e − x sin x therefore

∫

2 e − x cos x dx = − e − x cos x + sin x e − x and thus

Check: Let y =

=

1 −x e 2

∫e

−x

cos x dx = −

1 −x 1 e cos x + e − x sin x + c 2 2

(

) (

)

− e− x cos x e− x sin x 1 1 + + c , then y ′ = − − e − x cos x − e − x sin x + − e − x sin x + e − x cos x + 0 2 2 2 2 1 −x 1 −x 1 −x 1 −x 1 −x cos x + e sin x − e sin x + e cos x = e cos x + e cos x = e − x cos x 2 2 2 2 2

i. Given ∫ e −3 x sin 3x dx let u = sin 3x and dv = e −3 x dx then du = 3 cos 3x dx and ∫ dv = ∫ e −3 x dx which 1 3

implies v = − e −3 x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫e

−3 x

1 1 1 −3 x sin 3 x dx = sin 3 x ⋅ − e −3 x + ⋅ 3 cos 3 x dx = − e −3 x sin 3 x + e −3 x cos 3 x dx e 3 3 3

∫

To integrate dv = e −3 x dx

∫e

−3 x

∫e

−3 x

∫

(1 )

cos 3 x dx use the integration by parts formula again, i.e., let u = cos 3 x and

then du = − 3 sin 3x and

∫ dv = ∫ e

−3 x

1 dx which implies v = − e −3 x . Therefore, 3

1 1 1 −3 x cos 3 x dx = cos 3 x ⋅ − e −3 x − e ⋅ 3 sin 3 x dx = − e −3 x cos 3 x − e −3 x sin 3 x dx 3 3 3

∫

∫

(2)

Combining equations ( 1 ) and ( 2 ) together we have

∫e

−3 x

1 1 1 sin 3 x dx = − e −3 x sin 3 x + e −3 x cos 3 x dx = − e −3 x sin 3 x − e −3 x cos 3 x − e −3 x sin 3 x dx 3 3 3

Taking the integral

∫

∫e

Hamilton Education Guides

−3 x

∫

sin 3 x dx from the right hand side of the equation to the left hand side

15

Advanced Integration

1.1 Integration by Parts 1 3

1 3

we obtain ∫ e −3 x sin 3x dx + ∫ e −3 x sin 3x dx = − e −3 x sin 3x − e −3 x cos 3x therefore 1 1 2 e −3 x sin 3 x dx = − e −3 x sin 3 x − e −3 x cos 3 x and thus 3 3

∫

1 6

1 6

−3 x

sin 3 x dx = −

1 1 −3 x e sin 3 x − e − 3 x cos 3 x 6 6

(

)

1 − 3e −3 x sin 3 x + 3e −3 x cos 3 x 6 1 −3 x 1 1 1 sin 3 x − e −3 x cos 3 x + e −3 x cos 3 x + e −3 x sin 3 x e 2 2 2 2

Check: Let y = − e − 3 x sin 3x − e − 3 x cos 3x , then y ′ = −

(

∫e

)

1 − 3e −3 x cos 3 x − 3e −3 x sin 3 x + 0 = 6 1 −3 x 1 sin 3 x + e −3 x sin 3 x = e −3 x sin 3 x e 2 2

−

=

Example 1.1-4: Evaluate the following indefinite integrals: x

=

b.

∫ (x − 3)(3x − 1) dx = 3

c.

∫ (x + 1) csc

x dx =

e.

∫x

x − 5 dx =

f.

∫ ln (x

h.

∫ cos

−1

i.

∫ tan

a.

∫ 3 (x + 1)

d.

∫ x sec

g.

∫5e

1

x

2

4

dx

=

sin 3 x dx

3 x dx =

2

−1

2

x dx =

)

+1 dx = 5 x dx =

Solutions: a. Given

x

∫ 3 (x + 1)

4

dx =

1 x (x + 1)4 dx 3

∫

let u = x and dv = (x + 1)4 dx then du = dx and ∫ dv = ∫ (x + 1)4 dx

1 5

which implies v = (x + 1)5 . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain 1 x (x + 1)4 dx 3

∫

=

Check: Let y = +

x (x + 1)5 1 ⋅ − 3 5 3⋅5

x (x + 1)5 5 ( ) + ⋅ dx x 1 = ∫ 15

−

1 1 ⋅ (x + 1)5+1 + c 15 6

[

=

x ( x + 1) 5 1 ( x + 1) 6 + c − 15 90

]

5 x (x + 1)5 1 (x + 1)6 + c , then y ′ = 1 (x + 1)5 + 5 x(x + 1)4 − 6 (x + 1)5 + 0 = (x + 1) − 15 90 15 15 90

5 5x (x + 1)4 − (x + 1) = 5 x (x + 1)4 = x (x + 1)4 15 15 3 15

b. Given ∫ (x − 3)(3x − 1)3 dx let u = x − 3 and dv = (3x − 1)3 dx then du = dx and ∫ dv = ∫ (3x − 1)3 dx which implies v =

1 (3x − 1)4 . 12

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

(3x − 1) 3 ∫ (x − 3)(3x − 1) dx = (x − 3)⋅

4

12

=

( x − 3)(3 x − 1) 4 12

−

−

1 12

(x − 3) (3x − 1)4 4 ( ) 3 1 x − dx = ∫ 12

−

1 1 ⋅ (3 x − 1)4+1 + c 12 15

1 (3 x − 1) 5 + c 180

Hamilton Education Guides

16

Advanced Integration

Check: Let y = =

1.1 Integration by Parts

(x − 3)(3x − 1)4 12

−

[

]

1 (3x − 1)5 + c , then y ′ = 1 (3x − 1)4 + 12(x − 3)(x + 1)3 − 15 (3x − 1)4 + 0 180 12 180

3 1 (3x − 1)4 + 12(x − 3)(x + 1) − 1 (3x − 1)4 = (x − 3)(x + 1)3 12 12 12

c. Given ∫ (x + 1) csc 2 x dx let u = x + 1 and dv = csc 2 x dx then du = dx and ∫ dv = ∫ csc 2 x dx which implies v = − cot x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ (x + 1) csc

2

x dx = (x + 1) ⋅ − cot x + cot x dx = − ( x + 1) cot x + ln sin x + c

∫

[

]

Check: Let y = −(x + 1) cot x + ln sin x + c , then y ′ = − cot x − (x + 1) csc 2 x +

cos x + 0 = − cot x sin x

+ (x + 1) csc 2 x + cot x = (x + 1) csc 2 x

∫ x sec

d. Given

2

x dx let u = x and dv = sec 2 x dx then du = dx and

v = tan x . Using the integration by parts formula

∫ x sec

2

∫ dv = ∫ sec

∫ u dv = u v − ∫ v du

2

x dx which implies

we obtain

∫

x dx = x ⋅ tan x − tan x dx = x tan x − ln sec x + c

Check: Let y = x tan x − ln sec x + c , then y ′ = tan x + x sec 2 x −

sec x tan x + 0 = tan x + x sec 2 x − tan x sec x

= x sec 2 x

e. Given v=

∫x −

∫x

x − 5 dx let u = x and dv = x − 5 dx then du = dx and

3 2 ( x − 5) 2 3

∫ dv = ∫

x − 5 dx which implies

. Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

x − 5 dx = x ⋅

5 2 2 ⋅ ( x − 5) 2 + c 3 5

3 3 2 (x − 5) 2 − 2 (x − 5) 2 dx 3 3

∫

=

=

3 3 2 2 1 (x − 5) 2 +1 + c x ( x − 5) 2 − ⋅ 3 3 1+ 3 2

=

3 2 x ( x − 5) 2 3

3 5 4 2 x ( x − 5) 2 − ( x − 5) 2 + c 3 15

3 3 3 5 1 2 2 3 4 5 2 4 x (x − 5) 2 − (x − 5) 2 + c , then y ′ = (x − 5) 2 + ⋅ x(x − 5) 2 − ⋅ (x − 5) 2 + 0 3 3 2 15 2 15 3 1 3 3 1 2 2 = ( x − 5) 2 + x ( x − 5) 2 − ( x − 5) 2 = x ( x − 5) 2 = x x − 5 3 3

Check: Let y =

(

)

(

)

f. Given ∫ ln x 2 +1 dx let u = ln x 2 + 1 and dv = dx then du = v=x.

2x 2

x +1

dx

and ∫ dv = ∫ dx which implies

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

Hamilton Education Guides

17

Advanced Integration

∫ (

)

1.1 Integration by Parts

(

) ∫

2x

ln x 2 +1 dx = ln x 2 + 1 ⋅ x − x ⋅

(

)

= x ln x 2 + 1 − 2∫ dx + 2∫

1 2

x +1

(

2

x +1

dx

(

)

= x ln x 2 + 1 − 2∫

(

(

x2

)

= x ln x 2 + 1 − 2 x + 2 tan −1 x + c

dx

(

)

)

Check: Let y = x ln x 2 + 1 − 2 x + 2 tan −1 x + c , then y ′ = 1 ⋅ ln x 2 + 1 +

(

)

= ln x 2 + 1 +

g. Given

2x 2 2

x +1

1 e x sin 3 x dx 5

∫

−2+

) ∫

 1   dx dx = x ln x 2 + 1 − 2 1 − 2 x +1  x +1 2

(

2

)

= ln x 2 + 1 +

2

x +1

2x 2 − 2x 2 − 2 + 2 2

x +1

2x 2

x +1

⋅x−2+

(

)

2 1+ x 2

= ln x 2 + 1 +

+0

(

0

)

= ln x 2 + 1

2

x +1

let u = e x and dv = sin 3x dx then du = e x dx and ∫ dv = ∫ sin 3x dx which

1 3

implies v = − cos 3x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain 1 e x sin 3 x dx 5

∫

=

1 x 1 1 e ⋅ − cos 3 x + cos 3 x ⋅ e x dx 5 3 15

∫

= −

1 x 1 e cos 3 x + e x cos 3 x dx 15 15

(1 )

∫

To integrate ∫ e x cos 3x dx use the integration by parts method again, i.e., let u = e x and dv = cos 3 x dx

∫e

x

then du = e x dx and

∫ dv = ∫ cos 3x dx

1 3

which implies v = sin 3x . Therefore,

1 1 1 1 cos 3 x dx = e x ⋅ sin 3 x − sin 3 x ⋅ e x dx = e x sin 3 x − e x sin 3 x dx 3 3 3 3

∫

(2)

∫

Combining equations ( 1 ) and ( 2 ) together we have 1 e x sin 3 x dx 5

∫

= −

1 x 1 e x cos 3 x dx e cos 3 x + 15 15

Taking the integral − obtain = −

∫

1 e x sin 3 x dx 45

∫

1 1 e x sin 3 x dx + e x sin 3 x dx 5 45

∫

∫

1 x 1 x e cos 3 x + e sin 3 x + c 15 45

= −

1 1 x 1 x e x sin 3 x dx e sin 3 x − e cos 3 x + 45 45 15

∫

from the right hand side of the equation to the left side we = −

1 x 1 x e cos 3 x + e sin 3 x + c . 15 45

which implies

∫e

x

sin 3 x dx = −

Therefore,

2 e x sin 3 x dx 9

∫

3 x 1 x e cos 3 x + e sin 3 x + c 10 10

3 x 1 3 3 1 e cos 3 x + e x sin 3 x + c , then y ′ = − e x ⋅ cos 3 x + sin 3 x ⋅ 3 ⋅ e x + e x ⋅ sin 3 x 10 10 10 10 10 9 3 9 1 3 1 e x sin 3 x + cos 3 x ⋅ 3 ⋅ e x + 0 = − e x cos 3 x + e x sin 3 x + e x sin 3 x + e x cos 3 x = 10 10 10 10 10 10 1 10 x 9 +1 x + e x sin 3 x = e sin 3 x = e x sin 3 x e sin 3 x = 10 10 10

Check: Let y = −

h. Given ∫ cos −1 3x dx let u = cos −1 3x and dv = dx then du = −

Hamilton Education Guides

3 1 − 9x 2

dx and

∫ dv = ∫ dx

which

18

Advanced Integration

1.1 Integration by Parts

implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ cos

−1

∫

To integrate

implies dx = − 1

1 6

3x 1 − 9x

dw 18 x

= − ⋅ 2w 2 = −

2

∫ tan

−1

1 − 9x

2

3x

= x cos −1 3x + ∫

dx

1 − 9x 2

dx use the substitution method by letting w = 1 − 9 x 2 then

. Thus,

1 − 9x 2 3

3x

∫

dx =

1 − 9x 2

and

∫

∫

3x w

⋅−

dw 18 x

cos −1 3 x dx = x cos −1 3 x +

1 − 9x 2

Check: Let y = x cos −1 3x −

i. Given

3 dx

∫

3 x dx = cos −1 3 x ⋅ x + x ⋅

3

= −

1 6

= −

w

3x

∫

3x 1 − 9x 5

2

dx

1 + 25 x 2

1 6

∫

1

dw = −

1

w2

+

18 x 6 1 − 9x

2

which

−1 1 w 2 dw 6

∫

1 − 9 x2 +c 3

dx = x cos−1 3 x −

1 − 9x 2

+ c , then y ′ = cos −1 3 x −

5 x dx let u = tan −1 5 x and dv = dx then du =

∫

dw

dw = −18 x dx

+ 0 = cos −1 3 x

and ∫ dv = ∫ dx which

implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ tan

−1

∫

5 x dx = tan −1 5 x ⋅ x − x ⋅ 5x

To integrate

∫ 1 + 25x 2 dx

implies dx =

dw 50 x

∫ tan

−1

. Thus,

5 dx 1 + 25 x

2

= x tan −1 5 x − ∫

5x 1 + 25 x 2

dx

use the substitution method by letting w = 1 + 25 x 2 then 5x

5 x dw

5x

= x tan−1 5 x −

∫ 1 + 25x 2 dx = ∫ w ⋅ 50 x

5 x dx = x tan −1 5 x −

Check: Let y = x tan −1 5 x −

∫ 1 + 25x 2 dx

=

1 10

∫

dw w

=

1 ln w 10

=

dw = 50 x dx

1 ln 1 + 25 x 2 10

which

and

1 ln 1 + 25 x 2 + c 10

5x 1 50 x 1 − + 0 = tan −1 5 x ln 1 + 25 x 2 + c , then y ′ = tan −1 5 x + 2 10 2 10 1 + 25 x 1 + 25 x

Example 1.1-5: Evaluate the following indefinite integrals: a.

∫ sinh

d.

∫ cos 5x cos 7 x dx

−1

5 x dx =

=

b.

∫ x tan

e.

∫5e

−1

x dx =

x −x dx

=

c.

∫ sin x sin 7 x dx =

f.

∫ x sinh 3x dx =

Solutions: a. Given ∫ sinh −1 5 x dx let u = sinh −1 5 x and dv = dx then du =

5 1 + 25 x 2

dx and

∫ dv = ∫ x dx which

implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain Hamilton Education Guides

19

Advanced Integration

∫ sinh

−1

1.1 Integration by Parts

1 + 25 x

dw = 50 x dx which implies dx = 1 10

∫

1 1 w2

dw =

5 x dx

= x sinh −1 5 x − ∫

1 + 25 x 2

5 x dx

∫

To get the integral of

=

5 dx

∫

5 x dx = sinh −1 5 x ⋅ x − x ⋅

(1 )

1 + 25 x 2

use the substitution method by letting w = 1 + 25 x 2 then

2

dw 50 x

. Therefore,

∫

5 x dx 1 + 25 x

=

2

∫

5x

dw 50 x w ⋅

=

5 50

∫

1 w

dw

−1 1− 1 1 2 12 1 1 1 1 1 w 2 dw = w = w2 = w 2 = ⋅ 1 + 25 x 2 10 10 5 10 1 − 1 5 2

(2)

∫

Combining equations ( 1 ) and ( 2 ) together we have

∫

sinh −1 5 x dx = x sinh −1 5 x −

Check: Let y = x sinh −1 5 x − sinh −1 5 x +

b. Given

∫ x tan

−1

5x 1 + 25 x

2

5 x dx

∫

1 + 25 x 2

(

)

1 1 + 25 x 2 5 −

= x sinh −1 5 x −

1 2

5x 1 + 25 x

(

1 1 + 25 x 2 5

+ c , then y ′ = sinh −1 5 x +

)

1 2

+c

5x 1 + 25 x 2

−

50 x 10 1 + 25 x 2

+0

= sinh −1 5 x

2

x dx let u = tan −1 x and dv = x dx then du =

1 1+ x 2

dx

and ∫ dv = ∫ x dx which

1 2

implies v = x 2 . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ =

x tan −1 x dx = tan −1 x ⋅

x2 1 − 2 2

∫

x2 ⋅

1 1 2 1 1 dx x tan −1 x − dx + 2 1+ x 2 2 2

∫

Check: Let y =

∫

dx 1+ x

=

2

=

1 2 x2 1 1 1 x tan −1 x − dx = x 2 tan −1 x − 2 2 1+ x 2 2 2

∫





1

∫ 1 − 1 + x 2 dx

1 1 1 2 x tan −1 x − x + tan −1 x + c 2 2 2

1 1 x2 1 1 1 1 2 1 1 − + ⋅ +0 x tan −1 x − x + tan −1 x + c , then y ′ = ⋅ 2 x ⋅ tan −1 x + 2 2 1+ x 2 2 2 1+ x 2 2 2 2

1 x2 1 1 1 1 x 2 +1 1 1 1 x tan −1 x + ⋅ + ⋅ − = x tan −1 x + ⋅ − = x tan −1 x + − = x tan −1 x 2 1+ x 2 2 1+ x 2 2 2 1+ x 2 2 2 2

c. Given ∫ sin x sin 7 x dx let u = sin x and dv = sin 7 x dx then du = cos x dx and ∫ dv = ∫ sin 7 x dx which 1 7

implies v = − cos 7 x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain 1

1

1

1

∫ sin x sin 7 x dx = sin x ⋅ − 7 cos 7 x + 7 ∫ cos 7 x ⋅ cos x dx = − 7 sin x cos 7 x + 7 ∫ cos x ⋅ cos 7 x dx

(1 )

To integrate ∫ cos x ⋅ cos 7 x dx use the integration by parts method again, i.e., let u = cos x and Hamilton Education Guides

20

Advanced Integration

1.1 Integration by Parts 1

dv = cos 7 x dx then du = − sin x dx and 1

∫ dv = ∫ cos 7 x dx which implies v = 7 sin 7 x .

Therefore,

1 1 sin x sin 7 x dx cos x sin 7 x + 7 7

(2)

1

∫ cos x ⋅ cos 7 x dx = cos x ⋅ 7 sin 7 x + 7 ∫ sin 7 x ⋅ sin x dx =

∫

Combining equations ( 1 ) and ( 2 ) together we have 1

1

1

1

1

∫ sin x sin 7 x dx = − 7 sin x cos 7 x + 7 ∫ cos x ⋅ cos 7 x dx = − 7 sin x cos 7 x + 49 cos x sin 7 x + 49 ∫ sin x sin 7 x dx Taking the integral

∫ sin x sin 7 x dx

1 sin x sin 7 x dx 49

∫

to the left hand side and simplifying we have

1 49 1 49 sin x cos 7 x + ⋅ cos x sin 7 x + c 7 48 49 48

= − ⋅

Check: Let y = −

= −

7 1 sin x cos 7 x + cos x sin 7 x + c 48 48

7 49 1 7 cos x ⋅ cos 7 x + sin 7 x ⋅ sin x cos x sin 7 x + c , then y ′ = − sin x cos 7 x + 48 48 48 48

−

49 1 7 1 49 − 1 sin 7 x ⋅ sin x − sin x ⋅ sin 7 x = sin 7 x ⋅ sin x cos 7 x ⋅ cos x + 0 = sin x ⋅ sin 7 x + 48 48 48 48 48

=

48 sin 7 x ⋅ sin x = sin 7 x sin x 48

d. Given ∫ cos 5 x cos 7 x dx let u = cos 5 x and dv = cos 7 x dx then du = −5 sin 5 x dx and ∫ dv = ∫ cos 7 x dx 1 7

which implies v = sin 7 x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ cos 5x cos 7 x dx

1 7

= cos 5 x ⋅ sin 7 x +

5 sin 7 x ⋅ sin 5 x dx 7

∫

=

5 1 sin 5 x ⋅ sin 7 x dx cos 5 x sin 7 x + 7 7

(1 )

∫

To integrate ∫ sin 5 x ⋅ sin 7 x dx use the integration by parts method again, i.e., let u = sin 5 x and dv = sin 7 x dx

then du = 5 cos 5 x dx and 1

∫ dv = ∫ sin 7 x dx

1 7

which implies v = − cos 7 x . Therefore, 1

5

5

∫ sin 5x ⋅ sin 7 x dx = sin 5x ⋅ − 7 cos 7 x + 7 ∫ cos 7 x ⋅ cos 5x dx = − 7 sin 5x cos 7 x + 7 ∫ cos 5x cos 7 x dx

(2)

Combining equations ( 1 ) and ( 2 ) together we have

∫ cos 5x cos 7 x dx =

1 5 5 1 25 sin 5 x cos 7 x + cos 5 x cos 7 x dx cos 5 x sin 7 x + sin 5 x ⋅ sin 7 x dx = cos 5 x sin 7 x − 7 49 7 7 49

∫

Taking the integral

∫ cos 5x cos 7 x dx Check: Let y =

=

25 cos 5 x cos 7 x dx 49

∫

∫

to the left hand side and simplifying we have

5 49 1 49 ⋅ sin 5 x cos 7 x + c ⋅ cos 5 x sin 7 x − 49 24 7 24

=

7 5 cos 5 x sin 7 x − sin 5 x cos 7 x + c 24 24

7 5 35 49 cos 5 x sin 7 x − sin 5 x cos 7 x + c , then y ′ = − sin 5 x ⋅ sin 7 x + cos 7 x ⋅ cos 5 x 24 24 24 24

Hamilton Education Guides

21

Advanced Integration

e. Given

−

49 − 25 49 25 35 25 cos 5 x cos 7 x − cos 5 x cos 7 x = cos 5 x cos 7 x sin 5 x ⋅ sin 7 x + 0 = cos 5 x ⋅ cos 7 x + 24 24 24 24 24

=

24 cos 5 x cos 7 x = cos 5 x cos 7 x 24 x −x dx

∫5e

x −x e dx 5

=

1 5

x 5

and dv = e − x dx then du = dx and ∫ dv = ∫ e − x dx which implies

x dx ⋅ −e − x + e − x ⋅ 5 5

∫

Check: Let y = −

f. Given

let u =

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

v = −e − x .

∫

1.1 Integration by Parts

= −

xe − x e − x − +c 5 5

= −

e− x ( x + 1) + c 5

e − x xe − x e − x xe − x e − x xe − x + + +0 = − + c , then y ′ = − 5 5 5 5 5 5

∫ x sinh 3x dx let u = x

and dv = sinh 3x dx then du = dx and ∫ dv = ∫ sinh 3x dx which

1 3

implies v = cosh 3x dx . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ x sinh 3x dx

1 3

1 3

= x ⋅ cosh 3x − ∫ cosh 3x ⋅ dx =

1 1 x cosh 3 x − cosh 3 x ⋅ dx 3 3

∫

1 9

1 3

Check: Let y = x cosh 3x − sinh 3x + c , then y ′ = 1 + x sinh 3 x − cosh 3 x 3

=

1 1 x cosh 3 x − sinh 3 x + c 9 3

1 1 1 1 cosh 3 x + ⋅ 3 x sinh 3 x − ⋅ 3 cosh 3 x + 0 = cosh 3 x 3 9 3 3

= x sinh 3x

Section 1.1 Practice Problems – Integration by Parts 1. Evaluate the following integrals using the integration by parts method.

d.

∫ xe dx = ∫ x sin 5x dx =

g.

∫ cos ( ln x ) dx

j.

∫xe

a.

4x

− ax

=

dx =

x

f.

∫ ( 5 − x ) e dx = 3 3x ∫ x e dx =

=

i.

∫ ln x dx =

sin 3 x dx =

l.

∫e

c.

∫ arc tan x dx = −2 x ∫ e cos 3x dx =

b.

∫ 2 cos x dx

e.

∫x

h.

∫ 3 tan

k.

∫e

=

c.

3 − x dx =

x

x

−1

x dx

5x

5

x

cos 5 x dx =

2. Evaluate the following integrals using the integration by parts method. b.

d.

∫ x sec x dx = 3 ∫ sin 5x dx =

e.

∫ arc sin 3 y dy = 2 ∫ x cos x dx =

g.

∫ x ( 5x − 1)

h.

∫ x csc

2

x dx =

i.

∫ 3 cos

j.

∫ sinh

k.

∫ x sec

2

10 x dx =

l.

∫ 5 sinh 7 x dx

a.

2

−1

3

dx =

x dx =

Hamilton Education Guides

f.

2 x

−1

5 x dx

= =

22

Advanced Integration

1.2

1.2 Integration Using Trigonometric Substitution

Integration Using Trigonometric Substitution

Many integration problems involve radical expressions of the form a2 − b2 x2

a2 + b2 x2

b2 x2 − a2

In such instances we can use a trigonometric substitution by letting x=

a sin t b

x=

respectively to obtain

(1− sin t )

= a cos 2 t = a cos t

tan 2 t

= a 2 + a 2 tan 2 t = a

(1+ tan t )

= a sec 2 t = a sec t

sec 2 t − a 2

= a 2 sec 2 t − a 2 = a

( sec

= a tan 2 t = a tan t

a2 + b2 x2 =

a2 + b2 ⋅

b2 x2 − a2 =

b2 ⋅

b

a sec t b

= a 2 − a 2 sin 2 t = a

a2 − b2 ⋅

2

x=

sin 2 t

a2 − b2 x2 =

a2

a tan t b

a2 b

2

a2 b2

2

2

2

)

t −1

Notice that using trigonometric substitution result in elimination of the radical expression. This in effect reduces the difficulty of solving integrals with radical expressions. Reminder 1:

bx a b a a −1 b x x = tan t , then t = tan b a a −1 b x x = sec t , then t = sec b a

Given x = sin t , then t = sin −1

for −1 ≤ x ≤ 1 and −

Given

for all x and −

Given

π 2

π 2 t

≤t ≤

π 2

π 2

for x ≥ 1 or x ≤ −1 and 0 ≤ t

π 2

or π ≤ t

3π 2

Reminder 2: In solving this class of integrals the integrand in the original variable may be obtained by the use of a right triangle. For example, in a right triangle •

sin t =

opposite x = . hypotenuse a

Therefore, using the Pythagorean theorem, the adjacent side (w) is equal to

a = x 2 + w2 ; a 2 = x 2 + w2 ; w2 = a 2 − x 2 ; w = a 2 − x 2

•

cos t =

adjacent x = hypotenuse a

. Therefore, the opposite side (w) is equal to

a = x 2 + w2 ; a 2 = x 2 + w2 ; w2 = a 2 − x 2 ; w = a 2 − x 2

•

tan t =

opposite x = adjacent a

. Therefore, the hypotenuse (w) is equal to w = a 2 + x 2

•

cot t =

adjacent x = opposite a

. Therefore, the hypotenuse (w) is equal to w = a 2 + x 2

Hamilton Education Guides

23

Advanced Integration

•

sec t =

1.2 Integration Using Trigonometric Substitution

hypotenuse x = adjacent a

. Therefore, the opposite side (w) is equal to

x = a 2 + w2 ; x 2 = a 2 + w2 ; w2 = x 2 − a 2 ; w = x 2 − a 2

•

csc t =

hypotenuse x = opposite a

. Therefore, the adjacent side (w) is equal to

x = a 2 + w2 ; x 2 = a 2 + w2 ; w2 = x 2 − a 2 ; w = x 2 − a 2

Let’s integrate some integrals using the above trigonometric substitution method: Example 1.2-1: Use trigonometric substitution to evaluate the following indefinite integrals: a. d.

∫ x2

dx 4 − x2

dx

∫ ( 9 + x 2 )2

=

b.

=

e.

x 2 dx

∫

c.

∫

dx =

f.

∫ x4

25 − x 2 x2

∫

2

x −1

1+ x 2

=

x2

dx =

dx x 2 −1

=

Solutions: a. Given

∫ x2

(

dx 4− x

let x = 2 sin t , then dx = 2 cos t dt and 4 − x 2 = 4 − 4 sin 2 t = 4 1 − sin 2 t

2

)

= 4 cos 2 t = 2 cos t . Substituting these values back into the original integral we obtain:

∫ x2 =

dx 4 − x2

=

1 cos t − +c 4 sin t

2 cos t dt

∫ (2 sin t )2 ⋅ 2 cos t

=

4− x 2 2 x 2

1 − 4

+c

=

2 cos t

∫ 4 sin 2 t ⋅ 2 cos t dt

=

1

∫ 4 sin 2 t dt

 1 4− x2 1  4 − x 2 ⋅ 2/  −  + c =  x 4 2/ ⋅ x   

= −  4

1 csc 2 t dt 4

∫

=

1 4

= − cot t + c

 +  c 

Check: To check the answer we start with the solution and find its derivative. The derivative should match with the integrand, i.e., the algebraic expression inside the integral. Note that not all the steps in finding the derivative is given. At this level, it is expected that students are able to work through the details that are not shown (review differentiation techniques described in Chapters 2 and 3 in Calculus 1 book). Let y = −

1 4

4− x x

2

+c ,

1 then y ′ = −

=

4− x 2 2

Hamilton Education Guides

x

=

2 4− x

1 − 4

(

4− x 2 2

x

2

) = −

2 − 4−1 x 1 2 4− x − 4 x2

−2 x 2

⋅ x − 1⋅ 4 − x 2 x2

4

− x 2 − 4− x 2

− x 2 − 4− x 2 ⋅ 4− x 2

1 − 4

−2 x

+0

=

2

1 − x2 − 4 + x2 1 1 −4 = − = 4 x2 4 − x2 4 x2 4 − x2 x2 4 − x2

24

Advanced Integration

b. Given

1.2 Integration Using Trigonometric Substitution

(

x 2 dx

∫

25 − x

let x = 5 sin t , then dx = 5 cos t dt and 25 − x 2 = 25 − 25 sin 2 t = 25 1 − sin 2 t

2

)

= 25 cos 2 t = 5 cos t . Substituting these values back into the original integral we obtain: x 2 dx

∫

=

25 − x 2

25 sin 2 t ⋅ 5 cos t dt 5 cos t

∫

=

=

25 2

=

25  −1 x  25  x  25 − x 2  + c  sin −  2  5  2  25 

=

25 2

1+ x 2

∫

x

2

=

=

25 25 t − sin t cos t + c 2 2

then y ′ =

1  25 − x 2 x2 ⋅ −  − 2 2 25 − x 2 25 − x 2 5  5

   

25

=

−

2 25 − x 2

∫

25  −1 x  x  sin − 2  5 2

25 x x sin −1 − 25 − x 2 + c , 2 5 2

 50 − 4 x 2 −  2  4 25 − x

c. Given

∫

25  1   t − sin 2t  + c 2  2 

∫ ( 1 − cos 2t ) dt =

Check: Let y =

∫

1 − cos 2t 25 sin 2 t dt = 25 sin 2 t dt = 25 dt 1 2

25 − 2 x 2 2 25 − x 2

=

25  −1 x  25  x 25 − x 2 ⋅  sin − 5 5  2  5 2  

25 − x 2 + c

25 2

1 1−

x2 25

1 1 x − 2x 25 − x 2 + ⋅ − ⋅   5 2 2 25 − x 2 2  

25 − 25 + 2 x 2 2 25 − x 2

=

1+ x 2 =

dx let x = tan t , then dx = sec 2 t dt and

2x 2 2 25 − x 2

(

)

sec t ⋅ 1 + tan 2 t dt

∫

=

ln sec t + tan t +

2

tan t

=

1 cos t

∫ sin

2

t cos 2 t

∫

sec t + sec t tan 2 t

dt

2

tan t

dt

=

= ln sec t + tan t + ∫

∫

sec t tan 2 t 2

tan t cos 2 t 2

cos t sin t

dt

dt +

∫

sec t

1+ x 2 x2

∫ tan 2 t dt

=

   

Hamilton Education Guides

1 + x2 + x −

25 2 25 − x 2

25 − x 2

sec 2 t = sec t dx =

∫

sec t ⋅ sec 2 t dt tan 2 t sec t

∫ sec t dt + ∫ tan 2 t dt

= ln sec t + tan t + ∫

= ln sec t + tan t + ∫ sin −2 t cos t dt = ln sec t + tan t + sin −1 t + c = ln sec t + tan t + = ln sec t + tan t − csc t + c = ln

= x2

=

1+ tan 2 t =

Substituting these values back into the original integral we obtain =

)

(

  2 25 − x 2 − 2 x 2 25  =  −  2 2  2 25 − x  4 25 − x 

=

 +c  

cos t sin 2 t

dt

1 +c sin t

1 + x2 +c x

25

Advanced Integration

1.2 Integration Using Trigonometric Substitution

=

x

1 1+ x 2

+

dx

∫ ( 9 + x 2 )2

(

1

(

x 2 1+ x 2

(

x 2 1+ x 2

2 1+ x

2

⋅ x − 1+ x 2 x2

1

  

)

x2

2 x + 1 + x 2   2 2    x −1− x ⋅ − 2  2 1+ x 2 + x 2 1+ x 2  x 1+ x

) =

x 2 1+ x 2 + 1+ x 2

=

  1 2x ⋅ + 1 −   1+ x 2 + x  2 1+ x 2 

+ c , then y ′ =

  2 2   x − 1+ x − 2   x 1+ x 2  

 2x + 2 1 + x 2  ⋅ 2 1 + x + x  2 1 + x 2 1

=

d. Given

1+ x

1+ x 2 + x −

Check: Let y = ln

2

(

1+ x 2 1+ x 2

=

(

x 2 1+ x 2

)

)=

    

1 + x2 x2

let x = 3 tan t , then dx = 3 sec 2 t dt and 9 + x 2 = 9 + (3 tan t )2 = 9 + 9 tan 2 t

)

= 9 1 + tan 2 t = 9 sec 2 t Substituting these values back into the original integral we obtain dx

∫ ( 9 + x 2 )2 =

3 sec 2 t dt

∫ ( 9 sec 2 t )2

=

=

3 sec 2 t dt

∫ 81sec 2 t sec 2 t

=

1 27

dt

∫ sec 2 t

=

1 cos 2 t dt 27

∫

 1  1 1  (t + sin t cos t ) + c = 1  tan −1 x + x ⋅ 3  t + sin 2t  + c = 54  2 54 3 54  9 + x2 9 + x2 

Check: Let

+

= e. Given

∫

1  −1 x 3x  tan + 54  3 9 + x2

1 27 + 3 x 2 − 6 x 2 ⋅ 2 54 9 + x2

(

(

18

18 9 + x x2 2

x −1

)

)

2 2

=

=

  + c  1 ⋅ 54

then y ′ = 3

(9 + x ) 2

+

(

=

1 1 ⋅ 27 2

∫ (1 + cos 2t ) dt

  + c = 1  tan −1 x + 3 x  54  3 9+ x2 

)

1 9 1 1 1 3 9 + x 2 − 2 x ⋅ 3x + ⋅ ⋅ = ⋅ 54  x 2  54 54 2 2 31 + 9  9+ x 3 9 + x2  

1 27 − 3 x 2 ⋅ 2 54 9 + x2

(

)

=

(

1

(

18 9 + x 2

)

+

(

)

9 − x2

(

)

2

18 9 + x 2

=

)

9 + x2 + 9 − x2

(

18 9 + x 2

)

2

1

(9 + x )

2 2

x 2 −1 =

dx let x = sec t , then dx = sec t tan t and

Substituting these values back into the original integral we obtain

sec 2 t − 1 =

∫

x2 x 2 −1

tan 2 t = tan t

dx =

∫

sec 2 t sec t tan t dt tan t

= ∫ sec 3 t dt = ∫ sec 2 t sec t dt . Let u = sec t and dv = sec 2 t , then du = sec t tan t and v = tan t . Using the substitution formula uv − ∫ v du the integral ∫ sec 2 t sec t dt can be rewritten as

∫ sec

3

∫

∫

(

)

t dt = sec t tan t − sec t tan 2 t dt = sec t tan t − sec t sec 2 t − 1 dt = sec t tan t −

Hamilton Education Guides

  + c 

∫ ( sec

3

)

t − sec t dt

26

Advanced Integration

1.2 Integration Using Trigonometric Substitution

= sec t tan t − ∫ sec 3 t dt + ∫ sec t dt Note that ∫ sec 3 t dt = sec t tan t − ∫ sec 3 t dt + ∫ sec t dt therefore by moving − ∫ sec 3 t dt to the left hand side of the equality we obtain

∫ sec

∫

∫

∫

∫

t dt + sec 3 t dt = sec t tan t + sec t dt thus 2 sec 3 t dt = sec t tan t + sec t dt and

3

(

1 sec t tan t + sec t dt 2

∫

) = 12 ( sec t tan t + ln

1 2

)+ c

sec t + tan t

  Check: Let y =  x x 2 − 1 + ln x + x 2 − 1  + c , then y ′ = 

  2x  × 1 +   2  2 x −1 

= f. Given

(

=

dx x −1

t dt

1  2 2 x 2  1 1 x −1 + +   2 2 x 2 −1  2 x + x 2 −1 

)

=

   2 x −1 

1  4 x 2 − 2 + 2  2 x 2 −1 

1

x2 x 2 −1

x 2 −1 =

let x = sec t , then dx = sec t tan t dt and

2

3

 x 2 − 1 + ln x + x 2 − 1  + c 

2 x + x 2 − 1  1 1 2 x 2 −1 + 2x 2 1   + 2 x + x 2 −1 2 2 x 2 −1 2 x 2 −1

4x 2 1 4x 2 − 2 + 2 = = 2 2 x 2 −1 4 x 2 −1

∫ x4

1  x 2

=

∫ sec

sec 2 t − 1 =

tan 2 t = tan t

Substituting these values back into the original integral we obtain

∫ x4 =

dx

sec t tan t dt

∫ sec 4 t tan t

=

2

x −1

∫ (cos t − sin

2

)

t cos t dt =

∫ cos t dt − ∫ sin

 x 2 −1  x −1 ⋅ −  x2  x2  

x 2 − x 2 +1 x

2

∫ cos

3

t dt =

t cos t dt =

3

2

4

=

x − 1 1  x 2 − 1  ′ −   + c , then y = x x 3  

x 2 − x 2 +1

=

dt

∫ sec 3 t

2

Check: Let y =

=

=

2

x −1

Hamilton Education Guides

=

x 2 − x 2 +1 x 2 −1

x2

=

2

t cos t dt =

1 sin t − sin 3 t + c 3 2 x2 2

2 x −1

− x 2 −1 x2

2 ∫ (1 − sin t ) cos t dt 3

=

  x2 −1 1  x2 −1  −   + c 3 x x   2

3  x 2 − 1  −   ⋅ x 3  

 x 2 −1  1 ⋅ − 2   x 2 −1  x  x 2 x 2 −1

1 x2

∫ cos

=

1 x 2 x 2 −1

2 x2

2 x 2 −1

− x 2 −1 x2

−

x 2 −1 x 4 x 2 −1

1 x

4

x2 −1

27

Advanced Integration

1.2 Integration Using Trigonometric Substitution

Example 1.2-2: Use trigonometric substitution to evaluate the following indefinite integrals: a. d. g.

∫

dx

=

b.

∫x

2

dx =

e.

∫

a 2 − x 2 dx

=

h.

∫

9 − x 2 dx

(1 − )

3 x2 2

x2

∫

4+ x

2

x2

∫

9− x

dx

2

4 − x 2 dx

=

=

=

dx

c.

∫

f.

∫

i.

∫ x2

(

)

=

3

4 + 9x 2 2 x 2 − a 2 dx

1 4 + x2

=

dx

=

Solutions: a. Given

∫

dx

∫

(1 − )

3 x2 2

dx

∫

=

(1 − )

3 x2 2

cos t dt

(cos t )

Check: Let y = b. Given

∫x

2

(

4 1 − sin 2 t

∫x

2

let x = sin t , then dx = cos t dt and 1 − x 2 = 1− sin 2 t = cos 2 t . Therefore,

2

cos t

∫ cos 3 t dt

=

1

∫ cos 2 t dt

1⋅ 1 − x 2 −

x 1− x

4 − x 2 dx

)

=

3 2

+ c , then y ′ =

2

1− x

=

∫ sec

−2 x

2 1− x 2

2

t dt = tan t + c =

x

=

+c

1− x2

2− 2 x 2 + 2 x 2

⋅x

2 1− x 2

=

2

sin t +c cos t

1− x

=

2

(

2

2 1−

)

3 x2 2

=

1

(1 − x )

3

2 2

let x = 2 sin t , then dx = 2 cos t dt and 4 − x 2 = 4 − (2 sin t )2 = 4 − 4 sin 2 t

= 4 cos 2 t = 2 cos t . Therefore,

4 − x 2 dx

=

∫ 4 sin

)

(

= 4∫ 1 − cos 2 2t dt =

2

∫

t ⋅ 2 cos t ⋅2 cos t dt = 16 sin 2 t cos 2 t dt = 16

∫ 4 dt − 4∫ cos

1 2

2

2t dt =

1

∫ 4 dt − 4∫ 2 ( 1 + cos 4t ) dt

1

∫ 4 (1 − cos 2t ) (1 + cos 2t ) dt =

∫ 4 dt − ∫ 2 dt − 2∫ cos 4t dt

(

4 2

)

= 4t − 2t − sin 4t + c = 4t − 2t − sin 2t cos 2t + c = 2t − 2 (sin t cos t ) cos 2 t − sin 2 t + c x

x

= 2 sin −1 − 2  ⋅ 2 2 

4 − x2 2

Check: Let y = 2 sin −1

=

2 4 − x2

−

  4 − x2 x2   −1 x 2    4 − 4  + c = 2 sin 2 − 4 − x  

 2x − x 3   4 

x − 4 − x2 2

(

)(

 2  + c , then y ′ =   2 1−

− 2 x 2 + x 4 + 2 − 3x 2 4 − x 2

Hamilton Education Guides

4 4 − x2

)=

2 4 − x2

−

 2 x − x3   +c   4  

x2 4

−

(

− 2x 2x − x 3 8 4 − x2

) + (2 − 3x )(4 − x ) 2

1

2 2

− 2 x 2 + x 4 + 8 − 14 x 2 + 3 x 4 4 4 − x2

28

Advanced Integration

8 − 4 x 4 + 16 x 2 − 8

=

c. Given

∫

1.2 Integration Using Trigonometric Substitution

4 4− x dx

( 4 + 9x )

3

2

∫

=

2 2

(

dx 4 9

− 4 x 4 + 16 x 2

=

+ x2

4 4− x

)

3 2

2

∫ =

(

)

3 4 + 9x 2 2

1 12

dt

∫ sec t

∫

=

=

(

2 3

sec 2 t

)

3 4 sec 2 2

1 cos t dt 12

∫

dt

=

=

4 4− x

d. Given

∫

x 2 dx 4+ x

(

= 4 1 + tan 2 t

∫

x 2 dx 4+ x

2

=

∫

2

)

x

∫

2 3

3 42

)

4 4 + 9x 2

)

4 (4 − x ) 2

= x2 4 − x2

2 3

2× 3 sec 2

=

2



= 4 sec 2 t . Therefore,

sec 2 t

1 sin t + c 12

dt

=

t

1 12

2 sec 2 3

3x 4 + 9x 2

then y ′ =

t

∫ 2 4 3 sec 3 t dt +c

1⋅ 4 4 + 9 x − 4 ⋅ +c,

4− x

2

(

4 4 − x2 x2 4 − x2

=

2 3

=

2

Check: Let y =

4 − x2

⋅

2

2 3

= 4 + 9 ⋅ tan 2 t = 4 + 4 tan 2 t = 4 1 + tan 2 t dx

− 4 x 4 + 16 x 2

let x = tan t , then dx = sec 2 t dt and 4 + 9 x 2 = 4 + 9 ⋅  tan t 

(

4 9

=

(

=

dt

∫ 2 64 sec t

x

18 x

2 3

dt

∫ 8 sec t

)

(

4 4+9 x 2 −36 x 2

⋅x

=

)

=

+c

4 4 + 9x 2

2 4+9 x 2

16 4 + 9 x 2

2 3

(

4+9 x 2

16 4 + 9 x 2

)

=

1

( 4 + 9x )

3

2 2

let x = 2 tan t , then dx = 2 sec 2 t dt and 4 + x 2 = 4 + (2 tan t )2 = 4 + 4 tan 2 t = 4 sec 2 t = 2 sec t . Therefore,

4 tan 2 t ⋅ 2 sec 2 t dt 2 sec t

(

(

)

)

= 4∫ tan 2 t sec t dt = 4∫ sec 2 t − 1 sec t dt = 4∫ sec 3 t − sec t dt

= 4∫ sec 3 t dt − 4∫ sec t dt . To solve ∫ sec 3 t dt = ∫ sec 2 t ⋅ sec t dt use substitution method by letting u = sec t and dv = sec 2 t then du = sec t tan t dt and v = tan t . Using the substitution formula uv − ∫ v du we obtain ∫ sec 2 t ⋅ sec t dt = sec t tan t − ∫ tan t ⋅ sec t tan t dt

(

)

(

)

= sec t tan t − ∫ sec t tan 2 t dt = sec t tan t − ∫ sec t sec 2 t − 1 dt = sec t tan t − ∫ sec 3 t − sec t dt = sec t tan t − ∫ sec 3 t dt + ∫ sec t dt . Again at this point we know that ∫ sec 3 t dt = sec t tan t − ∫ sec 3 t dt + ∫ sec t dt bringing − ∫ sec 3 t dt into the left hand side of the equation we obtain ∫ sec 3 t dt + ∫ sec 3 t dt = sec t tan t + ∫ sec t dt . Therefore 2∫ sec 3 t dt = sec t tan t + ∫ sec t dt thus ∫ sec 3 t dt =

[

1 sec t tan t + sec t dt 2

∫

] = 12 [ sec t tan t + ln sec t + tan t

].

Now substituting this

value into 4∫ sec 3 t dt − 4∫ sec t dt we have

Hamilton Education Guides

29

Advanced Integration

∫

1.2 Integration Using Trigonometric Substitution

∫

4 sec 3 t dt − 4 sec t dt = 4 ⋅

1 sec t tan t + ln sec t + tan t 2

[

x 4 + x2 x + 4 + x2 1 2x 2 − 2 ln + c , then y ′ =  4 + x 2 + 2 2 2 2 4 + x2 

 2 1  2x −2 ⋅ ⋅ 1 +  x + 4 + x2 2  2 4 + x2  

∫

e. Given

− 4 4 + x2 a 2 − x 2 dx

(

=

a 2 − x 2 dx

)

=

Check: Let y =

=

) x +

4 + x 2  − 2 4 + x 2 − 2 x x − 4 + x2  ⋅  x + 4 + x2  4 + x2 x − 4 + x2    

(

=

− 4x 2 − 4 4 + x2

)

x2

=

4 + x2

let x = a sin t , then dx = a cos t dt and a 2 − x 2 = a 2 − a 2 sin 2 t

= a 2 cos 2 t = a cos t . Therefore,

∫

a cos t ⋅ a cos t dt =

2 a2 ( t + sin t cos t ) + c = a 2 2

=

 2 4 + x 2 + 2x −      x + 4 + x2  4 + x2  

4 + x 2  − 2 4 + x 2 − 2 x x4 − x3 4 + x2 − x3 4 + x2 − x2 4 + x2 x − 4 + x2  = ⋅  x + 4 + x2  4 + x2 − 4 4 + x2 x − 4 + x2    

x 4 − 4x 2 − x 4

= a 2 1− sin 2 t

∫

(

4 + x 2  − 2 4 + x 2 − 2 x 2 + x2  =  x + 4 + x2  4 + x2    

   

(2 + x ) x + 2

=

2     + 0 = 1  4 + 2x  2  4 + x2  

(2 + x ) x + 2

=

x 4+ x2 x + 4+ x2 4 + x2 x +c − 2 ln + +c = 2 2 2 2

4 + x2 x ⋅ − 2 ln 2 2

− 2 ln sec t + tan t + c = 2 ⋅

Check: Let y =

]

= 2 sec t tan t + 2 ln sec t + tan t − 4 ln sec t + tan t + c = 2 sec t tan t

− 4 ln sec t + tan t + c

=

= 2 [ sec t tan t + ln sec t + tan t

]− 4∫ sec t dt

∫

a 2 cos 2 t dt =

 2 2  sin −1 x + x ⋅ a − x  a a a 

a2 x x a2 − x2 sin −1 + +c, 2 a 2

a2 − x2 − x2 a2 a ⋅ + 2 a a2 − x2 2 a2 − x2

a2 − x2 2

a −x

Hamilton Education Guides

2

⋅

a2 − x2 2

a −x

2

=

a2 2

a2  1   t + sin 2t  + c 2  2 

 2  x a2 − x2 a −1 x +c + sin =  2  2 a  

then y ′ = a2 2 a2 − x2

=

∫ ( 1 + cos 2t ) dt =

a2 2 +

1 1−

( ax )2

a 2 − 2x 2

2

− x2 2

)

1 1 − 2x 2 + a2 − x2 + a 2 2 a2 − x2

=

2 a2 − x2

(a

⋅

2a 2 − 2 x 2 2 a2 − x2

a2 − x2

a −x

2

   + c 

= =

a2 − x2 a2 − x2

a2 − x2 30

Advanced Integration

f. Given

∫

1.2 Integration Using Trigonometric Substitution

x 2 − a 2 dx

=

a 2 tan 2 t

∫

x 2 − a 2 dx

∫ a tan t ⋅ a sec t tan t dt

=

=

 x a2  x x2 − a2 − ln + ⋅  a a 2 a 

=

x 2

x2 − a2 −

∫

∫a

  +c  

x2 − a2 a

) − a 2 ln =

x 2

x2 − a2 −

a2 1 ⋅ 2 x + x2 − a2

x2 − a2 + x2 2

2 x −a x2 9− x

2

dx

2

−

2x

x 2 −a 2

a2 2

2 x −a

=

2

)

∫

a2 (tan t sec t − ln sec t + tan t 2

sec t + tan t + c =

)+ c

x a2 x + x2 − a2 x2 − a2 − +c ln a 2 2

=

x 2

a2 ln x + x 2 − a 2 + c , 2

 ⋅ 1 +  2

(

∫

sec t tan 2 t dt = a 2 sec t sec 2 t − 1 dt = a 2 sec 3 t dt

2

a2 a2 ln a + c ln x + x 2 − a 2 + 2 2

Check: Let y =

g. Given

=

a2 (tan t sec t + ln sec t + tan t 2

∫

=

a 2 sec 2 t − a 2

= a tan t . Therefore,

− a 2 sec t dt =

−

x2 − a2 =

let x = a sec t , then dx = a sec t tan tdt and

x2 −a2 −

a2 ln x + x 2 − a 2 + c 2

1 then y ′ =  x 2 − a 2 + 2



  2 2  2 x −a  2x 2

 x2 − a2 + x2 a2 x + x2 − a2 1 +0 = − ⋅ ⋅  2 x + x2 − a2  2 x2 − a2 x2 − a2

2 x 2 − 2a 2 2

2 x −a

2

=

x2 − a2 2

x −a

2

x2 − a2

=

2

x −a

2

⋅

x2 − a2 2

x −a

2

let x = 3 sin t , then dx = 3 cos t dt and 9 − x 2 = 9 − 9 sin 2 t =

x2 − a2

=

(

9 1 − sin 2 t

)

= 9 cos 2 t = 3 cos t . Substituting these values back into the original integral we obtain:

∫

x 2 dx 9 − x2

=

∫

9 sin 2 t ⋅ 3 cos t dt 3 cos t

=

9 2

=

9  −1 x  9  x  9 − x2  + c −   sin 3 29 2 

Let

y=

∫

1 − cos 2t 9 sin 2 t dt = 9 sin 2 t dt = 9 dt 1 2

9 1   t − sin 2t  + c 2 2 

∫ ( 1 − cos 2t ) dt =

Check:

=

=

∫

=

9 9 t − sin t cos t + c 2 2

9  −1 x  9  x 9 − x 2 ⋅ −  sin 2 3  2  3 3 

=

 +c  

9  −1 x  x 9− x2 + c  sin − 2 3 2

x x 9 sin −1 − 9 − x2 + c , 2 3 2

Hamilton Education Guides

∫

then

y′

=

9 2

1 1−

x2 9

1 1 ⋅ − 3 2 

9 − x2 +

x ⋅   2 9 − x2 2  − 2x

31

Advanced Integration

 1  ⋅ − 9 − x 2 3 

9 2

=

1.2 Integration Using Trigonometric Substitution

    

  18 − 4 x 2 −  4 9 − x2 

h. Given

∫

9 − x 2 dx

=

9 cos 2 t

∫

9 − x 2 dx

=

9 − x2

3

2

9

=

2 9 − x2

9 − 2x 2

=

2 9 − x2

(

)

 2 2  2 9 − x − 2x − 2 9 − x 2  4 9 − x 2 9

=

9 − 9 + 2x 2 2 9 − x2

2x 2

=

9

=

2 9 − x2

x2

=

2 9 − x2

9 − x2

    

9 − x2

9 − 9 sin 2 t

=

(

9 1 − sin 2 t

=

)

= 9 cos t . Therefore,

∫ 3 cos t ⋅ 3 cos t dt = ∫ 9 cos

=

 9  −1 x x + ⋅ sin 2  3 3 

=

9 2

x 3

Check: Let y = sin −1 + 9 ⋅ 2 3

3 9− x

9 − x2

= i. Given

−

   2  2 9− x  x2

let x = 3 sin t , then dx = 3 cos t dt and

9 ( t + sin t cos t ) + c 2

=

−

9− x

2

1

∫ x2

2

2

2

9 − x2 − x2 9− x

2

9 − x2

=

dx

4 + x2

+

9 − x2

x

9− x

2

2

9 2

2

9− x

2

∫ ( 1 + cos 2t ) dt  9

9− x

=

2

+

9 1   t + sin 2t  + c 2 2 

=

 9− x2  +c 2  

x

x

=  sin −1 + 3 2 

9 2

then y ′ = 9

9 − x2

⋅

t dt =

9 − x 2  +c 3  

+c,

=

2

1 1−

( 3x )2

9 − 2x 2 2

9− x

(9 − x )

=

2

2

9 − x2

9− x

2

1 1 ⋅ + 3 2

9 − x2 +

9 + 9 − 2x 2 2

9− x

2

=

− 2x 2 2

9 − x2

18 − 2 x 2 2

9− x

2

=

(

2 9 − x2 2

9− x

)

2

9 − x2

=

let x = 2 tan t , then dx = 2 sec 2 t dt and 4 + x 2 = 4 + 4 tan 2 t = 2 1 + tan 2 t

= 2 sec 2 t = 2 sec t . Therefore, dx

∫ x2 =

1 4

4 + x2

∫

cos t u2

⋅

=

dx

du cos t

2 sec 2 t

∫ 4 tan 2 t ⋅ ( 2 sec t )

=

Check: Let y = −

1 4

1 4

1

∫ u2

du

4+ x x

2

dt

1 4

Hamilton Education Guides

4+ x 2

x

2

1 4

sec dt

∫ tan 2 t

1 4

= − u −1 + c = −

dt

1 +c 4u 2x

+c ,

1 then y ′ = −

2 4+ x

2

=

1 4

= −

∫

1 cos 2 t 1 ⋅ dt = 2 cos t sin t 4

1 +c 4 sin t

= −

x

= −

2

let u = sin t

4+ x2 +c x

1 4 1 4

cos t

∫ sin 2 t dt

2 x2

⋅ x − 1⋅ 4 + x 2

4

   x 2 − 4 + x 2 ⋅ 4 + x 2    

= −

=

2 4+ x

2

− 4 + x2 x2

x 2 − 4− x 2

= −

4+ x 2

4x

2

= −

−4 4x

2

4+ x

2

=

1 x

2

4 + x2 32

Advanced Integration

1.2 Integration Using Trigonometric Substitution

The following are additional standard forms of integration that have already been derived. Trigonometric substitution can be used in most of these cases in order to confirm the result. Table 1.2-1: Integration Formulas dx

1.

∫ a2 + x2

3.

∫ a2 − x2

5.

∫

7.

∫x

x 1 tan −1 + c a a

2.

∫ ( a 2 + x 2 )2

=

1 a+x x+a 1 ln +c ln +c = x−a 2a 2a a−x

4.

∫ ( a 2 − x 2 )2

6.

∫

a 2 + x 2 dx =

8.

∫

a2 + x2 dx = x

dx

dx 2

a +x

11.

∫x

13.

∫

15.

∫

17.

dx a2 + x2 dx

)

a2 + x2

−

a4 x sinh −1 + c a 8

∫

dx a −x

21.

∫

2

23.

∫x

25.

∫

27.

∫x ∫

(

dx = − sin −1

x2

∫x

2

2

dx x −a 2

2

x2 − a2 2

dx 2

x −a

2

dx 2ax − x

2

x a2 − x2 − +c a x

x + c = ln x + x 2 − a 2 + c a

= cosh −1

(

x 2x 2 − a 2 8

dx = cosh −1

=

)

a + a2 − x2 +c x

1 a

= − ln

x 2 − a 2 dx =

x

x +c a a4 x 1 sin −1 − x a 2 − x 2 a 2 − 2 x 2 + c 8 a 8

x 2 a 2 − x 2 dx =

a2 − x2

a + a2 + x2 +c x

1 a

= − ln = sin −1

a2 − x2

19.

29.

(

x a 2 + 2x 2 8

a2 + x2 x a2 + x2 +c dx = sinh −1 − a x x

∫

9.

x + c = ln x + a 2 + x 2 + c a

= sinh −1

2

a 2 + x 2 dx =

2

dx

=

x − a

)

x2 − a2 −

a4 x cosh −1 + c a 8

x2 − a2 +c x

1 x 1 a sec−1 + c = cos −1 +c a x a a  x−a +c  a 

= sin −1 

Hamilton Education Guides

dx

x2

= =

2a

(a

2

2a

2

(a

2

∫

12.

∫ x2

14.

∫

a 2 − x 2 dx =

16.

∫

a2 − x2 dx = x

18.

∫

20.

∫ x2

22.

∫

x 2 − a 2 dx =

24.

∫

x2 − a2 dx = x

26.

∫

28.

∫ x2

30.

∫

a2 + x2

x2 a2 − x2

a2 − x2

x2 2

x −a

2

x2 − a2

a 2 + x 2 dx =

1

tan −1

x +c a

dx

2a 2

∫ a2 − x2

a +c x

a2 + x2 a2x

+c

a2 x x a2 − x2 + sin −1 + c a 2 2

a 2 − x 2 − a ln

a2 − x2 a2x x 2

=

a + a2 − x2 +c x

x 1 a2 sin −1 − x a 2 − x 2 + c a 2 2

=

dx =

dx

)

+

1 2a 3

a2 x x a2 + x2 +c sinh −1 + 2 2 a

= −

dx =

dx

−x

2

x

)

+

a 2 + x 2 − a sinh −1

dx = −

dx

+x

2

x a2 x a2 + x2 + sinh −1 + c a 2 2

10.

a2 + x2

x

2

+c

x2 − a2 −

a2 x cosh −1 + c 2 a

x 2 − a 2 − a sec −1 a2 x x cosh −1 + 2 a 2

x2 − a2 a2x

x +c a x2 − a2 + c

+c

x 2 a2   a + x2 + ln  x + x 2 + a 2  + c 2 2  

33

Advanced Integration

1.2 Integration Using Trigonometric Substitution

Section 1.2 Practice Problems – Integration Using Trigonometric Substitution Evaluate the following indefinite integrals: a.

∫

dx x

2

16 − x 1

d.

∫ (49 + x2 )

g.

∫

=

2

dx = 2

36 − x 2 dx

=

Hamilton Education Guides

b.

e. h.

x2

∫ ∫ ∫

9− x

2

dx =

    x2 5 x +   dx  2   x −1 

(

dx

)

3

9 + 36 x 2 2

=

c.

=

f. i.

∫

dx x 9 + 4x

2

=

∫

x 2 − 25 dx

=

∫

9 − 4x 2 dx x

=

34

Advanced Integration

1.3

1.3 Integration by Partial Fractions

Integration by Partial Fractions

The primary objective of this section is to show that rational functions can be integrated by breaking them into simpler parts. A function F (x ) =

f (x ) , g (x )

where f (x ) and g (x ) are polynomials,

is referred to as a rational function. Depending on the degree of the f (x ) , the function F (x ) is either a proper or an improper rational fraction. •

A proper rational fraction is a fraction where the degree of the numerator f (x ) is less than the degree of the denominator g (x ) . For example, are proper rational fractions.

•

x +1 2

x − x−6

,

1 3

x −1

,

1 2

x + 2x − 3

, and

x 2

x +1

An improper rational fraction is a fraction where the degree of the numerator f (x ) is at least as large as the degree of the denominator g (x ) . In such cases, long division is used in order to reduce the fraction to the sum of a polynomial and a proper rational fraction. For example, x3 + 2

x2 − x − 6

,

x3

x 2 +1

, and

x 4 − x 3 − x −1 x3 − x2

are improper rational fractions. (See Section 6.3 of

Mastering Algebra – An Introduction for solved problems on long division.) A fraction, depending on its classification of the denominator, can be represented in four different cases. These cases are as follows: CASE I - The Denominator Has Distinct Linear Factors In this case the linear factors of the form ax + b appear only once in the denominator. To solve this class of rational fractions we equate each proper rational fraction with a single fraction of the form

A B , ax + b cx + d

,

C ex + f

, etc. The following examples show the steps as to how this class of

integrals are solved. Example 1.3-1: Evaluate the integral

x +1

∫ x 3 + x 2 − 6 x dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second - Factor the denominator x 3 + x 2 − 6 x into x(x − 2)(x + 3) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way: x +1 3

2

x + x − 6x

=

x +1 x(x − 2 )(x + 3)

=

A B C + + x x−2 x+3

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. x +1 x 3 + x 2 − 6x

Hamilton Education Guides

=

A (x − 2 )(x + 3) + Bx (x + 3) + Cx (x − 2 ) x(x − 2 )(x + 3)

35

Advanced Integration

x +1

1.3 Integration by Partial Fractions

) (

) (

(

)

= A x 2 + 3x − 2 x − 6 + B x 2 + 3x + C x 2 − 2 x = Ax 2 + Ax − 6 A + Bx 2 + 3Bx + Cx 2 − 2Cx x +1

= ( A + B + C )x 2 + ( A + 3B − 2C )x − 6 A therefore,

A+ B +C = 0

A + 3B − 2C = 1 1 6

which result in having A = − , B =

3 10

, and C = −

−6 A = 1

2 15

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x +1

A

B

1

C

1

3

1

2

1

∫ x 3 + x 2 − 6 x dx = ∫ x dx + ∫ x − 2 dx + ∫ x + 3 dx = − 6 ∫ x dx + 10 ∫ x − 2 dx − 15 ∫ x + 3 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. −

1 6

1

3

2

1

1

1

∫ x dx + 10 ∫ x − 2 dx − 15 ∫ x + 3 dx = − 6 ln

x +

3 2 ln x − 2 − ln x + 3 + c 10 15

Seventh - Check the answer by differentiating the solution. The result should match the integrand. 1 6

Let y = − ln x + = = =

2 3 ln x − 2 − ln x + 3 + c 15 10

− 150(x − 2 )(x + 3) + 270 x (x + 3) − 120 x (x − 2 ) 900 x (x − 2 )(x + 3)

(

(

900 x + 900

(

900 x 3 + x 2 − 6 x

)

=

(

900 x 3 + x 2 − 6 x

Example 1.3-2: Evaluate the integral

=

)

(

(

900 x x 3 + x 2

)

900(x + 1)

)

1 2 1 3 ⋅1 + 0 ⋅1 − ⋅ ⋅ 15 x + 3 10 x − 2

) − 6x)

(

− 150 x 2 + x − 6 + 270 x 2 + 3 x − 120 x 2 − 2 x

=

− 150 x 2 − 150 x + 900 + 270 x 2 + 810 x − 120 x 2 + 240 x 900 x 3 + x 2 − 6 x

1 1 6 x

, then y ′ = − ⋅ ⋅1 +

=

)

(− 150 + 270 − 120)x 2 + (− 150 + 810 + 240)x + 900

(

900 x 3 + x 2 − 6 x

)

x +1 3

x + x 2 − 6x dx

∫ x 2 + 3x + 2 .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second - Factor the denominator x 2 + 3x + 2 into (x + 1)(x + 2) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way: 1 2

x + 3x + 2

=

1

(x + 1)(x + 2)

=

A B + x +1 x + 2

Fourth - Solve for the constants A and B by equating coefficients of the like powers. 1 2

x + 3x + 2

Hamilton Education Guides

=

A (x + 2 ) + B (x + 1) (x + 1)(x + 2)

36

Advanced Integration

1.3 Integration by Partial Fractions

1 = A (x + 2 ) + B (x + 1) = Ax + 2 A + Bx + B 1 = ( A + B )x + (2 A + B ) therefore,

2A + B = 1

A+ B = 0

which result in having A = 1 and B = −1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 1

A

1

B

1

∫ x 2 + 3x + 2 dx = ∫ x + 1 dx + ∫ x + 2 dx = ∫ x + 1 dx − ∫ x + 2 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. 1

1

∫ x + 1 dx − ∫ x + 2 dx

= ln x + 1 − ln x + 2 + c

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = ln x + 1 − ln x + 2 + c , then y ′ = Example 1.3-3: Evaluate the integral

1 1 ⋅1 − ⋅1 + 0 x +1 x+2

=

(x + 2) − (x + 1) = (x + 1)(x + 2)

1

2

x + 3x + 2

x dx

∫ x 2 − 5x + 6 .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second - Factor the denominator x 2 − 5 x + 6 into (x − 2)(x − 3) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way: x 2

x − 5x + 6

=

x

(x − 2)(x − 3)

=

A B + x−2 x−3

Fourth - Solve for the constants A and B by equating coefficients of the like powers. x 2

x − 5x + 6 x

=

A (x − 3) + B (x − 2 ) (x − 2)(x − 3)

= A (x − 3) + B (x − 2) = Ax − 3 A + Bx − 2 B x

= ( A + B )x − (3 A + 2 B ) therefore,

A+ B =1

3 A + 2B = 0

which result in having A = −2 and B = 3 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

Hamilton Education Guides

37

Advanced Integration

1.3 Integration by Partial Fractions

x dx

A

2

B

3

∫ x 2 − 5x + 6 = ∫ x − 2 dx + ∫ x − 3 dx = − ∫ x − 2 dx + ∫ x − 3 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. −

2

3

∫ x − 2 dx + ∫ x − 3 dx = − 2 ln

x − 2 + 3 ln x − 3 + c

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = −2 ln x − 2 + 3 ln x − 3 + c , then y ′ = − 2 ⋅ =

−2 x + 6 + 3 x − 6

=

2

x − 5x + 6

1 1 ⋅1 + 3 ⋅ ⋅1 + 0 x−2 x−3

=

− 2(x − 3) + 3(x − 2 ) (x − 2)(x − 3)

x 2

x − 5x + 6

Example 1.3-4: Evaluate the integral

x 2 +1

∫ x 3 − x dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 3 − x into x x 2 − 1 = x(x − 1)(x + 1) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way: x 2 +1

=

x3 − x

x 2 +1 A B C = + + x(x − 1)(x + 1) x x −1 x +1

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. x 2 +1 3

x 2 +1

x −x

(

=

A (x − 1)(x + 1) + Bx (x + 1) + Cx (x − 1) x(x − 1)(x + 1)

) (

) (

)

= A x 2 + x − x − 1 + B x 2 + x + C x 2 − x = Ax 2 − A + Bx 2 + Bx + Cx 2 − Cx x 2 +1

= ( A + B + C )x 2 + (B − C )x − A therefore,

A+ B +C =1

−A = 1

B −C = 0

which result in having A = −1 , B = 1 , and C = 1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x 2 +1

A

B

C

1

1

1

∫ x 3 − x dx = ∫ x dx + ∫ x − 1 dx + ∫ x + 1 dx = − ∫ x dx + ∫ x − 1 dx + ∫ x + 1 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. −

1

1

1

∫ x dx + ∫ x − 1 dx + ∫ x + 1 dx

Hamilton Education Guides

= − ln x + ln x − 1 + ln x + 1 + c

38

Advanced Integration

1.3 Integration by Partial Fractions

Seventh - Check the answer by differentiating the solution. The result should match the integrand. 1 x

Let y = − ln x + ln x − 1 + ln x + 1 + c , then y ′ = − + =

− x 2 +1+ x 2 + x + x 2 − x

(

)

x x 2 −1

1 1 + +0 x −1 x +1

=

− (x − 1)(x + 1) + x (x + 1) + x (x − 1) x(x − 1)(x + 1)

x 2 +1

=

x3 − x

Example 1.3-5: Evaluate the integral

x−3

∫ x (x 2 + x − 2)

dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second - Factor the denominator x 2 + x − 2 into (x + 2)(x − 1) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way: x−3 x(x + 2 )(x − 1)

=

A B C + + x x + 2 x −1

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. x−3 x(x + 2 )(x − 1) x−3

(

=

) (

A (x + 2 )(x − 1) + Bx (x − 1) + Cx (x + 2) x(x + 2)(x − 1)

) (

)

= A x 2 + 2 x − x − 2 + B x 2 − x + C x 2 + 2 x = Ax 2 + Ax − 2 A + Bx 2 − Bx + Cx 2 + 2Cx x−3

= ( A + B + C )x 2 + ( A − B + 2C )x − 2 A therefore,

A+ B +C = 0

−2 A = −3

A − B + 2C = 1 3 2

5 6

which result in having A = , B = − , and C = −

2 3

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x−3

∫ x (x 2 + x − 2)

dx =

A

B

C

∫ x dx + ∫ x + 2 dx + ∫ x − 1 dx

=

3 2

1

5

1

2

1

∫ x dx − 6 ∫ x + 2 dx − 3 ∫ x − 1 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. 3 2

1

5

1

2

1

∫ x dx − 6 ∫ x + 2 dx − 3 ∫ x − 1 dx

=

3 5 2 ln x − ln x + 2 − ln x − 1 + c 2 6 3

Seventh - Check the answer by differentiating the solution. The result should match the 3 2

5 6

2 3

integrand.Let y = ln x − ln x + 2 − ln x − 1 + c , then y ′ = 9(x + 2 )(x − 1) − 5 x (x − 1) − 4 x (x + 2 ) 6 x (x + 2 )(x − 1)

Hamilton Education Guides

=

(

) ( ) ( 6 x (x + x − 2 )

5 1 2 1 3 1 ⋅1 + 0 ⋅ ⋅1 − ⋅ ⋅1 − ⋅ 3 x −1 6 x+2 2 x

9 x 2 + x − 2 − 5 x 2 − x − 4 x 2 + 2x 2

=

) = (9 − 5 − 4)x + (9 + 5 − 8)x − 18 6 x (x + x − 2 ) 2

2

39

Advanced Integration

=

(

6 x − 18

6x x 2 + x − 2

)

1.3 Integration by Partial Fractions

=

(

6(x − 3)

6x x 2 + x − 2

)

=

x−3

(

x x2 + x − 2

)

x3 + 2

∫ x 2 − x − 6 dx .

Example 1.3-6: Evaluate the integral

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. In this case the integrand is an improper rational fraction, i.e., the degree of the numerator is greater than the degree of the denominator. Applying the long division method we obtain x3 + 2 2

x − x−6

= (x + 1) +

7x + 8 2

x − x−6

To integrate the second term we proceed with the following steps: Second - Factor the denominator x 2 − x − 6 into (x − 3)(x + 2) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way: 7x + 8 2

x − x−6

=

7x + 8

(x − 3)(x + 2)

=

A B + x−3 x+ 2

Fourth - Solve for the constants A and B by equating coefficients of the like powers. 7x + 8 2

x − x−6 7x + 8 7x + 8

29 5

A ( x + 2 ) + B ( x − 3) (x − 3)(x + 2)

= Ax + 2 A + Bx − 3B

= ( A + B )x + (2 A − 3B ) therefore, 2 A − 3B = 8

A+ B = 7

which result in having A =

=

and B =

6 5

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x3 + 2

7x + 8

7x + 8

∫ x 2 − x − 6 dx = ∫ (x + 1)dx + ∫ x 2 − x − 6 dx = ∫ (x + 1)dx + ∫ (x − 3)(x + 2) dx = =

1 (x + 1)2 + 29 2 5

1

6

1 (x + 1)2 + 2

A

B

∫ x − 3 dx + ∫ x + 2 dx

1

∫ x − 3 dx + 5 ∫ x + 2 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. 1 (x + 1)2 + 29 2 5

1

6

1

∫ x − 3 dx + 5 ∫ x + 2 dx =

1 ( x + 1)2 + 29 ln x − 3 + 6 ln x + 2 + c 2 5 5

Seventh - Check the answer by differentiating the solution. The result should match the integrand.

Hamilton Education Guides

40

Advanced Integration 1 2

Let y = (x + 1)2 + = (x + 1) +

29 6 ln x − 3 + ln x + 2 + c 5 5

, then y ′ = (x + 1) +

6 1 29 1 ⋅1 + 0 ⋅1 + ⋅ ⋅ 5 x+2 5 x−3

5(x + 1)(x − 3)(x + 2 ) + 29(x + 2 ) + 6(x − 3) 6 1 29 1 = + ⋅ ⋅ 5 (x − 3)(x + 2 ) 5 x−3 5 x+ 2

5 x 3 + 10 = 5 (x − 3)(x + 2)

=

1.3 Integration by Partial Fractions

(

(

5 x3 + 2

)

5 x2 − x − 6

5 x 3 − 35 x − 30 + 29 x + 58 + 6 x − 18 5 (x − 3)(x + 2 )

x3 + 2

=

)

=

x2 − x − 6

Example 1.3-7: Evaluate the integral

1

∫ 49 − x 2 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second - Factor the denominator 49 − x 2 into (7 − x )(7 + x ) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way: 1 49 − x

2

=

1

(7 − x )(7 + x )

=

A B + 7−x 7+ x

Fourth - Solve for the constants A and B by equating coefficients of the like powers. 1 49 − x

2

=

A (7 + x ) + B (7 − x ) (7 − x )(7 + x )

1 = A (7 + x ) + B (7 − x ) = 7 A + Ax + 7 B − Bx 1 = ( A − B )x + (7 A + 7 B ) therefore, 7 A + 7B = 1

which result in having A =

1 14

A− B = 0

, and B =

1 14

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 1

A

B

∫ 49 − x 2 dx = ∫ 7 − x dx + ∫ 7 + x dx

=

1 14

1

1

1

∫ 7 − x dx + 14 ∫ 7 + x dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. 1 14

1

1

1

1

∫ 7 − x dx + 14 ∫ 7 + x dx = 14 ln

7− x +

1 ln 7 + x + c 14

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = =

(

1 1 ln 7 − x + ln 7 + x + c 14 14

7+7

14 49 + 7 x − 7 x − x 2

)

Hamilton Education Guides

=

(

14

14 49 − x 2

, then y ′ =

)

=

1 1 1 1 ⋅ + ⋅ +0 14 7 − x 14 7 + x

=

7+ x+7−x 14(7 − x )(7 + x )

1 49 − x 2

41

Advanced Integration

1.3 Integration by Partial Fractions

CASE II - The Denominator Has Repeated Linear Factors In this case each linear factor of the form ax + b appears n times in the denominator. To solve this class of rational fractions we equate each proper rational fraction, that appears n times in the denominator, with a sum of n partial fractions of the form

Mn M1 M2 + + ... + . The 2 ax + b (ax + b ) (ax + b )n

following examples show the steps as to how this class of integrals are solved. Example 1.3-8: Evaluate the integral

x+3

∫ x 3 − 2 x 2 + x dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 3 − 2 x 2 + x into x x 2 − 2 x + 1 = x(x − 1)2 . Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand can be represented in the following way: x+3 3

=

2

x − 2x + x

(

x+3

=

)

x x 2 − 2x +1

x+3

x(x − 1)

2

=

A B C + + x x − 1 (x − 1)2

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. x+3 x 3 − 2x 2 + x x+3

(

) (

=

A (x − 1)2 + Bx (x − 1) + Cx x(x − 1)(x − 1)2

)

= A x 2 − 2 x + 1 + B x 2 − x + Cx = Ax 2 − 2 Ax + A + Bx 2 − Bx + Cx x+3

= ( A + B )x 2 + (− 2 A − B + C )x + A therefore,

A+ B = 0

A=3

−2 A − B + C = 1

which result in having A = 3 , B = −3 , and C = 4 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. A

x+3

B

1

C

1

1

∫ x 3 − 2 x 2 + x dx = ∫ x dx + ∫ x − 1 dx + ∫ (x − 1)2 dx = 3∫ x dx − 3∫ x − 1 dx + 4∫ (x − 1)2 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. 1

1

1

∫ x dx − 3∫ x − 1 dx + 4∫ (x − 1)2 dx

3

= 3 ln x − 3 ln x − 1 −

4 +c x −1

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = 3 ln x − 3 ln x − 1 −

Hamilton Education Guides

4 +c x −1

1 x

, then y ′ = 3 ⋅ − 3 ⋅

3(x − 1)2 − 3 x (x − 1) + 4 x 1 4 + +0 = x − 1 (x − 1)2 x (x − 1)2

42

Advanced Integration

=

1.3 Integration by Partial Fractions

3x 2 − 6 x + 3 − 3x 2 + 3x + 4 x 3

x+3

=

2

x − 2x + x

3

x − 2x 2 + x

Example 1.3-9: Evaluate the integral

dx

∫ x3 − x2 .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second - Factor the denominator x 3 − x 2 into x 2 (x − 1) . Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand can be represented in the following way: 1 3

x −x

2

=

1

x

2

(x − 1)

A B C + + 2 x x x −1

=

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. 1 x3 − x2

(

=

Ax (x − 1) + B (x − 1) + Cx 2 x 2 (x − 1)

)

1 = A x 2 − x + B (x − 1) + Cx 2 = Ax 2 − Ax + Bx − B + Cx 2 1 = ( A + C )x 2 + (− A + B )x − B therefore, A+C = 0

−B = 1

−A + B = 0

which result in having A = −1 , B = −1 , and C = 1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. dx

∫ x3 − x2

=

A

B

1

C

1

1

∫ x dx + ∫ x 2 dx + ∫ x − 1 dx = − ∫ x dx − ∫ x 2 dx + ∫ x − 1 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. −

1

1

1

∫ x dx − ∫ x 2 dx + ∫ x − 1 dx = − ln

x +

1 + ln x − 1 + c x

=

1 + ln x − 1 − ln x + c x

Seventh - Check the answer by differentiating the solution. The result should match the integrand. − (x − 1) + x 2 − x (x − 1) 1 1 1 1 Let y = + ln x − 1 − ln x + c , then y ′ = − 2 + − +0 = 2 x

=

x

− x +1+ x 2 − x 2 + x 3

x −x

2

=

x

x

(x − 1)

1 3

x − x2

Example 1.3-10: Evaluate the integral

Hamilton Education Guides

x −1

5dx

∫ x 3 − 2x 2 + x . 43

Advanced Integration

1.3 Integration by Partial Fractions

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second – Factor the denominator x 3 − 2 x 2 + x into x(x − 1)2 . Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand can be represented in the following way: 5

x(x − 1)

A B C + + x x − 1 (x − 1)2

=

2

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. 5 x 3 − 2x 2 + x 5

(

A (x − 1)2 + Bx (x − 1) + Cx

=

) (

x(x − 1)2

)

= A x 2 − 2 x + 1 + B x 2 − x + Cx = Ax 2 − 2 Ax + A + Bx 2 − Bx + Cx 5

= ( A + B )x 2 + (− 2 A − B + C )x + A therefore,

A+ B = 0

−2 A − B + C = 0

A=5

which result in having A = 5 , B = −5 , and C = 5 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 5dx

∫ x 3 − 2x 2 + x

=

A

B

1

C

1

1

∫ x dx + ∫ x − 1 dx + ∫ (x − 1)2 dx = 5∫ x dx − 5∫ x − 1 dx + 5∫ (x − 1)2 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. 5

1

1

1

∫ x dx − 5∫ x − 1 dx + 5∫ (x − 1)2 dx = 5 ln x

− 5 ln x − 1 −

5 +c x −1

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = 5 ln x − 5 ln x − 1 − =

5(x − 1)2 − 5 x (x − 1) + 5 x x(x − 1)

2

=

5 +c x −1

(

1 x

, then y ′ = 5 ⋅ ⋅1 − 5 ⋅

)

5 x 2 − 2 x + 1 − 5x 2 + 5x + 5x 3

2

x − 2x + x

Example 1.3-11: Evaluate the integral

=

1 1 5 5 5 ⋅1 + 0 = − ⋅1 + 5 ⋅ + 2 1 x −1 − x x (x − 1) (x − 1)2 5 x 2 − 10 x + 5 − 5 x 2 + 5 x + 5 x 3

2

x − 2x + x

=

5 3

x − 2x 2 + x

x+6

∫ (x + 2)(x − 3)2 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

Hamilton Education Guides

44

Advanced Integration

1.3 Integration by Partial Fractions

Second – Factor the denominator. However, the denominator is already in its reduced form of (x + 2)(x − 3)2 . Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand can be represented in the following way: x+6

(x + 2)(x − 3)

=

2

A B C + + x + 2 x − 3 (x − 3)2

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. x+6

(x + 2)(x − 3)2 x+6

) (

(

A (x − 3)2 + B(x + 2)(x − 3) + C (x + 2)

=

(x + 2)(x − 3)2

)

= A x 2 − 6 x + 9 + B x 2 − x − 6 + C (x + 2) = Ax 2 − 6 Ax + 9 A + Bx 2 − Bx − 6 B + Cx + 2C = ( A + B )x 2 + (− 6 A − B + C )x + (9 A − 6 B + 2C ) therefore,

x+6 A+ B = 0

9 A − 6 B + 2C = 6

−6 A − B + C = 1

which result in having A =

4 25

, B=−

4 , 25

and C =

9 5

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x+6

A

B

C

∫ (x + 2)(x − 3)2 dx = ∫ x + 2 dx + ∫ x − 3 dx + ∫ (x − 3)2 dx =

4 25

1

4

1

9

1

∫ x + 2 dx − 25 ∫ x − 3 dx + 5 ∫ (x − 3)2 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. 4 25

1

4

1

9

1

∫ x + 2 dx − 25 ∫ x − 3 dx + 5 ∫ (x − 3)2 dx =

4 4 9 1 ln x + 2 − ln x − 3 − +c 25 25 5 ( x − 3)

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = = +

4 4 1 4 1 9 1 4 9 1 ⋅ − ⋅ + +0 ln x + 2 − ln x − 3 − + c , then y ′ = 25 x + 2 25 x − 3 5 (x − 3)2 25 25 5 (x − 3)

4(x − 3)2 − 4(x + 2 )(x − 3) + 45(x + 2 ) 25(x + 2 )(x − 3)2

4 x + 24 + 45 x + 90 25(x + 2 )(x − 3)

2

=

=

25 x + 150

25(x + 2 )(x − 3)

2

Example 1.3-12: Evaluate the integral

(

) (

)

4 x 2 − 6 x + 9 − 4 x 2 − x − 6 + 45(x + 2 )

=

25(x + 2 )(x − 3)2

25(x + 6 )

25(x + 2 )(x − 3)

2

=

=

4 x 2 − 24 x + 36 − 4 x 2 25(x + 2 )(x − 3)2

x+6

(x + 2)(x − 3)2

x+5

∫ x 3 + 4 x 2 + 4 x dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

Hamilton Education Guides

45

Advanced Integration

1.3 Integration by Partial Fractions

)

(

Second - Factor the denominator x 3 + 4 x 2 + 4 x into x x 2 + 4 x + 4 = x(x + 2)2 . Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand can be represented in the following way: x+5 3

2

x + 4x + 4x

=

(

x+5 2

x x + 4x + 4

=

)

x+5

x(x + 2 )

=

2

A B C + + x x + 2 ( x + 2 )2

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. x+5 x 3 + 4x 2 + 4x x+5

(

A (x + 2 )2 + Bx (x + 2 ) + Cx

=

x(x + 2 )(x + 2 )2

) (

)

= A x 2 + 4 x + 4 + B x 2 + 2 x + Cx = Ax 2 + 4 Ax + 4 A + Bx 2 + 2 Bx + Cx x+5

= ( A + B )x 2 + (4 A + 2 B + C )x + 4 A therefore,

A+ B = 0

4A = 5

4 A + 2B + C = 1 5 4

5 4

which result in having A = , B = − , and C = −

3 2

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. A

x+5

B

5

C

1

5

1

3

1

∫ x 3 + 4 x 2 + 4 x dx = ∫ x dx + ∫ x + 2 dx + ∫ (x + 2)2 dx = 4 ∫ x dx − 4 ∫ x + 2 dx − 2 ∫ (x + 2)2 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. 5 4

1

5

1

3

1

∫ x dx − 4 ∫ x + 2 dx − 2 ∫ (x + 2)2 dx =

1 5 3 5 +c ln x − ln x + 2 + ⋅ 2 x+2 4 4

Seventh - Check the answer by differentiating the solution. The result should match the integrand. 5 4

3 1 +c 2 x+2

5 4

Let y = ln x − ln x + 2 + ⋅ =

5 x 2 + 20 x + 20 − 5 x 2 − 10 x − 6 x

(

4 x 3 + 4x 2 + 4x

)

=

(

, then y ′ = 4 x + 20

4 x 3 + 4x 2 + 4x

Example 1.3-13: Evaluate the integral

)

5(x + 2 )2 − 5 x (x + 2 ) − 6 x 5 1 5 1 3 ⋅ − ⋅ − +0 = 4 x 4 x + 2 2(x + 2 )2 4 x ( x + 2 )2

=

4(x + 5)

(

4 x 3 + 4x 2 + 4x

)

=

x+5 3

x + 4x 2 + 4x

1

∫ x 5 + 2 x 4 + x 3 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 5 + 2 x 4 + x 3 into x 3 x 2 + 2 x + 1 = x 3 (x + 1)2 . Third - Write the linear factors in partial fraction form. Since both factors in the denominator are repeated, the integrand can be represented in the following way: Hamilton Education Guides

46

Advanced Integration

1.3 Integration by Partial Fractions

1 5

4

x + 2x + x

3

=

1

(

=

)

x 3 x 2 + 2x +1

1

x (x + 1) 3

A B C D E + + + + 2 3 x x x + 1 (x + 1)2 x

=

2

Fourth - Solve for the constants A , B , C , D , and E by equating coefficients of the like powers. 1 x 5 + 2x 4 + x 3

=

Ax 2 (x + 1)2 + Bx (x + 1)2 + C (x + 1)2 + Dx 3 (x + 1) + Ex 3 x 3 (x + 1)2

)

(

) (

(

)

1 = Ax 2 x 2 + 2 x + 1 + Bx x 2 + 2 x + 1 + C x 2 + 2 x + 1 + Dx 3 (x + 1) + Ex 3

1 = Ax 4 + 2 Ax 3 + Ax 2 + Bx 3 + 2 Bx 2 + Bx + Cx 2 + 2Cx + C + Dx 4 + Dx 3 + Ex 3 1 = ( A + D )x 4 + (2 A + B + D + E )x 3 + ( A + 2 B + C )x 2 + (B + 2C )x + C therefore, 2A + B + D + E = 0

A+ D = 0

A + 2B + C = 0

B + 2C = 0

C =1

which result in having A = 3 , B = −2 , C = 1 , D = −3 , and E = −1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 1

∫ x 5 + 2 x 4 + x 3 dx

1 x

= 3∫ dx − 2∫

1 x

2

dx +

1

1

1

∫ x 3 dx − 3∫ x + 1 dx − ∫ (x + 1)2 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. 1

1

1

1

1

∫ x dx − 2∫ x 2 dx + ∫ x 3 dx − 3∫ x + 1 dx − ∫ (x + 1)2 dx = 3 ln

3

x +

1 2 1 − − 3 ln x + 1 + +c x +1 x 2 x2

Seventh - Check the answer by differentiating the solution. The result should match the integrand. 2 x

Let y = 3 ln x + − =

+

1 2x

2

− 3 ln x + 1 +

1 +c x +1

3x 2 (x + 1)2 − 2 x(x + 1)2 + (x + 1)2 − 3 x 3 (x + 1) − x 3 x 3 (x + 1)2

− 3x 3 − x 3

(

)

x 3 x 2 + 2x + 1

=

1 x

, then y ′ = 3 ⋅ − =

2 x

2

+

1 x

3

− 3⋅

1 1 − +0 x + 1 (x + 1)2

3x 4 + 3x 2 + 6 x 3 − 2 x 3 − 2 x − 4 x 2 + x 2 + 1 + 2 x − 3x 4

(

)

x 3 x 2 + 2x + 1

1 5

x + 2x 4 + x 3

Example 1.3-14: Evaluate the integral

1

∫ x 4 − 6 x 3 + 9 x 2 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 4 − 6 x 3 + 9 x 2 into x 2 x 2 − 6 x + 9 = x 2 (x − 3)2 . Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand can be represented in the following way:

Hamilton Education Guides

47

Advanced Integration

1.3 Integration by Partial Fractions

1 4

3

x − 6x + 9x

2

=

1

(

x 2 x 2 − 6x + 9

=

)

1

x

2

(x − 3)

2

=

A B C D + + + 2 x x x − 3 (x − 3)2

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. 1 x 4 − 6x 3 + 9x 2

=

Ax (x − 3)2 + B (x − 3)2 + Cx 2 (x − 3) + Dx 2 x 2 (x − 3)2

(

) (

)

1 = Ax (x − 3)2 + B (x − 3)2 + Cx 2 (x − 3) + Dx 2 = Ax x 2 + 9 − 6 x + B x 2 + 9 − 6 x + Cx 2 (x − 3) + Dx 2

1 = Ax 3 + 9 Ax − 6 Ax 2 + Bx 2 + 9 B − 6 Bx + Cx 3 − 3Cx 2 + Dx 2 1 = ( A + C )x 3 + (− 6 A + B − 3C + D )x 2 + (9 A − 6 B )x + 9 B therefore, −6 A + B − 3C + D = 0

A+C = 0

which result in having A =

6 1 , B= 81 9

, C=−

6 , 81

9 A − 6B = 0

and D =

9B = 1

1 9

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. dx

∫ x 4 − 6x 3 + 9x 2

A

B

C

D

∫ x dx + ∫ x 2 dx + ∫ x − 3 dx + ∫ (x − 3)2 dx

=

=

6 dx 1 + 81 x 9

∫

6

dx

dx

1

dx

∫ x 2 − 81 ∫ x − 3 + 9 ∫ (x − 3)2

Sixth - Integrate each integral individually using integration methods learned in previous sections. 6 dx 1 + 81 x 9

∫

dx

6

dx

1

dx

∫ x 2 − 81 ∫ x − 3 + 9 ∫ (x − 3)2

=

1 6 6 1 +c − ln x − 3 − ln x − ( 9 x − 3) 9 x 81 81

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = =

=

6 1 1 1 6 1 6 1 6 1 + +0 ⋅ + − ⋅ ln x − − ln x − 3 − + c , then y ′ = 2 81 x − 3 9(x − 3)2 81 x 9 x 81 9 x 81 9(x − 3)

6 x(x − 3)2 + 9(x − 3)2 − 6 x 2 (x − 3) + 9 x 2 81x

2

(x − 3)

2

=

(

6 x 3 − 36 x 2 + 54 x + 9 x 2 − 54 x + 81 − 6 x 3 + 18 x 2 + 9 x 2

(

81x 2 x 2 − 6 x + 9

Hamilton Education Guides

)

) (

)

6 x x 2 − 6 x + 9 + 9 x 2 − 6 x + 9 − 6 x 3 + 18 x 2 + 9 x 2

(

81x 2 x 2 − 6 x + 9

=

(

81

81 x 4 − 6 x 3 + 9 x 2

)

)

=

1 4

x − 6x 3 + 9x 2

48

Advanced Integration

1.3 Integration by Partial Fractions

CASE III - The Denominator Has Distinct Quadratic Factors In this case the quadratic factors of the form ax 2 + bx + c appear only once in the denominator and are irreducible. To solve this class of rational fractions we equate each proper rational fraction with a single fraction of the form

Ax + B 2

ax + bx + c

Cx + D

,

2

cx + dx + e

,

Ex + F

2

ex + fx + g

, etc. The following

examples show the steps as to how this class of integrals are solved. x2 − x + 3

∫

Example 1.3-15: Evaluate the integral

dx .

x3 + x

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 3 + x into x x 2 + 1 . Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way: x2 − x + 3 3

x +x

=

x2 − x + 3

(

)

x x 2 +1

=

A Bx + C + x x 2 +1

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. x2 − x + 3 3

x +x x2 − x + 3

(

=

(

)

A x 2 + 1 + (Bx + C ) x

(

)

x x 2 +1

)

= A x 2 +1 + (Bx + C ) x = Ax 2 + A + Bx 2 + Cx

x2 − x + 3

= ( A + B )x 2 + Cx + A therefore,

A+ B =1

C = −1

A=3

which result in having A = 3 , B = −2 , and C = −1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

∫

x2 − x + 3 3

x +x

dx =

A

1

Bx + C

−2 x − 1

1

2x

1

∫ x dx + ∫ x 2 + 1 dx = 3∫ x dx + ∫ x 2 + 1 dx = 3∫ x dx − ∫ x 2 + 1 dx − ∫ x 2 + 1 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. To solve the second integral let u = x 2 + 1 . 1

2x

1

1

2 x du

1

∫ x dx − ∫ x 2 + 1 dx − ∫ x 2 + 1 dx = 3∫ x dx − ∫ u ⋅ 2 x − ∫ x 2 + 1 dx

3

1 x

1 u

= 3∫ dx − ∫ du − ∫

1 2

x +1

dx

= 3 ln x − ln u − tan −1 x + c = 3 ln x − ln x 2 + 1 − tan −1 x + c

Hamilton Education Guides

49

Advanced Integration

1.3 Integration by Partial Fractions

Seventh - Check the answer by differentiating the solution. The result should match the integrand. 1 x

−1

Let y = 3 ln x − ln x 2 + 1 − tan −1 x + c , then y ′ = 3 ⋅ + =

(

)

3 x 2 +1 − 2x 2 − x

(

)

x x 2 +1

=

3x 2 + 3 − 2 x 2 − x x3 + x

Example 1.3-16: Evaluate the integral

=

2

x +1

⋅ 2x −

1

1+ x

2

+0

=

3 2x 1 − − x x 2 +1 x 2 +1

x2 − x + 3 x3 + x 1

∫ x 3 + 25x dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 3 + 25 x into x x 2 + 25 . Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way: 1 3

x + 25 x

=

1

(

x x 2 + 25

A Bx + C + x x 2 + 25

=

)

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. 1

=

3

x + 25 x

(

(

)

A x 2 + 25 + (Bx + C ) x

(

x x 2 + 25

)

)

1 = A x 2 + 25 + (Bx + C ) x = Ax 2 + 25 A + Bx 2 + Cx 1 = ( A + B )x 2 + Cx + 25 A therefore, 25 A = 1

C=0

which result in having A =

1 25

, B=−

1 , 25

A+ B = 0

and C = 0

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 1

A

Bx + C

∫ x 3 + 25x dx = ∫ x dx + ∫ x 2 + 25 dx

=

1 25

1

1

x

∫ x dx − 25 ∫ x 2 + 25 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. To solve the second integral let u = x 2 + 25 . 1 25

=

1

1

x

∫ x dx − 25 ∫ x 2 + 25 dx =

1 25

1

1

x du

∫ x dx − 25 ∫ u ⋅ 2 x

=

1 25

1

1

1

∫ x dx − 50 ∫ u du =

1 1 ln x − ln u + c 25 50

1 1 ln x − ln x 2 + 25 + c 25 50

Hamilton Education Guides

50

Advanced Integration

1.3 Integration by Partial Fractions

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y =

(x

=

2

1 1 ln x − ln x 2 + 25 + c 25 50

)

+ 25 − x 2

(

25 x x 2 + 25

)

=

x 2 + 25 − x 2

(

25 x 3 + 25 x

, then y ′ =

=

)

(

25

25 x 3 + 25 x

1 1 1 1 ⋅ 2x + 0 ⋅ − ⋅ 2 25 x 50 x + 25

=

)

=

1 − 25 x

x

(

25 x 2 + 25

)

1 3

x + 25 x

1

∫ x 4 + 16 x 2 dx .

Example 1.3-17: Evaluate the integral

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 4 + 16x 2 into x 2 x 2 + 16 . Third - Write the factors in partial fraction form. Since the factors in the denominator are in quadratic form, the integrand can be represented in the following way: 1 4

x + 16 x

=

2

1

(

x 2 x 2 + 16

=

)

Ax + B x

2

+

Cx + D x 2 + 16

Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers. 1 x 4 + 16 x 2

(

=

( Ax + B ) (x 2 + 16)+ x 2 (Cx + D )

(

x 2 x 2 + 16

)

)

1 = ( Ax + B ) x 2 + 16 + x 2 (Cx + D ) = Ax 3 + 16 Ax + Bx 2 + 16 B + Cx 3 + Dx 2 1 = ( A + C )x 3 + (B + D )x 2 + 16 Ax + 16 B therefore, B+D =0

A+C = 0

which result in having A = 0 , B =

1 16

16 A = 0

, C = 0 , and D = −

16 B = 1

1 16

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 1

∫ x 4 + 16 x 2 dx = ∫

Ax + B x

2

dx +

B

Cx + D

D

∫ x 2 + 16 dx = ∫ x 2 dx + ∫ x 2 + 16 dx

=

1 16

1

1

1

∫ x 2 dx − 16 ∫ x 2 + 16 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. 1 16

=

1

1

1

∫ x 2 dx − 16 ∫ x 2 + 16 dx

=

x 1 1 1 ⋅− − ⋅ 4 tan −1 + c 4 16 x 256

Hamilton Education Guides

1 16

= −

1

1

1

∫ x 2 dx − 16 ∫ 16(x 2 + 1 ) 16

dx

=

1 16

1

1

1

∫ x 2 dx − 256 ∫ (x 2 + 1 )

dx

16

1 1 x − tan −1 + c 16 x 64 4

51

Advanced Integration

1.3 Integration by Partial Fractions

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = − =

1 16 x

2

−

x 1 1 − tan −1 + c 4 16 x 64

(

1

16 16 + x 2

=

)

1

, then y ′ =

(16 + x )− x 16 x (16 + x ) 2

2

2

16 x

=

2

2

−

1 1 1 ⋅ ⋅ +0 2 64 1 + x 4

=

16

16 + x 2 − x 2

(

16 x 2 16 + x 2

)

=

(

1 16 x

2

16 1 ⋅ 256 16 + x 2

−

16

16 x 2 16 + x 2

)

=

1 4

x + 16 x 2

1

∫ x 3 − 8 dx .

Example 1.3-18: Evaluate the integral

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 3 − 8 into (x − 2) x 2 + 2 x + 4 . Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way: 1 3

x −8

=

1

(x − 2) (x 2 + 2 x + 4)

=

A Bx + C + 2 x − 2 x + 2x + 4

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. 1 3

x −8

(

=

(

)

A x 2 + 2 x + 4 + (Bx + C )(x − 2 )

(x − 2) (x 2 + 2 x + 4)

)

1 = A x 2 + 2 x + 4 + (Bx + C )(x − 2 ) = Ax 2 + 2 Ax + 4 A + Bx 2 − 2 Bx + Cx − 2C 1 = ( A + B )x 2 + (2 A − 2 B + C )x + (4 A − 2C ) therefore, A+ B = 0

4 A − 2C = 1

2 A − 2B + C = 0

which result in having A =

1 12

, B=−

1 , 12

and C = −

1 3

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 1

∫ x3 − 8

dx

=

∫

A dx + x−2

Bx + C

∫ x 2 + 2x + 4

dx

=

1 12

∫

1 dx + x−2

∫

1 x−1 − 12 3 dx 2 x + 2x + 4

=

1 12

1

1

x+4

∫ x − 2 dx − 12 ∫ x 2 + 2 x + 4 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. To solve the second integral let u = x 2 + 2 x + 4 , then x+4

can be rewritten as x + 4 = (x + 1) + 3 =

Hamilton Education Guides

1 (2 x + 2) + 3 . 2

du = 2x + 2 dx

and dx =

du . 2x + 2

Also,

Therefore,

52

Advanced Integration

1 12

∫

1 1 dx − 12 x−2

1.3 Integration by Partial Fractions

x+4

∫ x 2 + 2x + 4

1

1

1

1

1 12

dx =

2x + 2

∫

(x + 1) + 3

1 1 dx − 12 x−2

∫ x 2 + 2x + 4

3

1

1 12

dx =

1

∫

1

1 1 dx − x−2 12

∫

1 2

(2 x + 2) + 3

x 2 + 2x + 4

2 x + 2 du 3 ⋅ − 2 x + 2 12 u

dx

1

=

1 12

∫ x − 2 dx − 24 ∫ x 2 + 2 x + 4 dx − 12 ∫ x 2 + 2 x + 4 dx = 12 ∫ x − 2 dx − 24 ∫

=

1 12

∫ x − 2 dx − 24 ∫ u du − 4 ∫ (x + 1)2 + 3 dx = 12 ln

=

1 1 3 3 x +1 x +1 1 1 ln x − 2 − ln x 2 + 2 x + 4 − tan −1 +c ln x 2 + 2 x + 4 − +c = ln x − 2 − tan −1 24 4⋅3 12 12 24 12 3 3

1

1

1

1

1

x−2 −

∫ (x + 1)2 + 3 dx

1 1 x +1 ln u − tan −1 +c 24 4 3 3

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = y′

=

1 1 3 x +1 ln x − 2 − ln x 2 + 2 x + 4 − tan −1 + c , then 12 24 12 3

1 1 1 1 3 ⋅ − ⋅ ⋅ (2 x + 2 ) − ⋅ 12 x − 2 24 x 2 + 2 x + 4 12

−

3 3 3 ⋅ ⋅ 2 3 12 3 + (x + 1)

=

1 + 12(x − 2 )

=

(

1 − 12(x − 2 )

−x − 4

12 x 2 + 2 x + 4

4+8

(

=

12(x − 2 ) x + 2 x + 4 2

)

=

)

=

(x

1 1 +  x +1   3

x +1

(

12 x 2 + 2 x + 4

) 12(x − 2 ) (x

)

−

⋅

2

(1⋅ 3 )− 0 ⋅ (x + 1) + 0 = 1 − x + 1 12(x − 2 ) ( 3 )2 12(x 2 + 2 x + 4 ) 3

(

12 x 2 + 2 x + 4

+ 2 x + 4 + (− x − 4 )(x − 2 )

2

(

2

+ 2x + 4

12

12(x − 2 ) x + 2 x + 4 2

Example 1.3-19: Evaluate the integral

)

=

)

3

=

=

)

1 + 12(x − 2 )

(

−x −1− 3

12 x 2 + 2 x + 4

)

x 2 + 2x + 4 − x 2 + 2x − 4x + 8 12(x − 2 ) x 2 + 2 x + 4

(

)

1

1

2

2

x + 2x + 4x − 2x − 4x − 8

=

3

x −8

1

∫ x 3 + 8 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 3 + 8 into (x + 2) x 2 − 2 x + 4 . Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way: 1 3

x +8

=

1

(x + 2) (x 2 − 2 x + 4)

=

A Bx + C + 2 x + 2 x − 2x + 4

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. 1 3

x +8

Hamilton Education Guides

=

(

)

A x 2 − 2 x + 4 + (Bx + C )(x + 2 )

(x + 2) (x 2 − 2 x + 4)

53

Advanced Integration

1.3 Integration by Partial Fractions

)

(

1 = A x 2 − 2 x + 4 + (Bx + C )(x + 2 ) = Ax 2 − 2 Ax + 4 A + Bx 2 + 2 Bx + Cx + 2C 1 = ( A + B )x 2 + (− 2 A + 2 B + C )x + (4 A + 2C ) therefore, A+ B = 0

−2 A + 2 B + C = 0 1 12

which result in having A =

, B=−

1 , 12

and C =

4 A + 2C = 1

1 3

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 1

∫ x3 + 8

dx

=

∫

A dx + x+2

Bx + C

∫ x 2 − 2x + 4

dx

=

1 12

1 dx + x+2

∫

1 x+ 1 − 12 3

1

1

1

x−4

∫ x 2 − 2 x + 4 dx = 12 ∫ x + 2 dx − 12 ∫ x 2 − 2 x + 4 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. To solve the second integral let u = x 2 − 2 x + 4 , then x−4 1 12

∫

can be rewritten as x − 4 = (x − 1) − 3 = 1 1 dx − 12 x+2

x−4

∫ x 2 − 2x + 4

dx =

1 12

∫

1 (2 x − 2) − 3 . 2

du = 2x − 2 dx

and dx =

du . 2x − 2

Also,

Therefore,

(x − 1) − 3

1 1 dx − x+2 12

∫ x 2 − 2x + 4

3

1

dx =

1 12

∫

1

1

1 1 dx − x+2 12

∫

1 2

(2 x − 2) − 3

dx

x 2 − 2x + 4

2 x − 2 du 3 ⋅ + u 2 x + 2 12

1

=

1 12

∫ x + 2 dx − 24 ∫ x 2 − 2 x + 4 dx + 12 ∫ x 2 − 2 x + 4 dx = 12 ∫ x + 2 dx − 24 ∫

=

1 12

∫ x + 2 dx − 24 ∫ u du + 4 ∫ (x − 1)2 + 3 dx = 12 ln

=

1 1 3 1 1 3 x −1 x −1 ln x + 2 − ln x 2 − 2 x + 4 + tan −1 ln x + 2 − ln x 2 − 2 x + 4 + tan −1 +c = +c 12 24 4⋅3 12 24 12 3 3

1

1

1

1

2x − 2

1

1

1

1

1

x+2 −

∫ (x − 1)2 + 3 dx

x −1 1 1 +c tan −1 ln u + 24 4 3 3

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = y′

=

1 1 3 x −1 ln x + 2 − ln x 2 − 2 x + 4 + tan −1 + c , then 12 24 12 3

1 1 1 1 3 ⋅ − ⋅ ⋅ (2 x − 2 ) + ⋅ 2 12 x + 2 24 x − 2 x + 4 12

+

3 3 3 ⋅ ⋅ 12 3 + (x − 1)2 3

=

1 + 12(x + 2 )

(

=

1 − 12(x + 2 )

−x + 4

12 x 2 − 2 x + 4

Hamilton Education Guides

)

=

(x

(

1 1 +  x −1   3

x −1

12 x 2 − 2 x + 4

) 12(x + 2 ) (x

2

)

+

2

⋅

(1⋅ 3 )− 0 ⋅ (x − 1) + 0 = 1 − x − 1 12(x + 2 ) ( 3 )2 12(x 2 − 2 x + 4 )

(

12 x 2 − 2 x + 4

− 2 x + 4 + (− x + 4 )(x + 2 ) 2

3

− 2x + 4

)

=

)

=

1 + 12(x + 2 )

(

−x +1+ 3

12 x 2 − 2 x + 4

)

x 2 − 2x + 4 − x 2 − 2x + 4x + 8

(

12(x + 2 ) x 2 − 2 x + 4

)

54

Advanced Integration

=

1.3 Integration by Partial Fractions

4+8

(

12(x + 2 ) x 2 − 2 x + 4

=

)

(

12

12(x + 2 ) x 2 − 2 x + 4

Example 1.3-20: Evaluate the integral

1

=

)

3

2

2

x − 2x + 4x + 2x − 4x + 8

=

1 3

x +8

x2

∫ 16 − x 4 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)(

)

(

)

Second - Factor the denominator 16 − x 4 into 4 − x 2 4 + x 2 = (2 − x )(2 + x ) 4 + x 2 . Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way: x2 16 − x

=

4

x2

A B Cx + D + + 2 − x 2 + x 4 + x2

=

(2 − x )(2 + x ) (4 + x 2 )

Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers. x2 16 − x x2 x2

x2

4

(

)

(

)

A(2 + x ) 4 − x 2 + B(2 − x ) 4 − x 2 + (2 − x )(2 + x )(Cx + D )

=

(

)

(2 − x )(2 + x ) (4 + x 2 )

(

)

= A(2 + x ) 4 + x 2 + B(2 − x ) 4 + x 2 + (2 − x )(2 + x )(Cx + D )

= 8 A + 2 Ax 2 + 4 Ax + Ax 3 + 8B + 2 Bx 2 − 4 Bx − Bx 3 + 4Cx + 4 D − Cx 3 − Dx 2

= ( A − B − C )x 3 + (2 A + 2 B − D )x 2 + (4 A − 4 B + 4C )x + (8 A + 8B + 4 D ) therefore, 2 A + 2B − D = 1

A− B −C = 0

1 8

4 A − 4 B + 4C = 0

1 8

which result in having A = , B = , C = 0 , and D = −

8 A + 8B + 4 D = 0

1 2

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x2

A

B

1

Cx + D

1

1

1

1

1

∫ 16 − x 4 dx = ∫ 2 − x dx + ∫ 2 + x dx + ∫ 4 + x 2 dx = 8 ∫ 2 − x dx + 8 ∫ 2 + x dx − 2 ∫ 4 + x 2 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. 1

1

1

1

1

1

1

1

1

1

1

1 8

∫ 2 − x dx + 8 ∫ 2 + x dx − 2 ∫ 4 + x 2 dx = 8 ∫ 2 − x dx + 8 ∫ 2 + x dx − 2 ∫ 2 2 + x 2 dx

=

1 1 1 1 x ln 2 − x + ln 2 + x − ⋅ tan −1 + c 8 8 2 2 2

=

1 1 1 x ln 2 − x + ln 2 + x − tan −1 + c 8 8 4 2

Seventh - Check the answer by differentiating the solution. The result should match the integrand.

Hamilton Education Guides

55

Advanced Integration

1.3 Integration by Partial Fractions 1 8

1 8

1 4

1 1 1 1 1 1 1 ⋅ + ⋅ − ⋅ ⋅ +0 2 8 2 − x 8 2 + x 4 1+ x 2

x 2

Let y = ln 2 − x + ln 2 + x − tan −1 + c , then y ′ = =

+

1 1 − + 8(2 − x ) 8(2 + x )

4

(

) (2 x =

8 4 + x2

2 x 2 − 4 x − x 3 − 16 + 4 x 2

(

)(

8 4 − x2 4 + x2

=

)

2

(2 + x ) (4 + x 2 )+ (2 − x ) (4 + x 2 )− 4(2 − x )(2 + x ) = 8(2 − x )(2 + x ) (4 + x 2 )

)

+ 2 x 2 + 4 x 2 + (8 + 8 − 16 )

(

8 16 − x 4

Example 1.3-21: Evaluate the integral

=

)

(

8x 2

8 16 − x 4

)

4

8 + 2x 2 + 4x + x 3 + 8

(

)(

8 4 − x2 4 + x2

)

x2

=

16 − x 4

5

∫ x 4 − 1 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)(

)

(

)

Second - Factor the denominator x 4 − 1 into x 2 − 1 x 2 + 1 = (x − 1)(x + 1) x 2 + 1 . Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way: 5 4

x −1

=

5

=

(x − 1)(x + 1) (x 2 + 1)

A B Cx + D + + x −1 x +1 x 2 +1

Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers. 5

=

4

x −1 5 5 5

(

)

)

(

A(x + 1) x 2 + 1 + B(x − 1) x 2 + 1 + (x − 1)(x + 1)(Cx + D )

(

(x − 1)(x + 1) (x 2 + 1)

(

)

)

= A(x + 1) x 2 + 1 + B(x − 1) x 2 + 1 + (x − 1)(x + 1)(Cx + D )

= Ax 3 + Ax 2 + Ax + A + Bx 3 − Bx 2 + Bx − B + Cx 3 + Dx 2 − Cx − D

= ( A + B + C )x 3 + ( A − B + D )x 2 + ( A + B − C )x + ( A − B − D ) therefore,

A+ B +C = 0

A− B + D = 0

5 4

A+ B −C = 0

5 4

which result in having A = , B = − , C = 0 , and D = −

A− B − D = 5

5 2

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 5

A

B

∫ x 4 − 1 dx = ∫ x − 1 dx + ∫ x + 1 dx + ∫

Cx + D 2

x +1

dx

=

5 4

1

5

1

5

1

∫ x − 1 dx − 4 ∫ x + 1 dx − 2 ∫ x 2 + 1 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections.

Hamilton Education Guides

56

Advanced Integration

5 4

1

5

1.3 Integration by Partial Fractions

1

5

1

∫ x − 1 dx − 4 ∫ x + 1 dx − 2 ∫ x 2 + 1 dx

5 5 5 ln x − 1 − ln x + 1 − tan −1 x + c 4 4 2

=

Seventh - Check the answer by differentiating the solution. The result should match the integrand. 5 4

5 4

5 2

5 1 5 1 5 1 ⋅ − ⋅ − ⋅ ⋅1 + 0 2 4 x −1 4 x +1 2 x +1

Let y = ln x − 1 − ln x + 1 − tan −1 x + c , then y ′ = =

5 5 − − 4(x − 1) 4(x + 1)

5

(

(

2

)(

4 x −1 4 + x

2

)

2 x 2 +1

+ 5 x 2 − 5 x + 5 − 10 x 2 + 10

)

=

(

(

)

(

)

5(x + 1) x 2 + 1 − 5(x − 1) x 2 + 1 − 10(x − 1)(x + 1)

=

(

)

4(x − 1)(x + 1) x 2 + 1

20 4

)

4 x −1

5x 3 + 5x 2 + 5x + 5 − 5x 3

(

)(

4 x 2 −1 4 + x 2

)

5

=

Example 1.3-22: Evaluate the integral

=

4

x −1 1

∫ x 3 − 64 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 3 − 64 into (x − 4) x 2 + 4 x + 16 . Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way: 1

=

3

x − 64

1

(x − 4) (x 2 + 4 x + 16)

=

A Bx + C + 2 x − 4 x + 4 x + 16

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. 1 3

x − 64

=

(

)

(

)

A x 2 + 4 x + 16 + (Bx + C )(x − 4 )

(x − 4) (x 2 + 4 x + 16)

1 = A x 2 + 4 x + 16 + (Bx + C )(x − 4 ) = Ax 2 + 4 Ax + 16 A + Bx 2 − 4 Bx + Cx − 4C 1 = ( A + B )x 2 + (4 A − 4 B + C )x + (16 A − 4C ) therefore, 4 A − 4B + C = 0

A+ B = 0

which result in having A =

1 48

, B=−

1 , 48

and C = −

16 A − 4C = 1

1 6

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 1

∫ x 3 − 64

dx =

∫

A dx + x−4

Bx + C

∫ x 2 + 4 x + 16

dx =

1 48

∫

1 dx + x−4

1 x− 1 − 48 6

∫ x 2 + 4 x + 16 dx =

1 48

1

1

x +8

∫ x − 4 dx − 48 ∫ x 2 + 4 x + 16 dx

Sixth - Integrate each integral individually using integration methods learned in previous

Hamilton Education Guides

57

Advanced Integration

1.3 Integration by Partial Fractions

sections. To solve the second integral let u = x 2 + 4 x + 16 , then Also, x + 8 can be rewritten as x + 8 = (x + 2) + 6 = 1 48

∫

1 1 dx − 48 x−4

∫

1

1

1

1

1 dx = 2 48 x + 4 x + 16 x +8

∫

1 1 dx − 48 x−4

1 (2 x + 4) + 6 . 2

∫

du = 2x + 4 dx

du . 2x + 4

Therefore,

(x + 2 ) + 6

1 dx = 2 48 x + 4 x + 16

1 48

and dx =

∫

1

1

1 1 dx − x−4 48

1 2

(2 x + 4) + 6

∫ x 2 + 4 x + 16 dx

6 2 x + 4 du ⋅ − u 2 x + 4 48

1

=

1 48

∫ x − 4 dx − 96 ∫ x 2 + 4 x + 16 dx − 48 ∫ x 2 + 4 x + 16 dx =

=

1 48

∫ x − 4 dx − 96 ∫ u du − 8 ∫ (x + 2)2 + 12 dx

=

12 1 1 12 x+2 x+2 1 1 ln x − 4 − ln x 2 + 4 x + 16 − tan −1 +c ln x − 4 − ln x 2 + 4 x + 16 − tan −1 +c = 96 48 96 96 8 ⋅12 48 12 12

2x + 4

1

1

1

6

1

∫ x − 4 dx − 96 ∫

∫ (x + 2)2 + 12 dx

1 1 1 x+2 ln x − 4 − ln u − tan −1 +c 48 96 8 12 12

=

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y =

1 1 12 x+2 ln x − 4 − ln x 2 + 4 x + 16 − tan −1 + c , then 48 96 96 12

12 1 1 1 1 ⋅ (2 x + 4 ) − ⋅ ⋅ − ⋅ 96 48 x − 4 96 x 2 + 4 x + 16

y′

=

−

12 12 12 ⋅ ⋅ 2 96 12 + (x + 2 ) 12

=

1 + 48(x − 4 )

=

(

=

−x − 8

48 x 2 + 4 x + 16

16 + 32

(

48(x − 4 ) x 2 + 4 x + 16

Hamilton Education Guides

)

)

=

1 − 48(x − 4 )

=

(x

(

1 +  x + 2   12 

x+2

48 x 2 + 4 x + 16

) 48(x − 4 ) (x

2

1

)

⋅

2

−

(1⋅ 12 )− 0 ⋅ (x + 2) + 0 = ( 12 )2

(

48 x 2 + 4 x + 16

+ 4 x + 16 + (− x − 8)(x − 4 )

(

2

+ 4 x + 16

48

48(x − 4 ) x 2 + 4 x + 16

)

=

6

)

=

1 + 48(x − 4 )

=

)

1 − 48(x − 4 )

2

48 x + 4 x + 16

−x − 2 − 6

48 x 2 + 4 x + 16

)

x 2 + 4 x + 16 − x 2 + 4 x − 8 x + 32

(

48(x − 4 ) x 2 + 4 x + 16

1 3

(

(

x+2 2

2

x + 4 x + 16 x − 4 x − 16 x − 64

=

)

1 3

x − 64

58

)

Advanced Integration

1.3 Integration by Partial Fractions

CASE IV - The Denominator Has Repeated Quadratic Factors In this case each irreducible quadratic factor of the form ax 2 + bx + c appears n times in the denominator. To solve this class of rational fractions we equate each proper rational fraction, that appears n times in the denominator, with a sum of n partial fractions of the form M 1 x + N1 2

ax + bx + c

+

M 2x + N2

( ax

2

+ bx + c

)

2

+ ... +

M n x + Nn

( ax

2

+ bx + c

this class of integrals are solved.

Example 1.3-23: Evaluate the integral

)

n

. The following examples show the steps as to how

x2

∫ x 4 + 2 x 2 + 1 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

2

Second - Factor the denominator x 4 + 2 x 2 + 1 into x 2 + 1 . Third - Write the factors in partial fraction form. Since the quadratic form in the denominator is repeated, the integrand can be represented in the following way: x2 4

2

x + 2x + 1

=

x2

(x + 1)

2

2

=

Ax + B

+

2

x +1

Cx + D

(x + 1)

2

2

Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers. x2 x 4 + 2x 2 + 1 x2

(

=

( Ax + B ) (x 2 + 1)+ Cx + D

(x + 1) 2

)

2

= ( Ax + B ) x 2 + 1 + Cx + D = Ax 3 + Ax + Bx 2 + B + Cx + D x2

= Ax 3 + Bx 2 + ( A + C )x + (B + D ) therefore, B =1

A=0

A+C = 0

B+D =0

which result in having A = 0 , B = 1 , C = 0 , and D = −1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x2

Ax + B

0 +1

Cx + D

1

0 −1

1

∫ x 4 + 2 x 2 + 1 dx = ∫ x 2 + 1 dx + ∫ (x 2 + 1)2 dx = ∫ x 2 + 1 dx + ∫ (x 2 + 1)2 dx = ∫ x 2 + 1 dx − ∫ (x 2 + 1)2 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. To solve the second integral let x = tan t , then 1

∫ x 2 +1

dx −

1

∫ (x 2 + 1)2

Hamilton Education Guides

dx = arc tan x −

sec 2 t

∫ (tan 2 t + 1)2

dt

dx = sec 2 t dt

= arc tan x − ∫

which implies dx = sec 2 t dt .

sec 2 t

(sec t ) 2

2

dt

= arc tan x − ∫

sec 2 t sec 4 t

dt

59

Advanced Integration

= arc tan x − ∫

1.3 Integration by Partial Fractions

1 2

sec t

= arc tan x − ∫ cos 2 t dt = tan −1 x −

dt

1 2

1 2

1 2

= tan −1 x − ( t + sin t cos t ) + c = tan −1 x − tan −1 x − ⋅

1 2

∫ ( 1 + cos 2t ) dt

x

1

⋅

1+ x 2

1 1 = tan −1 x −  t + sin 2t  + c

+c

1+ x 2

2

2

=



x 1 +c tan −1 x − 2 2 x2 + 1

(

)

Or, we could use the already derived integration formulas by using Tables 1.2-1 and 1.4-3. Note – Since the objective of this section is to teach the process for solving integrals using the Partial Fractions method, in the remaining example problems, we will use the already derived integration formulas summarized primarily in Tables 1.2-1 and 1.4-3 in order to solve this class of problems. Seventh - Check the answer by differentiating the solution. The result should match the integrand. 1 2

Let y = tan −1 x − =

(

1

)

2 x 2 +1

−

x

(

(

)

2 x 2 +1

− x 2 +1

)

2 x 2 +1

2

=

(

x 2 +1+ x 2 −1

(

=

)

2 x 2 +1

2

Example 1.3-24: Evaluate the integral

)

1 1 1 1⋅ x 2 + 1 − 2 x ⋅ x 1 +0 = ⋅ − ⋅ 2 2 x 2 +1 2 x 2 +1 2 x 2 +1

+ c , then y ′ =

(

2x 2

)

2 x 2 +1

2

=

(

x2

=

(x + 1)

2

2

)

(

)

−

x 2 − 2x 2 + 1

(

)

2 x 2 +1

2

x2 x 4 + 2x 2 + 1

x 2 +1

∫ x 4 + 8x 2 + 16 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

2

Second - Factor the denominator x 4 + 8 x 2 + 16 into x 2 + 4 . Third - Write the factors in partial fraction form. Since the quadratic form in the denominator is repeated, the integrand can be represented in the following way: x 2 +1 4

2

x + 8 x + 16

x 2 +1

=

(x

2

+4

)

2

=

Ax + B 2

x +4

+

Cx + D

(x

2

+4

)

2

Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers. x 2 +1 x 4 + 8 x 2 + 16 x 2 +1

(

( Ax + B ) (x 2 + 4)+ Cx + D

(x

2

+4

)

2

= ( Ax + B ) x 2 + 4 + Cx + D = Ax 3 + 4 Ax + Bx 2 + 4 B + Cx + D x 2 +1

A=0

)

=

= Ax 3 + Bx 2 + (4 A + C )x + (4 B + D ) therefore, B =1

4A + C = 0

4B + D = 1

which result in having A = 0 , B = 1 , C = 0 , and D = −3 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants

Hamilton Education Guides

60

Advanced Integration

1.3 Integration by Partial Fractions

with their specific values. x 2 +1

Ax + B

0 +1

Cx + D

0−3

3

1

∫ x 4 + 8x 2 + 16 dx = ∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx = ∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx = ∫ x 2 + 4 dx − ∫ (x 2 + 4)2 dx Sixth - Integrate each integral individually by using Tables 1.2-1 and 1.4-3. 1

3

∫ x 2 + 4 dx − ∫ (x 2 + 4)2 dx

=

1 x 3 x tan −1 − tan −1 − 2 2 16 2

(

3x 2

8 x +4

+c =

)

x 5 tan −1 − 2 16

(

3x 2

8 x +4

)

+c

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = =

(

x 5 tan −1 − 16 2

5

8 x2 + 4

)

(

3x

8 x2 + 4

)

3 x 2 − 2x 2 + 4 − ⋅ 2 8 x2 + 4

(

)

+ c , then y ′ =

=

(

5

8 x2 + 4

Example 1.3-25: Evaluate the integral

)

5 ⋅ 16

1

()

x 2 2

3 − x2 + 4 − ⋅ 2 8 x2 + 4

(

(

)

1 3 1⋅ x 2 + 4 − 2 x ⋅ x − ⋅ +0 2 2 8 2 +1 x +4 ⋅

=

)

(

)

5 x 2 + 20 + 3 x 2 − 12

(

8 x2 + 4

)

2

=

x 2 +1

(x

2

+4

)

2

=

x 2 +1 x 4 + 8 x 2 + 16

1

∫ x4 + 10 x2 + 25 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is a rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

2

Second - Factor the denominator x 4 + 10 x 2 + 25 into x 2 + 5 . Third - Write the factors in partial fraction form. Since the quadratic form in the denominator is repeated, the integrand can be represented in the following way: 1 4

=

2

x + 10 x + 25

(x

1 2

+5

)

2

Ax + B

=

+

2

x +5

Cx + D

(x

2

+5

)

2

Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers. 1 x 4 + 10 x 2 + 25

(

)

=

( Ax + B ) (x 2 + 5)+ Cx + D

(x

2

+5

)

2

1 = ( Ax + B ) x 2 + 5 + Cx + D = Ax 3 + 5 Ax + Bx 2 + 5 B + Cx + D 1 = Ax 3 + Bx 2 + (5 A + C )x + (5B + D ) therefore, A=0

B=0

5A + C = 0

5B + D = 1

which result in having A = 0 , B = 0 , C = 0 , and D = 1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

Hamilton Education Guides

61

Advanced Integration

1.3 Integration by Partial Fractions

1

Ax + B

0+0

Cx + D

0 +1

1

∫ x4 + 10 x2 + 25 dx = ∫ x 2 + 5 dx + ∫ (x 2 + 5)2 dx = ∫ x 2 + 5 dx + ∫ (x 2 + 5)2 dx = ∫ (x 2 + 5)2 dx Sixth - Integrate each integral individually by using Tables 1.2-1 and 1.4-3. 1

1

∫ (x 2 + 5)2 dx = 10

5

x

tan −1

5

+

(

x

10 x 2 + 5

)

+c

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y =

=

(

1 10 5

5

50 x 2 + 5

)

tan −1

+

x

+

5

x

(

10 x 2 + 5

1 x 2 − 2x 2 + 5 ⋅ 2 10 x2 + 5

)

(

+ c , then y ′ =

)

=

(

1

10 x 2 + 5

Example 1.3-26: Evaluate the integral

)

+

1

1

⋅

10 5   

− x2 + 5

(

10 x 2 + 5

)

2

2

 +1  

x 5

⋅

1

+

5

x2 + 5− x2 + 5

=

(

10 x 2 + 5

(

)

1⋅ x 2 + 5 − 2 x ⋅ x

)

2

(

10 x 2 + 5

=

(x

)

2

1 2

+5

)

2

+0

=

1 4

x + 10 x 2 + 25

x3

∫ x 4 + 4 x 2 + 4 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is a rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

2

Second - Factor the denominator x 4 + 4 x 2 + 4 into x 2 + 2 . Third - Write the factors in partial fraction form. Since the quadratic form in the denominator is repeated, the integrand can be represented in the following way: x3 4

2

x + 4x + 4

=

(x

x3 2

+2

)

2

=

Ax + B 2

x +2

+

Cx + D

(x

2

+2

)

2

Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers. x3 x 4 + 4x 2 + 4 x3

(

)

=

( Ax + B ) (x 2 + 2)+ Cx + D

(x

2

+2

)

2

= ( Ax + B ) x 2 + 2 + Cx + D = Ax 3 + 2 Ax + Bx 2 + 2 B + Cx + D x3

A =1

= Ax 3 + Bx 2 + (2 A + C )x + (2 B + D ) therefore, 2A + C = 0

B=0

2B + D = 0

which result in having A = 1 , B = 0 , C = −2 , and D = 0 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x3

Ax + B

Cx + D

x+0

−2 x + 0

x

2x

∫ x 4 + 4 x 2 + 4 dx = ∫ x 2 + 2 dx + ∫ (x 2 + 2)2 dx = ∫ x 2 + 2 dx + ∫ (x 2 + 2)2 dx = ∫ x 2 + 2 dx − ∫ (x 2 + 2)2 dx Hamilton Education Guides

62

Advanced Integration

1.3 Integration by Partial Fractions

Sixth - Integrate each integral individually by using Tables 1.2-1 and 1.4-3. 2x

x

x 1

1

x

∫ x 2 + 2 dx − ∫ (x 2 + 2)2 dx = ∫ u ⋅ 2 x du − 2∫ u 2 ⋅ 2 x du =

1 1 ln u − u − 2+1 + c 2 − 2 +1

=

1 ln u + u −1 + c 2

=

1 2

=

1

1

2

∫ u du − 2 ∫ u 2 du

1 1 ln u + + c 2 u

=

1 2

1

∫ u du − ∫ u

−2

du

1 1 ln x 2 + 2 + +c 2 2 x +2

=

Seventh - Check the answer by differentiating the solution. The result should match the integrand. 1 2

1

Let y = ln x 2 + 2 + =

x 3 + 2x − 2x

(x

2

+2

)

2

=

(x

2

x +2

x3 2

+2

+c ,

=

)

2

then y ′ =

1 2x ⋅ − 2 x2 + 2

(x

2x 2

+2

)

2

+0 =

x 2

x +2

−

(x

2x 2

+2

)

2

=

(

)

x x 2 + 2 − 2x

(x

2

+2

)

2

x3 x 4 + 4x 2 + 4

Example 1.3-27: Evaluate the integral

2x 2 + x + 7

∫ x 4 + 8x 2 + 16 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is a rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

2

Second - Factor the denominator x 4 + 8 x 2 + 16 into x 2 + 4 . Third - Write the factors in partial fraction form. Since the quadratic form in the denominator is repeated, the integrand can be represented in the following way: 2x 2 + x + 7 4

=

2

x + 8 x + 16

2x 2 + x + 7

(x

2

+4

)

2

=

Ax + B 2

x +4

+

Cx + D

(x

2

+4

)

2

Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers. 2x 2 + x + 7 x 4 + 8 x 2 + 16 2x 2 + x + 7

(

=

)

( Ax + B ) (x 2 + 4)+ Cx + D

(x

2

+4

)

2

= ( Ax + B ) x 2 + 4 + Cx + D = Ax 3 + 4 Ax + Bx 2 + 4 B + Cx + D

2x 2 + x + 7

= Ax 3 + Bx 2 + (4 A + C )x + (4 B + D ) therefore,

B=2

A=0

4A + C = 1

4B + D = 7

which result in having A = 0 , B = 2 , C = 1 , and D = −1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 2x 2 + x + 7

Ax + B

Cx + D

0+2

x −1

2

x −1

∫ x 4 + 8x 2 + 16 dx = ∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx = ∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx = ∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx =

2

x

1

∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx − ∫ (x 2 + 4)2 dx

Hamilton Education Guides

63

Advanced Integration

1.3 Integration by Partial Fractions

Sixth - Integrate each integral individually by using Tables 1.2-1 and 1.4-3. 2

1

x

∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx − ∫ (x 2 + 4)2 dx 1 2

1 2

x 1 1 1 u − 2+1 − tan −1 − 2 2 − 2 +1 16

x 2

= 2 ⋅ tan −1 + ⋅ x 2

= tan −1 −

(

1 2

2 x +4

)

−

x 1 tan −1 − 16 2

(

x 2

= 2 ⋅ tan −1 + ∫

x 2

8 x +4

x

(

8 x2 + 4

)

+c =

)

x u

2

⋅

x x 1 1 +c du − tan −1 − 2 8 x2 + 4 16 2x

(

+ c = tan −1

15 x tan −1 + 16 2

(

)

x 1 x 1 − − tan −1 − 2 2 2u 16 −4 − x 2

8 x +4

)

+c =

x

(

8 x2 + 4

x 15 tan −1 − 16 2

)

(

+c x+4 2

8 x +4

)

+c

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = =

15 x tan −1 − 16 2

15

(

2

8 x +4

)

+ c , then y ′ =

1 x 2 − 2 x 2 − 8x + 4 − ⋅ 2 8 x2 + 4

( ) ( 8 (2 x 2 + x + 7 ) 2x = = 2 8(x 2 + 4 ) (x 8 x2 + 4

x+4

)

2

+ x+7

2

+4

)

2

=

=

(

15

8 x2 + 4

)

(

)

15 1 1 1 ⋅ x 2 + 4 − 2 x ⋅ (x + 4) ⋅ ⋅ − +0 2 16 x 2 + 1 2 2 8 x +4 2

()

+

x 2 + 8x − 4

(

8 x2 + 4

)

2

=

(

)

15 x 2 + 60 + x 2 + 8 x − 4

(

8 x2 + 4

=

)

2

16 x 2 + 8 x + 56

(

8 x2 + 4

)

2

2x 2 + x + 7 x 4 + 8 x + 16

Section 1.3 Practice Problems – Integration by Partial Fractions Evaluate the following integrals: a.

dx

∫ x 2 + 5x + 6 x+5

=

d.

∫ x 3 + 2x 2 + x

g.

∫ x 3 − 1 dx =

dx

=

1

Hamilton Education Guides

x 2 +1

b.

∫ x 3 − 4 x dx =

e.

1

∫ x 3 − 2x 2 + x

h.

∫ x4 − 1 dx =

1

dx

c. =

1

∫ 36 − x 2 dx = x2 + 3

f.

∫ x 2 − 1 dx =

i.

∫ x 3 + 64 dx =

1

64

Advanced Integration

1.4

1.4 Integration of Hyperbolic Functions

Integration of Hyperbolic Functions

In the following examples we will solve problems using the formulas below: Table 1.4-1: Integration Formulas for Hyperbolic Functions 1.

∫ sinh x dx = cosh x + c

2.

∫ cosh x dx = sinh x + c

4.

∫ coth x dx = ln

5.

∫ sec h x dx = sin

7.

∫ tanh x sec h x dx = − sec h x + c

8.

∫ coth x csc h x dx = − csc h x + c

sinh x + c

sinh 2 x x + +c 4 2

10.

∫ cosh

2

x dx =

13.

∫ sec h

2

x dx = tanh x + c

11.

∫ tanh

14.

∫ csc h

2

2

−1

(tanh x ) + c

x dx = x − tanh x + c

3.

∫ tanh x dx = ln cosh x + c

6.

∫ csc h x dx = ln

9.

∫ sinh

12.

2

∫ coth

x dx = 2

tanh

x +c 2

sinh 2 x x − +c 4 2

x dx = x − coth x + c

x dx = − coth x + c

Additionally, the following formulas, similar to the trigonometric functions, hold for the hyperbolic functions: 1. Unit Formulas tanh h 2 x + sec h 2 x = 1

cosh 2 x − sinh 2 x = 1

coth h 2 x − csc h 2 x = 1

2. Addition Formulas sinh (x ± y ) = sinh x cosh y ± cosh x sinh y tanh (x ± y ) =

tanh x ± tanh y 1 ± tanh x tanh y

cosh (x ± y ) = cosh x cosh y ± sinh h x sinh y coth (x ± y ) =

coth x coth y ± 1 coth y ± coth x

3. Half Angle Formulas sinh

1 x= 2

cosh x − 1 2

cosh

1 x= 2

cosh x + 1 2

tanh

1 x= 2

cosh x − 1 cosh x + 1

tanh

1 x 2

sinh x cosh x + 1

=

=

x 0

sinh

cosh x − 1 1 x=− 2 2

x 0

x 0

sinh

cosh x − 1 1 x=− 2 cosh x + 1

x 0

or,

cosh x − 1 sinh x

4. Double Angle Formulas sinh 2 x = 2 sinh x cosh x tanh 2 x =

cosh 2 x = cosh 2 x + sinh 2 x = 2 cosh 2 x − 1 = 1+ 2 sinh 2 x

2 tanh x 1 + tanh 2 x

Hamilton Education Guides

65

Advanced Integration

1.4 Integration of Hyperbolic Functions

Also, the hyperbolic functions are defined by

(

2

) (e (e

1 2

(e

sinh x =

1 x e − e −x 2

tanh x =

sinh x 2 = cosh x 1

1

sec h x =

1 = cosh x

x x

− e −x + e −x

) )

1 x

+e

−x

)

=

=

e x + e −x

2 e +e

2

) (e (e

1 2

(e

1 x e + e −x 2

coth x =

cosh x 2 = sinh x 1

1

e x − e −x

x

(

cosh x =

csc h x =

−x

1 = sinh x

x x

+ e −x − e −x

1 x

−e

−x

) )

=

=

)

e x + e −x e x − e −x

2 x

e − e −x

Also note that the negative argument of the hyperbolic functions is equal to the following: sinh (− x ) = − sinh x

tanh (− x ) = − tanh x

coth (− x ) = − coth x

csc h (− x ) = − csc h x

cosh (− x ) = cosh x

sec h (− x ) = sec h x

Finally, we need to know how to differentiate the hyperbolic functions (addressed in Chapter 3, Section 3.4 of Calculus 1 book) in order to check the answer to the given integrals below. The derivatives of hyperbolic functions are repeated here and are as follows: Table 1.4-2: Differentiation Formulas for Hyperbolic Functions d sinh u dx d cosh u dx d tanh u dx

d coth u dx d sec h u dx d csc h u dx

du dx du sinh u ⋅ dx du sec h 2 u ⋅ dx

= cosh u ⋅ = =

= − csc h 2 u ⋅

du dx

du dx du − csc h u coth u ⋅ dx

= − sec h u tanh u ⋅ =

Let’s integrate some hyperbolic functions using the above integration formulas. Example 1.4-1: Evaluate the following integrals: 1

a.

∫ sinh 5x dx =

b.

∫ sinh 6 x dx

d.

∫ cosh 5 x dx =

1

e.

∫ (sinh 4 x + cosh 2 x ) dx

g.

∫ csc h

2

h.

∫ csc h

k.

∫x

j.

5 x dx =

2 2 ∫ x sec h (x + 5) dx =

2

2

=

1 x dx 4

=

=

csc h 2 x 3 dx =

c.

∫ cosh 7 x dx =

f.

∫ csc h 8x dx =

i.

∫x

l.

∫ 2 x csc h

2

sec h 2 x 3 dx = 2 2

x dx =

Solutions: a. Given ∫ sinh 5 x dx let u = 5 x , then du

∫ sinh 5x dx = ∫ sinh u ⋅ 5 Hamilton Education Guides

=

du d = 5x dx dx

1 sinh u du 5

∫

=

= 5 which implies dx =

1 cosh u + c 5

=

du 5

. Therefore,

1 cosh 5 x + c 5

66

Advanced Integration

1.4 Integration of Hyperbolic Functions 1 5

Check: Let y = cosh 5 x + c , then y ′ =

1 d d ⋅ sinh 5 x + c 5 dx dx

5 ⋅ sinh 5 x = sinh 5 x 5 x du d x x = sinh dx let u = , then 6 dx dx 6 6

1 d ⋅ sinh 5 x ⋅ 5x + 0 5 dx

=

=

1 ⋅ sinh 5 x ⋅ 5 5

=

∫

b. Given x

∫ sinh 6 dx = ∫ sinh u ⋅ 6 du

1 6

=

which implies dx = 6 du . Therefore,

= 6∫ sinh u du = 6 cosh u + c = 6 cosh

x +c 6

d x d x x 1 x d x Check: Let y = 6 cosh + c , then y ′ = 6 ⋅ cosh + c = 6 ⋅ sinh ⋅ + 0 = 6 ⋅ sinh ⋅ 6

=

6 x ⋅ sinh 6 6

=

6

dx

6 dx 6

dx

6 6

x sinh 6

c. Given ∫ cosh 7 x dx let u = 7 x , then du

∫ cosh 7 x dx = ∫ cosh u ⋅ 7

=

1 7

du d = 7x dx dx

1 cosh u du 7

∫

Check: Let y = sinh 7 x + c , then y ′ =

=

= 7 which implies dx =

1 sinh u + c 7

. Therefore,

1 sinh 7 x + c 7

=

1 d d ⋅ sinh 7 x + c 7 dx dx

7 ⋅ cosh 7 x = cosh 7 x 7 x x du d x cosh dx let u = , then = 5 5 dx dx 5

du 7

1 d 7x + 0 ⋅ cosh 7 x ⋅ 7 dx

=

=

1 ⋅ cosh 7 x ⋅ 7 7

= d. Given

∫ x

∫ cosh 5 dx

=

∫ cosh u ⋅ 5 du x 5

5 x ⋅ cosh 5 5

= cosh

1 5

which implies dx = 5 du . Therefore,

= 5∫ cosh u du = 5 sinh u + c = 5 sinh

Check: Let y = 5 sinh + c , then y ′ = 5 ⋅ =

=

d x d sinh + c 5 dx dx

x +c 5 x d x +0 5 dx 5

= 5 ⋅ cosh ⋅

x 1 5 5

= 5 ⋅ cosh ⋅

x 5

e. Given ∫ (sinh 4 x + cosh 2 x ) dx = ∫ sinh 4 x dx + ∫ cosh 2 x dx let: a. u = 4 x , then

du du d du = 4x ; = 4 ; du = 4dx ; dx = 4 dx dx dx

b. v = 2 x , then

dv dv dv d = 2x ; = 2 ; dv = 2dx ; dx = 2 dx dx dx

Therefore, =

and

.

du

dv

∫ sinh 4 x dx + ∫ cosh 2 x dx = ∫ sinh u ⋅ 4 + ∫ cosh v ⋅ 2

1 1 cosh u + c1 + sinh v + c 2 4 2

Hamilton Education Guides

=

1 1 cosh 4 x + sinh 2 x + c1 + c 2 4 2

=

=

1 1 sinh u du + cosh v dv 4 2

∫

∫

1 1 cosh 4 x + sinh 2 x + c 4 2

67

Advanced Integration

1.4 Integration of Hyperbolic Functions

1 4

1 2

Check: Let y = cosh 4 x + sinh 2 x + c then y ′ =

1 d 1 d d ⋅ cosh 4 x + ⋅ sinh 2 x + c 4 dx 2 dx dx

4 2 1 d 2 x + 0 = ⋅ sinh 4 x + ⋅ cosh 2 x = sinh 4 x + cosh 2 x ⋅ cosh 2 x ⋅ 4 2 2 dx du d du csc h 8 x dx let u = 8 x , then = 8 x = 8 which implies du = 8dx ; dx = dx dx 8

=

d 1 4x ⋅ sinh 4 x ⋅ dx 4

+

∫

f. Given

du

∫ csc h 8x dx = ∫ csc h u ⋅ 8

1 csc h u du 8

∫

=

1 8

(

)

1 1 ⋅ ⋅ sec h 2 4 x ⋅ 4 + 0 8 tanh 4 x cosh 2 4 x − sinh 2 4 x

=

1 ⋅ 2

=

1 sinh 8 x

cosh 2 4 x sinh 4 x cosh 4 x

1 = ⋅ 2

2

5 x dx

=

2

1 4 sec h 4 x ⋅ 8 tanh 4 x

1 cosh 2 4 x sinh 4 x cosh 4 x

1 8x ln tanh +c 8 2

=

=

=

2

1 sec h 4 x ⋅ 2 tanh 4 x

=

1 ln tanh 4 x + c 8

d 1 1 ⋅ ⋅ (tanh 4 x ) + 0 8 tanh 4 x dx

=

2

1 1 − tanh 4 x ⋅ tanh 4 x 2

=

1 ⋅ 2

2 1 − sinh 2 4 x

cosh 4 x sinh 4 x cosh 4 x

cosh 4 x 1 1 1 ⋅ = = 2 cosh 2 4 x ⋅ sinh 4 x 2 cosh 4 x ⋅ sinh 4 x sinh 2 ⋅ 4 x

= csc h 8 x

g. Given ∫ csc h 2 5 x dx let u = 5 x , then

∫ csc h

=

=

1 d d ⋅ ln tanh 4 x + c 8 dx dx

Check: Let y = ln tanh 4 x + c , then y ′ = =

1 u ln tanh +c 8 2

=

. Therefore,

∫ csc h

2

u⋅

du 5

du du d du = 5 ; du = 5dx ; dx = = 5x ; dx 5 dx dx

∫

1 5

1 5

1 = − coth u + c = − coth 5 x + c

1 csc h 2 u du 5

=

5

1 d (coth 5 x )⋅ d 5 x + d c dx dx 5 dx

Check: Let y = − coth 5 x + c , then y ′ = − ⋅ 5 ⋅ csc h 2 5 x = csc h 2 5 x 5 1 1 csc h 2 x dx let u = x , 4 4

. Therefore,

1 5

(

)

= − ⋅ − csc h 2 5 x ⋅ 5 + 0

=

∫

h. Given

∫ csc h

2

1 x dx 4

=

∫ csc h

2

u ⋅ 4du

then

du d x du 1 ; = = dx dx 4 dx 4

1 4

= 4∫ csc h 2 u du = − 4 coth u + c = − 4 coth x + c

x 4

Check: Let y = −4 coth + c , then y ′ = − 4 ⋅ = i. Given

∫x

2

4 x csc h 2 4 4

= csc h 2

x 4

= csc h 2

sec h 2 x 3 dx let u = x 3 , then

Hamilton Education Guides

; 4du = dx ; dx = 4du . Therefore,

x d d c coth + dx 4 dx

x d x +0 4 dx 2

= − 4 ⋅ − csc h 2 ⋅

x 1 4 4

= 4 csc h 2 ⋅

1 x 4

du du d 3 du = x ; = 3x 2 ; du = 3 x 2 dx ; dx = dx dx dx 3x 2

. Therefore,

68

Advanced Integration

∫x

2

1.4 Integration of Hyperbolic Functions

sec h 2 x 3 dx =

∫x

2

sec h 2 u ⋅

du 3x

=

2/

1 sec h 2 u du 3

∫

1 3

d 1 d c ⋅ tanh x 3 + dx 3 dx

Check: Let y = tanh x 3 + c , then y ′ = 3x 2 ⋅ sec h 2 x 3 3

=

2 2 ∫ x sec h (x + 5) dx

1 2

=

∫ x sec h

(

2

u⋅

du 2x

=

(

du d 2 = x +5 dx dx

1 sec h 2 u du 2

∫

)

∫x

k. Given

∫x

2

(

)

1 ⋅ sec h 2 x 2 + 5 ⋅ 2 x 2 2

csc h 2 x 3 dx let

csc h 2 x 3 dx =

∫x

2

=

1 tanh u + c 2

(

)

(

)

(

du 3x

2

=

1 csc h 2 u du 3

∫

∫ 2 x csc h

x dx =

1 tanh x 2 + 5 + c 2

=

1 d ⋅ sec h 2 x 2 + 5 ⋅ x2 + 5 + 0 2 dx

)

(

) (

)

. Therefore,

1 3

1 3

= − ⋅ − csc h 2 x 3 ⋅

3x 2 ⋅ csc h 2 x 3 = x 2 csc h 2 x 3 3 du du d 2 du . u = x 2 , then = 2 x ; du = 2 x dx ; dx = = x ; 2x dx dx dx

1 ⋅ csc h 2 x 3 ⋅ 3 x 2 3

2 2

)

=

3

1 d d coth x 3 + c dx 3 dx

l. Given ∫ 2 x csc h 2 x 2 dx let

(

Therefore,

1 = − coth u + c = − coth x 3 + c

Check: Let y = − coth x 3 + c , then y ′ = − ⋅ =

1 ⋅ sec h 2 x 3 ⋅ 3 x 2 3

=

) ; dudx = 2 x ; du = 2 x dx ; dx = du2 x .

2x ⋅ sec h 2 x 2 + 5 = x sec h 2 x 2 + 5 2 du du d 3 du u = x 3 , then = = 3x 2 ; du = 3 x 2 dx ; dx = x ; dx dx dx 3x 2

=

csc h 2 u ⋅

1 3

1 tanh x 3 + c 3

1 d 3 ⋅ sec h 2 x 3 ⋅ x +0 3 dx

1 d d ⋅ tanh x 2 + 5 + c 2 dx dx

Check: Let y = tanh x 2 + 5 + c , then y ′ = =

=

=

= x 2 sec h 2 x 3

2 2 2 ∫ x sec h (x + 5) dx let u = x + 5 , then

j. Given

1 tanh u + c 3

=

d 3 x +0 dx

=

∫ 2 x csc h

2

u⋅

du 2x

=

∫ csc h

Check: Let y = − coth x 2 + c , then y ′ = −

2

u du

Therefore,

= − coth u + c = − coth x 2 + c

d d coth x 2 + c dx dx

= csc h 2 x 2 ⋅

d 2 x +0 dx

= csc h 2 x 2 ⋅ 2 x

= 2 x csc h2 x 2 Example 1.4-2: Evaluate the following indefinite integrals: a.

∫ sec h

d.

∫ sinh

g.

∫

j.

∫ sec h

m.

5 x dx =

b.

∫x

(x + 1) cosh (x + 1) dx =

e.

∫ cosh

h.

∫

k.

∫ csc h 7 x coth 7 x dx

n.

∫ sinh

2

5

cosh 4 2

∫ cosh

x x sinh dx 2 2

=

(5 x − 1) dx = 3

x dx 5

=

Hamilton Education Guides

2

sinh x 3 dx = 5

x sinh x dx =

x 2 cosh x 3 dx =

3

x dx =

=

c.

∫ x csc h x

f.

∫ cosh

5

2

dx =

5 x sinh 5 x dx =

e 2x sec h e 2 x dx 3

i.

∫

l.

∫ tanh

o.

∫ 2 sinh x dx =

2

=

10 x dx =

x

69

Advanced Integration

1.4 Integration of Hyperbolic Functions

Solutions: du du d du = 5x ; = 5 ; du = 5 dx ; dx = dx dx dx 5

a. Given ∫ sec h 2 5 x dx let u = 5 x , then

∫ sec h

2

5 x dx =

∫ sec h

2

u⋅

du 5

=

∫x

2

Thus,

∫x

2

=

1 tanh 5 x + c 5

1 d d ⋅ tanh 5 x + c 5 dx dx

=

1 sec h 2 5 x ⋅ 5 + 0 5

∫

1 5

b. Given

1 tanh u + c 5

1 sec h 2 u du 5

Check: Let y = tanh 5 x + c , then y ′ =

∫x

2

sinh u ⋅

du 3x

2

1 sinh u du 3

∫

=

1 3

Check: Let y = cosh x 3 + c , then y ′ =

c. Given

∫

=

5 sec h 2 5 x 5

=

= sec h 2 5 x

du du d 3 sinh x 3 dx let u = x 3 , then . = x = 3x 2 which implies du = 3 x 2 dx ; dx = dx dx 3x 2

sinh x 3 dx =

=

. Therefore,

3x 2 ⋅ sinh x3 3

∫ x csc h x

x csc h x 2 dx =

dx let u = x 2 , then

∫

x csc h u ⋅

Check: Let y = ln tanh

1 1 = ⋅ 2 tanh

x2 2

1 cosh u + c 3

d 1 d c ⋅ cosh x 3 + dx 3 dx

=

1 cosh x 3 + c 3

=

d 3 1 ⋅ sinh x 3 ⋅ x +0 dx 3

=

1 ⋅ sinh x3 ⋅ 3 x 2 3

= x 2 sinh x3

2

1 2

=

du 2x

=

1 2

∫

du du d 2 du . = 2 x ; du = 2 x dx ; dx = = x ; 2x dx dx dx

csc h u du

=

1 u ln tanh +c 2 2

=

Therefore,

x2 1 +c ln tanh 2 2

1 1 1 d x2 x2 d + c , then y ′ = ⋅ ln tanh + c = ⋅ 2 tanh 2 2 dx 2 dx 2 x2 2

x2 d x2 1 sec h ⋅ sec h 2 ⋅ +0 = ⋅ 2 dx 2 2 tanh

x2 2

2x ⋅ 2

=

2 x2

x 1 − tanh 2 ⋅ 2 2 tanh x

x2 2

⋅

d x2 tanh +0 2 dx

x = ⋅ 2

2

1−

sinh 2 x

2

2 2 x2 cosh 2 x2

sinh

2

2 cosh x 2

=

x ⋅ 2

2 2 cosh 2 x − sinh 2 x 2 2 2 cosh 2 x 2 x2

sinh

cosh

=

x sinh x 2

1

=

2 x2

x ⋅ 2

2 cosh 2 x 2 x2

sinh

cosh

2

= x csc h x

2 x2

=

x ⋅ 2

2

cosh cosh 2

Hamilton Education Guides

2

x x ⋅ sinh 2 2

x

= 2 cosh

2

2

x x ⋅ sinh 2 2

=

x 2

sinh 2 ⋅ x2

2

d. Given ∫ sinh 5 (x + 1) cosh (x + 1) dx let u = sinh (x + 1) , then du = cosh (x + 1) ⋅ dx ; dx =

2

x2 2

du . cosh (x + 1)

du d du = sinh (x + 1) ; = cosh (x + 1) ; dx dx dx

Therefore,

70

Advanced Integration

∫ sinh

5

1.4 Integration of Hyperbolic Functions

(x + 1) cosh (x + 1) dx =

∫u

cosh (x + 1) ⋅

5

du cosh (x + 1)

∫u

x sinh x dx =

5

sinh x ⋅

du sinh x

∫u

=

1 6

5

du =

Check: Let y = g. Given ∫ cosh 4

∫ cosh

4

∫u

5 x sin 5 5 x dx =

5

sinh 5 x ⋅

1 cosh 6 5 x + c , 30

x x sinh dx 2 2

x x sinh dx 2 2

=

∫u

2 5

4

du 5 sinh 5 x

sinh

x 2

∫x

2

∫x

2

x 2 du ⋅ 2 sinh x

2

∫x

2

cosh u ⋅

du 3x

2

=

∫

∫

e 2x sec h e 2 x dx 3

=

∫

1 2x du e sec h u ⋅ 3 2e 2 x

(

)

Hamilton Education Guides

=

6 cosh 5 x sinh x 6

=

1 1 6 ⋅ u +c 5 6

=

Therefore,

= cosh 5 x sinh x

; dx =

du . 5 sinh 5 x

Thus,

1 cosh 6 5 x + c 30

x du d x du 1 = sinh = cosh ; 2 dx 2 2 dx dx 2 5 u +c 5

=

=

1 ⋅ sinh u + c 3

1 ⋅ cosh x 3 ⋅ 3 x 2 + c 3

2du

; dx =

sin

=

=

x 2

. Therefore,

x 2 cosh 5 + c 5 2

=

10 x x cosh 4 sin 2 2 10

= cosh 4

=

1 sec h u du 6

∫

=

1 sinh x 3 + c 3

3 2 x cosh x 3 + 0 3

1 sin −1 (tanh u ) + c 6

1 6 1 − tanh 2 e 2 x

⋅

x x sinh 2 2

. Therefore,

= x 2 cosh x 3

du du d 2 x du = 2e 2 x ; du = 2e 2 x ⋅ dx ; dx = = e ; dx dx dx 2e 2 x

1 Check: Let y = sin −1 tanh e 2 x + c , then y ′ = 6

∫

1 cosh u du 3

let u = e 2 x , then

∫

du . sinh x

; dx =

du du d 3 du = = 3x 2 ; du = 3 x 2 ⋅ dx ; dx = x ; dx dx dx 3x 2

1 3

e 2x sec h e 2 x dx 3

= sinh 5 (x + 1) cosh (x + 1)

1 cosh 6 x + c 6

=

x x 1 2 ⋅ 5 cosh 4 ⋅ sinh ⋅ + 0 5 2 2 2

Check: Let y = sinh x 3 + c , then y ′ = i. Given

1 u 5 du 5

= 2∫ u 4 du =

x 2

cosh x 3 dx let u = x 3 , then

cosh x 3 dx =

=

, then

Check: Let y = cosh 5 + c , then y ′ = h. Given

1 6 u +c 6

1 sinh 6 ( x + 1) + c 6

=

6 cosh 5 5 x ⋅ 5 sinh 5 x 30 +0 = cosh 5 5 x sinh 5 x = cosh 5 hx sinh 5 x 30 30

then y ′ =

let u = cosh

1 6 u +c 6

du du d = cosh 5 x ; = 5 sinh 5 x dx dx dx

f. Given ∫ cosh 5 5 x sinh 5 x dx let u = cosh 5 x , then 5

du =

1 ⋅ 6 cosh 5 x ⋅ sinh x + 0 6

Check: Let y = cosh 6 x + c , then y ′ =

∫ cosh

5

du d du = cosh x ; = sinh x dx dx dx

e. Given ∫ cosh 5 x sinh x dx let u = cosh x , then 5

∫u

1 ⋅ 6 sinh 5 (x + 1) ⋅ cosh (x + 1) + 0 6

1 6

Check: Let y = sinh 6 (x + 1) + c , then y ′ =

∫ cosh

=

=

d tanh e 2 x + 0 dx

. Thus,

(

)

1 sin −1 tanh e 2 x + c 6

=

sec h 2 e 2 x 6 sec h 2 e 2 x

⋅

d 2x e dx

71

Advanced Integration

=

1.4 Integration of Hyperbolic Functions

sec h 2 e 2 x

6 sec h e 2 x

⋅ 2e 2 x

2e 2 x sec h 2 e 2 x ⋅ 6 sec h e 2 x

=

du d (5 x − 1) ; du = 5 ; du = 5dx ; dx = du = dx 5 dx dx

j. Given ∫ sec h 2 (5 x − 1) dx let u = 5 x − 1 , then

∫ sec h

2

(5 x − 1) dx =

∫ sec h

2

u⋅

du 5

1 sec h 2 u du 5

∫

=

1 5

Check: Let y = tanh (5 x − 1) + c , then y ′ = 5 ⋅ sec h 2 (5 x − 1) 5

∫ csc h u coth u

=

=

1 tanh u + c 5

d 1 ⋅ sec h 2 (5 x − 1) ⋅ (5 x − 1) + 0 dx 5

du 7

=

1 csc h u coth u du 7

∫

∫ tanh

2

Check: Let y =

∫ tanh

2

(

∫ cosh

3

x dx 5

= ∫ cosh 2

x 5

= ∫ cosh dx + ∫ sinh 2 x

du 10

=

∫ cosh 5 dx + ∫ sinh

2

∫

2

x x cosh dx 5 5

x x cosh dx 5 5

=

=

then y ′ =

) = 1 − sec h 10 x = tanh

x x cosh dx 5 5

7

Hamilton Education Guides

d 7x + 0 dx

=

1 ⋅ csc h 7 x coth 7 x ⋅ 7 7

2

dw cosh u

1 (u − tanh u ) + c 10

=

. Therefore,

1 (10 x − tanh 10 x ) + c 10

1  d  2 10 − sec h 10 x ⋅ 10 x  + 0 10  dx 

=

(

1 10 − sec h 2 10 x ⋅10 10

)

10 x

x x x x x = ∫ 1 + sinh 2  cosh dx = ∫  cosh + sinh 2 cosh  dx 5



let u =

x 5

, then

5

du d x = dx dx 5

∫ cosh u 5du + ∫ sinh

To solve the second integral let w = sinh u , then = 5 sinh u + 5∫ w 2 cosh u ⋅

1 7

du du d du = 10 x ; = 10 ; du = 10 dx ; dx = dx dx dx 10

1 tanh 2 u du 10

1 (10 x − tanh 10 x ) + c , 10

10 1 − sec h 2 10 x 10

m.

u⋅

. Therefore,

= csc h 7 x coth 7 x

10 x dx let u = 10 x , then

10 x dx =

1 ⋅ sec h 2 (5 x − 1) ⋅ 5 5

1 = − csc h u + c = − csc h 7 x + c

1 7

7 ⋅ csc h 7 x coth 7 x 7 2

=

du du du d = 7 ; du = 7 dx ; dx = = 7x ; dx dx dx 7

1 7

∫ tanh

1 tanh (5 x − 1) + c 5

=

Check: Let y = − csc h 7 x + c , then y ′ = − ⋅ − csc h 7 x coth 7 x ⋅

l. Given

. Therefore,

= sec h 2 (5 x − 1)

k. Given ∫ csc h 7 x coth 7 x dx let u = 7 x , then

∫ csc h 7 x coth 7 x dx

e 2x ⋅ sec h e 2 x 3

=

2



;

5

5

5

du 1 1 = ; du = dx ; dx = 5 du . dx 5 5

Therefore,

∫

u cosh u 5du = 5 sinh u + 5 sinh 2 u cosh u du

dw dw dw d = = cosh u ; dx = sinh u ; cosh u dx dx dx 5 3

= 5 sinh u + 5∫ w 2 dw = 5 sinh u + w 3 + c =

thus,

5 x x sinh 3 + 5 sinh + c 3 5 5

72

Advanced Integration

1.4 Integration of Hyperbolic Functions x 5

5 3

x 1 5 5

x 5

5 3

x 1 5 5

x 5

Check: Let y = 5 sinh + sinh 3 + c , then y ′ = 5 ⋅ cosh ⋅ + ⋅ 3 sinh 2 ⋅ cosh ⋅ + 0 = n.

∫ sinh

3

5 x 15 x x ⋅ cosh + ⋅ sinh 2 ⋅ cosh 5 15 5 5 5

x dx =

∫ sinh

2

x sinh x dx =

x x x x x = cosh 1 + sinh 2  = cosh ⋅ cosh 2 = cosh 3 5

5

∫ ( cosh

2

)

x − 1 sinh x dx =

5

∫ ( cosh

2

5

5

)

x sinh x − sinh x dx

= ∫ cosh 2 x sinh x dx − ∫ sinh x dx = ∫ cosh 2 x sinh x dx − cosh x . To solve the first integral let u = cosh x , then

du du d du . = cosh x ; = sinh x ; dx = sinh x dx dx dx

∫ cosh

2

x sinh x dx − cosh x

∫u

=

2

sinh x ⋅

Therefore,

du − cosh x sinh x

=

∫u

2

du − cosh x

=

1 3 u − cosh x + c 3

cosh 3 x − cosh x + c 3

=

1 3

Check: Let y = cosh 3 x − cosh + c , then y ′ =

(

)

1 ⋅ 3 cosh 2 x ⋅ sinh x − sinh x + 0 3

= cosh 2 x ⋅ sinh x − sinh x

= sinh x cosh 2 x − 1 = sinh x ⋅ sinh 2 x = sinh 3 x o. Given

1

x

∫ 2 sinh x dx = 2 ∫ x sinh x dx let u = x

and dv = sinh x dx then du = dx and ∫ dv = ∫ sinh x dx

which implies v = cosh x . Using the substitution by parts formula ∫ u dv = u v − ∫ v du we obtain 1 2

∫ x sinh x dx =

1 1 x cosh x − cosh x dx 2 2

∫

1 2

1 1 x cosh x − sinh x + c 2 2

=

1 2

Check: Let y = x cosh x − sinh x + c , then y ′ = −

cosh x 2

=

x sinh x 2

=

1 (cosh x + x sinh x ) − 1 cosh x + 0 2 2

cosh x x sinh x + 2 2

=

1 x sinh x 2

Example 1.4-3: Evaluate the following indefinite integrals: a.

∫ tanh

8

x sec h 2 x dx =

b.

∫ tanh

d.

∫ coth

5

3 x csc h 2 3 x dx =

e.

g.

∫ sec h 5x tanh 5x dx =

h.

j.

∫x

k.

2

coth x 3 dx =

Hamilton Education Guides

(x + 3) sec h 2 (x + 3) dx =

c.

∫ coth

∫ tanh 5x dx =

f.

∫ 2 tanh 3 dx =

∫ sec h 2 tanh 2 dx =

x

i.

∫ csc h

∫x

l.

∫

2

5

x

sec h 5 x 3 dx =

3

x csc h 2 x dx = x

2

( 1 − 2 x ) dx =

sec h x x

dx =

73

Advanced Integration

1.4 Integration of Hyperbolic Functions

Solutions: a.

∫ tanh

8

x sec h 2 x dx let u = tanh x , then

∫ tanh

8

x sec h 2 x dx =

∫u

8

du d = tanh x dx dx

du

⋅ sec h 2 x ⋅

2

sec h x

1 9

Check: Let y = tanh 9 x + c then y ′ = b.

∫ tanh

5

(x + 3) sec h 2 (x + 3) dx let

; du = sec h 2 (x + 3) dx ; dx = =

∫u

5

du =

1 5+1 u +c 5 +1

sec h

2

(x + 3)

1 6 u +c 6

=

1 6

2

csc h x

. Therefore,

∫ coth

3

du =

1 8+1 u +c 8 +1

=

5

=

. Thus,

sec h 2 x

1 tanh 9 x + c 9

= (tanh x )8 sec h 2 x = tanh 8 x sec h 2 x

du d tanh (x + 3) = dx dx

∫ tanh

1 9 u +c 9

du

;

du = sec h 2 (x + 3) c ; dx

(x + 3) sec h 2 (x + 3) dx =

∫u

5

⋅ sec 2 (x + 3) ⋅

du

sec (x + 3) 2

1 tanh 6 ( x + 3 ) + c 6

1 ⋅ 6 [ tanh (x + 3) ]6−1 ⋅ sec h 2 (x + 3) + 0 6

c. Given ∫ coth 3 x csc h 2 x dx let u = coth x , then du

; du = sec h 2 x dx ; dx =

then

. Thus,

Check: Let y = tanh 6 (x + 3) + c then y ′ =

; dx = −

8

du = sec h 2 x dx

1 ⋅ 9 (tanh x )9−1 ⋅ sec h 2 x + 0 9

u = tanh (x + 3) ,

du

=

∫u

=

;

du d = coth x dx dx

x csc h 2 x dx =

∫u

3

;

= tanh 5 (x + 3) sec h 2 (x + 3)

du = − csc h 2 x c ; du = − csc h 2 x dx dx

⋅ csc h 2 x ⋅

−du 2

csc h x

= − ∫ u 3 du =

−1 3+1 u +c 3 +1

1 4

1 = − u 4 + c = − coth 4 x + c 4

1 4

1 4

Check: Let y = − coth 4 x + c then y ′ = − ⋅ 4(coth x )4−1 ⋅ − csc h 2 x + 0 = coth 3 x csc h 2 x d. Given ∫ coth 5 3x csc h 2 3x dx let u = coth 3x , then ; dx = −

du 2

3 csc h 3 x

. Therefore,

1 1 5+1 u +c 3 5 +1

= − ⋅

Check: Let y = − e. Given

= −

1 6 u +c 18

1 coth 6 3 x + c 18

∫ coth = −

3 x csc h 2 3 x dx =

∫u

5

;

du = −3 csc h 2 3 x c ; du = −3 csc h 2 3 x dx dx

⋅ csc h 2 3 x ⋅

−du 2

3 csc h 3 x

= −

1 u 5 du 3

∫

1 coth 6 3 x + c 18

then y ′ = −

∫ tanh 5x dx let u = 5x , then

Hamilton Education Guides

5

du d = coth 3 x dx dx

1 ⋅ 6 (coth 3 x )6−1 ⋅ 3 ⋅ − csc 2 3 x + 0 18

du d = 5x dx dx

;

du =5 dx

; du = 5 dx ; dx =

du 5

= coth 5 3x csc h 2 3x . Therefore,

74

Advanced Integration

1.4 Integration of Hyperbolic Functions

du

∫ tanh 5x dx = ∫ tanh u ⋅ 5

=

1 tanh u du 5

∫

1 ln cosh u + c 5

=

=

1 ln cosh 5 x + c 5

1 1 Check: Let y = ln cosh 5 x + c then y ′ = ⋅

1 ⋅ sinh 5 x ⋅ 5 + 0 5 cosh 5 x

5

x

x

du d x = dx dx 3

∫ 2 tanh 3 dx let u = 3 , then

f. Given

x

∫ 2 tanh 3 dx

du 1 = dx 3

x 3

du

=

1 5

x x tanh dx 2 2

let u =

x 2

, then

x

∫ sec h 2 tanh 2 dx = ∫ sec h u ⋅ tanh u ⋅ 2du

du d x = dx dx 2

∫ csc h

2

u⋅−

du 2

= −

1 2

∫x

2

∫x

2

coth x 3 dx let u = x 3 , then

coth x 3 dx =

∫x

2

⋅ coth u ⋅

du 3x

2

=

Hamilton Education Guides

du 1 = dx 2

= 2 tanh

; du = 5dx ; dx =

du 5

x 3

. Therefore, 1 5

1 = − sec h u + c = − sec h 5 x + c 5

=

5 sec h 5 x tanh 5 x 5

= sec h 5 x tanh 5 x

; 2du = dx ; dx = 2du . Therefore,

x x 1 tanh ⋅ + 0 2 2 2

du d = 1− 2 x dx dx

;

=

du = −2 dx

2 x x ⋅ sec h tanh 2 2 2

x +c 2

= sec h

; du = −2dx ; dx =

du . −2

x x tanh 2 2

Thus,

∫

1 ⋅ − csc h 2 ( 1 − 2 x ) ⋅ −2 + 0 2

du d 3 = x dx dx

1 coth u ⋅ du 3

∫

1 1 Check: Let y = ln sinh x 3 + c then y ′ = ⋅ 3

;

6 x tanh 3 3

1 1 1 csc h 2 u du = coth u + c = coth ( 1 − 2 x ) + c 2 2 2

Check: Let y = coth ( 1 − 2 x ) + c then y ′ = j. Given

du =5 dx

=

x +c 3

= 2∫ sec h u tanh u du = − 2 sec h u + c = − 2 sec h

i. Given ∫ csc h 2 ( 1 − 2 x ) dx let u = 1− 2 x , then

(1 − 2 x ) dx =

;

∫

x 2

2

x 1 ⋅ +0 3 3

1 sec h u tanh u du 5

Check: Let y = −2 sec h + c then y ′ = − 2 ⋅ − sec h

∫ csc h

⋅ sinh

1 ⋅ sec h 5 x tanh 5 x ⋅ 5 + 0 5

Check: Let y = − sec h 5 x + c , then y ′ = h. Given ∫ sec h

x 3

du d = 5x dx dx

g. Given ∫ sec h 5 x tanh 5 x dx let u = 5 x , then

∫ sec h 5x tanh 5x dx = ∫ sec h u ⋅ tanh u ⋅ 5

1 cosh

= tanh 5 x

; 3du = dx ; dx = 3du . Therefore,

= 2∫ tanh u ⋅ 3du = 6∫ tanh u du = 6 ln cosh u + c = 6 ln cosh

Check: Let y = 6 ln cosh + c , then y ′ = 6 ⋅

x

;

5 sinh 5 x ⋅ 5 cosh 5 x

=

;

du = 3x 2 dx

= 1

3 sinh x

= csc h 2 ( 1 − 2 x )

; du = 3x 2 dx ; dx =

1 ln sinh u + c 3

=

⋅ cosh x3 ⋅ 3 x 2 + 0 3

du 3x 2

. Therefore,

1 ln sinh x 3 + c 3

=

3 x 2 cosh x 3 ⋅ = x 2 coth x 3 3 sinh x 3

75

Advanced Integration

∫x

k. Given

∫x

2

2

∫x

sec h 5 x 3 dx =

=

∫

l. Given

15 sec h 5 x

du 15 x

)

⋅15 x 2

3

=

sec h u x

Check: Let y = 2 sin

−1

=

15 1 − tanh 2 5 x 3

=

sec h

⋅

1 x2

1 1 2x 2

. Thus,

(

)

1 sin −1 tanh 5 x 3 + c 15

sec h 2 5 x 3 d d ⋅ 5x 3 tanh 5 x 3 + 0 = dx 2 3 dx 15 sec h 5 x

du d 12 du 1 − 12 1 ; dx = 2 x du . Therefore, = x ; = x = dx 2 dx dx 2 x

(

)

1

2

then y ′ = 1

2

=

15x 2

= 2∫ sec h u ⋅ du = 2 sin −1 (tanh u ) + c = 2 sin −1 tanh x + c

⋅ 2 x du

( tanh x )+ c

1

⋅

=

du

= x 2 sec h 5 x 3

1

1 − tanh 2 x 2

2 sec h 2 x 2

; du = 15 x 2 dx ; dx =

1 sin −1 (tanh u ) + c 15

1

then y ′ =

15 x 2 sec h 2 5 x 3 ⋅ 15 sec h 5 x 3

1

∫

∫

du = 15x 2 dx

;

1 sec h u ⋅ du 15

=

2

dx let u = x 2 , then

x

x

⋅ sec h u ⋅

(

sec h 2 5 x 3

dx =

2

1 sin −1 tanh 5 x 3 + c 15

sec h x

sec h x

du d 5x 3 = dx dx

sec h 5 x 3 dx let u = 5x 3 , then

Check: Let y =

∫

1.4 Integration of Hyperbolic Functions

1 2x 2

⋅

sec h 2 x 2 1 x2

sec h

1

=

sec h x 2 1 x2

=

1 d 2 sec h 2 x 2 d 12 ⋅ tanh x 2 + 0 = ⋅ x 1 dx dx 2 2 sec h x

sec h x x

Example 1.4-4: Evaluate the following indefinite integrals: 5

dx

a.

∫ sec h

d.

∫ cosh 7 x dx =

g.

∫ 2 sinh

j.

∫

x2

5

x

3

=

1

e

cosh x

3

3x 3x csc h dx 2 2

sinh

=

x dx = 3

x dx

b.

∫ sinh x 2

e.

∫

h.

3x ∫ ( cosh x sec h x + e ) dx

k.

∫

=

c.

cosh 5 x + sinh 5 x dx sinh 5 x

= =

e tanh 5 x sec h 2 5 x dx =

∫x

2

sinh x dx =

 1 + cosh x   dx sinh x 

f.

∫ 

i.

∫e

l.

∫

sinh 8 x

cosh 8 x dx =

1 coth 7 x

e3

=

csc h 2 7 x dx

=

Solutions: a. Given ∫ sec h 5 x 2

∫

dx

5

sec h x 2

5

x

3

=

dx 5

x3

∫

sec h u ⋅

Hamilton Education Guides

2

let u = x 5 , then 5

5 x 3 du 5

2 x

3

=

2 du d 52 du 2 − 53 5 = x = = x ; ; dx = 5 x 3 du . Therefore, 5 3 dx 5 2 dx dx 5 x 5 sec h u ⋅ du 2

∫

=

5 sin −1 (tanh u ) + c 2

=

5 5   sin −1  tanh x 2  + c 2  

76

Advanced Integration

1.4 Integration of Hyperbolic Functions

5 Check: Let y = sin −1  tanh 5 x 2  + c , then y ′ = 2   2

5 sec h 2 x 5

=

⋅

2

2 sec h x 5

b. Given

x dx

∫ sinh x 2

x dx

∫ sinh x 2

=

∫

=

2

=

3

3

5x 5

2

1 2

=

2

=

sec h x 5

∫

du sinh u

=

∫

x = ⋅ 2

1−

sinh

cosh

∫x

2

sinh 2 ⋅ x2 2

=

sinh

cosh

2

x

= c. Given

2 x2

x = ⋅ 2

2 2 cosh 2 x − sinh 2 x 2 2 2 cosh 2 x 2 x2 2 x2 2

5

du d 2 = x dx dx

1 csc h u du 2

1 1 1 x2 + c , then y ′ = ⋅ Check: Let y = ln tanh 2 tanh 2 2 2 sinh 2 x 2 2 cosh 2 x 2 x2

5

x5

dx let u = x 2 , then

x2 2

=

;

x3

du = 2x dx

; du = 2 x dx ; dx =

1 u ln tanh +c 2 2

=

2 x2 2

x 2 2x 2 x sec h ⋅ sec h ⋅ = ⋅ 2 2 4 tanh

cosh 2 x sinh cosh

2

2 x2 2 x2

=

du . 2x

Thus,

x2 1 +c ln tanh 2 2

=

2

1

x ⋅ 2

2 d 5 sec h 2 x 5 d 52 ⋅ tanh x 5 + 0 = ⋅ x 2 dx dx 2 5 2 sec h x

sec h x 2

=

3

sec h x 5

10 x 5

∫ x csc h x

x du ⋅ sinh u 2 x

⋅

sec h 2 x 5

2

2 1 − tanh 2 x 5 2

2

10

2

5

=

x2 2

x2

cosh 2 x ⋅ 2 cosh 2 x 2 ⋅ sinh x 2

2

=

2

2

2 x2

x 1 − tanh 2 ⋅ 2 2 tanh x

x 2 cosh

2

x 2

⋅ sinh

x2 2

2

x sinh x 2

sinh x dx let u = x 2 and dv = sinh x dx then du = 2 x dx and

∫ dv = ∫ sinh x dx which implies

implies v = cosh x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫x

2

∫

(1 )

∫

sinh x dx = x 2 ⋅ cosh x − cosh x ⋅ 2 x dx = x 2 cosh x − 2 x cosh x dx

To integrate

∫ x cosh x dx

let u = x and dv = cosh x dx then du = dx and ∫ dv = ∫ cosh x dx which

implies v = sinh x . Using the integration by parts formula again we have

∫ x cosh x dx

(2)

= x ⋅ sinh x − ∫ sinh x ⋅ dx = x sinh x − cosh x + c

Combining equations ( 1 ) and ( 2 ) together we obtain

∫x

2

(

)

sinh x dx = x 2 cosh x − 2 x cosh x dx = x 2 cosh x − 2 (x sinh x − cosh x + c ) = x 2 + 2 cosh x − 2 x sinh x + c

∫

( ) = (x + 2)sinh h x − 2 sinh x = x

(

)

Check: Let y = x 2 + 2 cosh x − 2 x sinh x + c , then y ′ = 2 x cosh x + x 2 + 2 sinh h x − 2 sinh x − 2 x cosh x + 0 2

Hamilton Education Guides

2

sinh h x + 2 sinh x − 2 sinh x

= x 2 sinh h x

77

Advanced Integration

d. Given

1.4 Integration of Hyperbolic Functions

1

∫ cosh 7 x dx = ∫ sec h 7 x dx let u = 7 x , then

1

∫ cosh 7 x dx = ∫ sec h 7 x dx

=

du

∫ sec h u ⋅ 7

1 sec h u du 7

∫

=

e. Given

∫

;

du =5 dx

∫

cosh 5 x + sinh 5 x dx sinh 5 x

=

1 ln sinh u + x + c 5

 cosh hx  + 1 dx  sinh 5 x 

= ∫ 

; du = 5dx ; dx =  cosh hx

du 5

; du = 7dx ; dx =

1 sin −1 (tanh u ) + c 7

=

du 7

. Thus,

1 sin −1 (tanh 7 x ) + c 7

⋅

 1 + cosh x   dx sinh x 

∫ 

=



tanh

x 2

1

du

∫  sinh 5x + 1 dx = ∫ coth 5x dx + ∫ dx = ∫ coth u ⋅ 5x + x = 5 ∫ coth u du + x

=

1 ln sinh 5 x + x + c 5

=



cosh x 

1

∫  sinh x + sinh x  dx

Check: Let y = ln tanh

1

= ∫ (coth 5 x + 1) dx = ∫ coth 5 x dx + ∫ dx let u = 5 x , then

. Therefore,

1 cosh 5 x ⋅ 5 ⋅ +1+ 0 5 sinh 5 x

1 5

=

du =7 dx

sec h 2 7 x d d ⋅ 7x tanh 7 x + 0 = dx dx 2 2 7 sec h 7 x 7 1 − tanh 7 x

Check: Let y = ln sinh 5 x + x + c , then y ′ = f.

;

7 sec h 2 7 x sec h 2 7 x 1 ⋅7 = ⋅ = sec h 7 x = 7 sec h 7 x 7 sec h 7 x cosh 7 x

cosh 5 x + sinh 5 x dx sinh 5 x

du d = 5x dx dx

=

1

1 7

Check: Let y = sin −1 (tanh 7 x ) + c , then y ′ = =

du d = 7x dx dx

=

x + ln sinh x + c , 2

∫ csc h x dx + ∫ coth x dx then y ′ =

1 1 ⋅ sec h 2 2x ⋅ + ⋅ cosh x = 2 sinh x

1 tanh

sec h 2 2x 2 tanh 2x

x 2

⋅

=

= ln tanh

d tanh dx

cosh x + = sinh x

5 cosh 5 x ⋅ +1 5 sinh 5 x

x 2

+

= coth 5 x + 1

x + ln sinh x + c 2

1 d ⋅ sinh x + 0 sinh x dx

2 x 1 1 − tanh 2 ⋅ 2 tanh 2x

cosh x 1 + = ⋅ sinh x 2

1−

sinh 2 x

2

cosh 2 x

sinh x

2

cosh x

2

cosh 2 x − sinh 2 x 2

+

=

g.

∫

2 sinh

cosh x sinh x

=

cosh 2 x

1 ⋅ 2

2 sinh x 2 cosh x 2

1 2 cosh 2x ⋅ sinh

3x 3x csc h dx 2 2

Hamilton Education Guides

2

x 2

cosh x + sinh x

= 2∫ sinh

+

cosh x sinh x

=

1 cosh 2 x

cosh 2x cosh x cosh x 1 1 2 ⋅ + ⋅ + = 2 sinh 2x sinh x 2 cosh 2 x ⋅ sinh x sinh x 2 2 cosh x

2

=

1 sinh 2 ⋅ 2x

3x 1 dx ⋅ 2 sinh 3 x 2

cosh x + sinh x

= 2∫

=

sinh 32x

sinh 32x

cosh x 1 + sinh x sinh x

dx

=

1+ cosh x sinh x

= 2∫ dx = 2 x + c

78

2

Advanced Integration

1.4 Integration of Hyperbolic Functions

Check: Let y = 2 x + c , then y ′ = 2 ⋅ x1−1 + 0 = 2 ⋅ x 0 = 2 h.

3x ∫ ( cosh x sec h x + e ) dx



1

∫  cosh x ⋅ cosh x + e

=

  dx 

3x

=

3x 3x ∫ (1 + e ) dx = ∫ dx + ∫ e dx =

1 3

1 3

x+

1 3x e +c 3

3 3

Check: Let y = x + e 3 x + c , then y ′ = x1−1 + ⋅ e 3 x ⋅ 3 + 0 = x 0 + ⋅ e 3 x = 1+ e 3 x du d = sinh 8 x dx dx

i. Given ∫ e sinh 8 x cosh 8 x dx let u = sinh 8 x , then

∫

∫

e sinh 8 x cosh 8 x dx =

e u cosh 8 x ⋅

du 8 cosh 8 x

1 8

Check: Let y = e sinh 8 x + c , then y ′ = j. Given ∫ e

∫

e

cosh x

3

cosh x

3

sinh

sinh

x dx = 3

Check: Let y = 3e

eu du 8

∫

=

∫e

du 5 sec h 2 5 x

tanh 5 x

∫

x 3du ⋅ 3 sinh x 3

e u sinh

cosh x

+ c , then y ′ = 3e

3

1 coth 7 x

∫e

u

sec h 2 5 x ⋅

csc h 2 7 x dx

7 3

; du = − csc h 2 7 x dx ; dx = −

∫

=

du 8 cosh 8 x

. Thus,

1 sinh 8 x e +c 8

8 sinh 8 x cosh 8 x ⋅e 8

= e sinh 8 x cosh 8 x

cosh x

3

⋅ sinh

cosh x 3

+c

cosh x x 3 cosh 3x x x 1 3 sinh ⋅ +0 = e ⋅ sinh = e 3 3 3 3 3

du d = tanh 5 x dx dx

;

du = 5 sec h 2 5 x dx

; du = 5 sec h 2 5 x dx

. Therefore, du 2

5 sec h 5 x

1 5

1 coth 7 x

=

1 u e +c 8

=

= 3∫ e u du = 3e u + c = 3e

Check: Let y = e tanh 5 x + c , then y ′ =

e3

∫

; dx =

x x du d x x du 1 3du dx let u = cosh , then . Therefore, = cosh ; = sinh ; dx = 3 dx dx 3 3 3 dx 3 sin 3x

sec h 2 5 x dx =

l. Given ∫ e 3

du = 8 cosh 8 x dx

1 u e du 8

1 sinh 8 x ⋅ cosh 8 x ⋅ 8 + 0 ⋅e 8

k. Given ∫ e tanh 5 x sec h 2 5 x dx let u = tanh 5 x , then ; dx =

=

;

csc h 2 7 x dx

3 7

=

1 u e du 5

∫

1 3

3du 7 csc h 2 7 x

=

1 u e +c 5

1 tanh 5 x e ⋅ sec h 2 5 x ⋅ 5 + 0 5

let u = coth 7 x , then

e u csc h 2 7 x ⋅

1 coth 7 x

Check: Let y = − e 3

Hamilton Education Guides

∫

=

=

=

1 tanh 5 x e +c 5

5 tanh 5 x e sec h 2 5 x 5

du d 1 = coth 7 x dx dx 3

;

= e tanh 5 x sec h 2 5 x

7 du = − csc h 2 7 x 3 dx

. Therefore, −3du 7 csc h 2 7 x

+ c , then y ′ = −

= −

3 u e du 7

∫

3 7

1 coth 7 x

3 = − eu + c = − e 3 7

+c

1 coth 7 x 3 13 coth 7 x 1 e ⋅ − csc h 2 7 x ⋅ 7 + 0 = e 3 csc h 2 7 x 7 3

79

Advanced Integration

•

1.4 Integration of Hyperbolic Functions

To integrate even powers of sinh x and cosh x use the following identities: sinh 2 x =

cosh 2 x − sinh 2 x = 1

1 ( cosh 2 x − 1) 2

cosh 2 x =

1 ( cosh 2 x + 1) 2

To integrate odd powers of sin x and cos x use the following equalities:

∫ sinh

2 n +1

∫ cosh •

•

2 n +1

x dx =

∫ cosh

2n

x sinh x dx =

2n

n n 2 2 ∫ ( sinh x ) sinh x dx = ∫ ( cosh x − 1) sinh x dx

x cosh x dx =

( let u = cosh x )

n n 2 2 ∫ ( cosh x ) cosh x dx = ∫ (1 + sinh x ) cosh x dx

sinh x sinh y

=

1 [ cosh (x + y ) − cosh (x − y ) ] 2

cosh x cosh y

=

1 [ cosh (x + y ) + cosh (x − y ) ] 2

sinh x cosh y

=

1 [ sinh (x + y ) + sinh (x − y ) ] 2

To integrate tanh n x , set

( let u = sinh x )

(

)

(

)

= tanh n −2 x tanh 2 x = tanh n −2 x 1 − sec h 2 x = tanh n −2 x − tanh n −2 x sec h 2 x

To integrate coth n x , set coth n x

•

∫ sinh

To integrate products of sinh x , sinh y , cosh x , and cosh y use the identities below:

tanh n x

•

x dx =

= coth n −2 x coth 2 x = coth n −2 x 1 + csc h 2 x = coth n −2 x + coth n −2 x csc h 2 x

To integrate sec h n x For even powers, set

sec h n x

(

= sec h n −2 x sec h 2 x = 1 − tanh 2 x

)

n−2 2

sec h 2 x

For odd powers change the integrand to a product of even and odd functions, i.e., write

∫ sec h •

3

x dx as

∫ sec h

2

x sec h x dx (see Example 1.4-6, problem letter h).

To integrate csc h n x For even powers, set

csc h n x

(

)

= csc h n −2 x csc h 2 x = coth 2 x − 1

n−2 2

csc h 2 x

For odd powers change the integrand to a product of even and odd functions, i.e., write

∫ csc h

3

x dx as

∫ csc h

2

x csc h x dx (see Example 1.4-6, problem letter i).

In the following examples we use the above general rules in order to solve integral of products and powers of hyperbolic functions: Example 1.4-5: Evaluate the following indefinite integrals: a. d.

∫ sinh 5x cosh 7 x dx = ∫ sinh 3x sinh 5x dx =

Hamilton Education Guides

b. e.

∫ sinh x cosh x dx = ∫ cosh 3x cosh 5x dx =

c. f.

∫ cosh 3x cosh 2 x dx = 5 ∫ sinh x dx = 80

Advanced Integration

g.

∫ sinh

j.

∫ sinh

3

7

1.4 Integration of Hyperbolic Functions

x dx =

h.

∫ cosh

x dx =

k.

∫ sec h

5

4

x dx =

i.

∫ tanh

4

x dx =

x dx =

l.

∫ cosh

3

x dx =

Solutions: a.

∫ sinh 5x cosh 7 x dx

1

∫ 2 [ sinh (5 + 7)x + sinh (5 − 7)x ] dx 1

=

1 2

=

1 1 cosh 12 x − cosh 2 x + c 24 4

=

1

1 1 sinh 12 x − sinh 2 x 2 2

1 1 sinh 12 x + sinh (− 2 x ) 2 2

=

1

1 sinh 2 x dx 2

∫

=

1 1 ⋅ cosh 2 x 2 2

=

1 4

1

∫ 2 [ sinh (2 x ) + sinh (0 x ) ] dx

1 ⋅ 2 sinh 2 x + 0 4

=

1 sinh 2 x 2

1

1 cosh x dx 2

∫

=

Check: Let y =

=

=

1 1 1 ⋅ sinh 5 x + sinh x + c 2 5 2

=

1 1 cosh ( 3 + 2 )x + cosh ( 3 − 2 )x 2 2

1

∫ sinh 3x sinh 5x dx = ∫ 2 [ cosh (3 + 5)x − cosh (3 − 5)x ] dx =

1 2

=

1 1 sinh 8 x − sinh 2 x + c 16 4

∫ (cosh 8x − cosh 2 x ) dx

Hamilton Education Guides

=

=

1 1 cosh 8 x dx − cosh 2 x dx 2 2

∫

= sinh 5 x cosh 7 x

∫ (sinh 2 x + 0) dx

=

= sinh x cosh x

1 cosh 5 x dx 2

∫

1 1 sinh 5 x + sinh x + c 10 2

1 1 1 1 ⋅ 5 cosh 5 x + ⋅ cosh x + 0 sinh 5 x + sinh x + c , then y ′ = 10 2 10 2

=

1 2

=

1 ⋅ 2 sinh x cosh x 2

1

1 1 cosh 5 x + cosh x 2 2

12 2 sinh 12 x − sinh 2 x 24 4

1 1 sinh (5 + 7 )x + sinh (5 − 7 )x 2 2

=

∫ cosh 3x cosh 2 x dx = ∫ 2 [ cosh (3 + 2)x + cosh (3 − 2)x ] dx = ∫ 2 (cosh 5x + cosh x ) dx +

d.

=

=

1 cosh 2 x + c 4

Check: Let y = cosh 2 x + c , then y ′ =

c.

1 1 1 1 ⋅ cosh 12 x − ⋅ cosh 2 x + c 2 12 2 2

=

1 1 1 1 ⋅12 sinh 12 x − ⋅ 2 sinh 2 x + 0 cosh 12 x − cosh 2 x + c , then y ′ = 24 24 4 4

∫ sinh x cosh x dx = ∫ 2 [ sinh ( 1 + 1)x + sinh ( 1 − 1)x ] dx =

1

∫ 2 [ sinh ( 12 x ) + sinh (− 2 x ) ] dx

=

∫ (sinh 12 x − sinh 2 x ) dx = 2 ∫ sinh 12 x dx − 2 ∫ sinh 2 x dx

Check: Let y =

b.

=

∫

=

5 1 cosh 5 x + cosh x 10 2

= cosh 3x cosh 2 x

1

∫ 2 [ cosh (8x ) − cosh (− 2 x ) ] dx =

1 1 1 1 ⋅ sinh 8 x − ⋅ sinh 2 x + c 2 8 2 2

81

Advanced Integration

Check: Let y =

= e.

1 1 1 1 sinh 8 x − sinh 2 x + c , then y ′ = ⋅ cosh 8 x ⋅ 8 − ⋅ cosh 2 x ⋅ 2 + 0 16 4 16 4

1 1 cosh 8 x − cosh 2 x 2 2

∫ cosh 3x cosh 5x dx 1

=

1 1 sinh 8 x + sinh 2 x + c 16 4

∫ sinh

5

∫ sinh

4

∫ sinh

∫ ( cosh

∫ (u

4

x dx =

∫

=

1 [ cosh (8 x ) + cosh (− 2 x ) ] 2

x sinh x dx =

du = sinh x dx

;

=

∫

2

)

x −1

∫u

− 2u 2 + 1 du =

4

1 5

du . sinh x

)

)

sinh x dx

∫

∫ (u

=

∫

2

du − 2 u 2 du + du =

−1

2

1 1 1 1 ⋅ sinh 8 x + ⋅ sinh 2 x + c 2 8 2 2

=

=

2 8 cosh 8 x + cosh 2 x 4 16

1 [ cosh (3 + 5)x + cosh (3 − 5)x ] 2

=

= cosh 3x cosh 5 x

Let u = cosh x , then

Therefore, =

∫ (u

1 5 2 3 u − u +u +c 5 3

Check: Let y = cosh 5 x − cosh 3 x + cosh x + c , then y ′ =

(

= sinh 3x sinh 5 x

1

sinh x dx

2 3

2 8 cosh 8 x − cosh 2 x 4 16

∫ 2 [ cosh (8x ) + cosh (− 2 x ) ] dx

2 2 2 2 ∫ ( sinh x ) sinh x dx = ∫ ( cosh x − 1 ) sinh x dx .

; du = sinh x dx ; dx = 2

=

1 [ cosh (3 + 5)x − cosh (3 − 5)x ] 2

=

1 1 1 1 sinh 8 x + sinh 2 x + c , then y ′ = ⋅ cosh 8 x ⋅ 8 + ⋅ cosh 2 x ⋅ 2 + 0 16 4 4 16

du d cosh x = dx dx 5

=

1 1 cosh 8 x dx + cosh 2 x dx 2 2

=

1 1 cosh 8 x + cosh 2 x 2 2

x dx =

1 [ cosh (8 x ) − cosh (− 2 x ) ] 2

1

∫ 2 (cosh 8x + cosh 2 x ) dx

=

=

∫ 2 [ cosh (3 + 5)x + cosh (3 − 5)x ] dx

=

=

Check: Let y =

f.

1.4 Integration of Hyperbolic Functions

4

=

)

− 2u 2 + 1 sinh x ⋅

du sinh x

1 2 cosh 5 x − cosh 3 x + cosh x + c 5 3

1 2 ⋅ 5 cosh 4 x ⋅ sinh x − ⋅ 3 cosh 2 x ⋅ sinh x + sinh x 5 3

(

)

= sinh x cosh 4 x − 2 sinh x cosh 2 x + sinh x = sinh x cosh 4 x − 2 cosh 2 x + 1 = sinh x cosh 2 x − 1

(

= sinh x sinh 2 x

∫ sinh

g. ;

3

; du = sinh x dx ; dx =

du . sinh x

∫ sinh

2

∫ ( cosh =

2

2

= sinh x sinh 4 x = sinh 5 x

∫ ( cosh

x dx =

1 3 u −u +c 3

2

x sinh x dx =

x dx =

du = sinh x dx

∫ sinh =

3

)

)

)

x − 1 sinh x dx =

2

)

x − 1 sinh x dx . Let u = cosh x , then

du d = cosh x dx dx

Therefore,

∫ (u

2

)

− 1 sinh x ⋅

du sinh x

=

∫ (u

2

)

− 1 du =

∫u

2

∫

du − du

1 cosh 3 x − cosh x + c 3

Hamilton Education Guides

82

Advanced Integration

1.4 Integration of Hyperbolic Functions 1 3

Check: Let y = cosh 3 x − cosh x + c , then y ′ =

(

)

1 ⋅ 3 cosh 2 x ⋅ sinh x − sinh x + 0 3

= sinh x cosh 2 x − sinh x

= sinh x cosh 2 x − 1 = sinh x sinh 2 x = sinh 3 x h.

∫ cosh then

x dx =

∫ cosh

du d = sinh x dx dx

∫ cosh =

5

∫u

5

4

x dx =

4

x cosh x dx =

du = cosh x dx

;

2 2 2 2 ∫ ( cosh x ) cosh x dx = ∫ (1 + sinh x ) cos x dx .

; du = cosh x dx ; dx =

2 2 ∫ (1 + sinh x ) cosh x dx

∫

∫

du + 2 u 2 du + du =

du cosh x

1 5 2 3 u + u +u +c 5 3

=

=

2 2 ∫ (1 + u ) du

∫ (u

=

4

)

+ 2u 2 + 1 du

1 2 sinh 5 x + sinh 3 x + sinh x + c 5 3

2 3

1 5

. Therefore,

du 2 2 ∫ (1 + u ) cosh x ⋅ cosh x

=

Let u = sinh x ,

Check: Let y = sinh 5 x + sinh 3 x + sinh x + c , then y ′ =

(

6 5 sinh 4 x ⋅ cosh x + sinh 2 x ⋅ cosh x + cosh x + 0 3 5

)

(

= cosh x sinh 4 x + 2 cosh x sinh 2 x + cosh x = cosh x sinh 4 x + 2 sinh 2 x + 1 = cosh x 1 + sinh 2 x

(

= cosh x cosh 2 x i.

∫ tanh =

4

x dx =

∫ tanh

2

)

2

2 2 2 2 2 ∫ tanh x (1 − sec h x ) dx = ∫ tanh x dx − ∫ tanh x sec h x dx

2 2 2 ∫ (1 − sec h x ) dx − ∫ tanh x sec h x dx

= − ∫ tanh 2 x sec h 2 x dx − ∫ sec h 2 x dx + ∫ dx . To solve the first

du d tanh x = dx dx

integral let u = tanh x , then

4

x dx =

∫ tanh

2

;

du

∫

− tanh 2 x sec h 2 x dx = − u 2 sec h 2 x ⋅

∫ tanh

2

= cosh x cosh 4 x = cosh 5 x

x tanh 2 x dx =

∫

)

2

sec h x

du = sec h 2 x dx

; du = sec h 2 x dx ; dx =

sec h 2 x

. Thus,

1 3

1 3

= − ∫ u 2 du = − u 3 = − tanh 3 x . Therefore,

∫

∫

∫

x tanh 2 x dx = − tanh 2 x sec h 2 x dx − sec h 2 x dx + dx = −

1 3

du

1 tanh 3 x − tanh x + x + c 3

Check: Let y = − tanh 3 x − tanh x + x + c , then y ′ = − tanh 2 x sec h 2 x − sec h 2 x + 1

(

)

)(

(

)

= sec h 2 x − tanh 2 x − 1 + 1 = 1 − tanh 2 x − tanh 2 x − 1 + 1 = − tanh 2 x − 1 + tanh 4 x + tanh 2 x + 1

(

)

= − tanh 2 x + tanh 2 x + (− 1 + 1) + tanh 4 x = tanh 4 x j.

∫ sinh

7

x dx =

∫ sinh

6

x sinh x dx =

du d = cosh x dx dx

;

∫ sinh

∫ ( cosh

7

x dx =

du = sinh x dx

Hamilton Education Guides

2

x −1

3 3 2 2 ∫ ( sinh x ) sinh x dx = ∫ ( cosh x − 1 ) sinh x dx .

; du = sinh x dx ; dx =

)

3

sinh x dx

=

∫ (u

2

du . sinh x

−1

)

3

Let u = cosh x , then

Therefore,

sinh x dx

=

∫ (u

6

)

− 3u 4 + 3u 2 − 1 sinh x ⋅

du sinh x

83

Advanced Integration

1.4 Integration of Hyperbolic Functions

=

∫ (u

)

=

3 1 cosh 7 x − cosh 5 x + cosh 3 x − cosh x + c 5 7

∫u

− 3u 4 + 3u 2 − 1 du =

6

6

∫

∫

∫

du − 3 u 4 du + 3 u 2 du − du =

1 1 1 7 u − 3⋅ u 5 + 3⋅ u 3 − u + c 3 5 7

3 5

1 7

Check: Let y = cosh 7 x − cosh 5 x + cosh 3 x − cosh x + c , then y ′ = + 3 cosh 2 x ⋅ sinh x − sinh x + 0

15 7 cosh 6 x ⋅ sinh x − cosh 4 x ⋅ sinh x 5 7

= sinh x cosh 6 x − 3 sinh x cosh 4 x + 3 sinh x cosh 2 x − sinh x

(

)

(

= sinh x cosh 6 x − 3 cosh 4 x + 3 cosh 2 x − 1 = sinh x cosh 2 x − 1

)

3

(

= sinh x sinh 2 x

)

3

= sinh x sinh 6 x = sinh 7 x

∫ sec h

k.

=

4

∫ sec h

x dx =

2

x sec h 2 x dx =

2 2 2 2 2 ∫ sec h x (1 − tanh x ) dx = ∫ sec h x dx − ∫ tanh x sec h x dx

2 2 2 2 2 2 ∫ (1 − tanh x ) dx − ∫ tanh x sec h x dx = ∫ dx − ∫ tanh x dx − ∫ tanh x sec h x dx .

du d tanh x = dx dx

integral let u = tanh x , then

∫

∫

− tanh 2 x sec h 2 x dx = − u 2 sec h 2 x ⋅

∫ sec h

4

∫ sec h

x dx =

2

x sec h 2 x dx =

;

du 2

sec h x

du = sec h 2 x dx

To solve the third

; du = sec h 2 x dx ; dx =

du sec h 2 x

. Therefore,

1 3

1 3

= − ∫ u 2 du = − u 3 = − tanh 3 x . Thus,

∫ dx − ∫ tanh

2

∫

x dx − tanh 2 x sec h 2 x dx =

∫ dx − ∫ tanh

2

x dx −

1 tanh 3 x 3

1 3

1 1 = x − (x − tanh x ) − tanh 3 x + c = − tanh 3 x + tanh x + (x − x ) + c = − tanh 3 x + tanh x + c 3

3

1 3

3 3

Check: Let y = − tanh 3 x + tanh x + c , then y ′ = − tanh 2 x ⋅ sec h 2 x + sec h 2 x = − tanh 2 x sec h 2 x + sec h 2 x

(

)

= sec h 2 x 1 − tanh 2 x = sec h 2 x sec h 2 x = sec h 4 x l.

∫ cosh ;

3

du = cosh x dx

∫ cosh =

3

2 2 ∫ ( cosh x ) cosh x dx = ∫ (1 + sinh x ) cosh x dx .

x dx =

; du = cosh x dx ; dx =

x dx =

1 3 u +u +c 3

du cosh x

Let u = sinh x , then

. Therefore,

du 2 2 ∫ (1 + sinh x ) cosh x dx = ∫ (1 + u )cosh x ⋅ cosh x

=

du d = sinh x dx dx

=

2 ∫ (1 + u ) du

=

∫u

2

∫

du + du

1 sinh 3 x + sinh x + c 3 1 3

Check: Let y = sinh 3 x + sinh x + c , then y ′ =

(

)

(

)

1 ⋅ 3 sinh 2 x ⋅ cosh x + cosh x + 0 3

= cosh x sinh 2 x + cosh x

= cosh x 1 + sinh 2 x = cosh x cosh 2 x = cosh 3 x

Hamilton Education Guides

84

Advanced Integration

1.4 Integration of Hyperbolic Functions

Example 1.4-6: Evaluate the following indefinite integrals: b.

d.

∫ cosh x dx = 3 ∫ tanh x dx =

g.

∫ coth

a.

4

3

x dx =

e.

∫ cosh 5x dx = 4 ∫ coth x dx =

f.

∫ sinh x dx = 6 ∫ tanh x dx =

h.

∫ sec h

i.

∫ csc h

2

3

4

c.

x dx =

3

x dx =

Solutions: a.

∫

cosh 4 xdx =

∫ (cosh x ) 2

2

dx =

∫

2

1 1   2 (cosh 2 x + 1 ) dx = 4  

=

2 1 1 dx + cosh 2 x dx cosh 2 2 x dx + 4 4 4

+

1 1 cosh 4 x dx + sinh 2 x 4 8

∫

∫

∫

∫

Check: Let y =

=

x 1 + 4 4

∫ (cosh 2 x + 1 ) dx 2

1

∫ (1 + cosh

1 4

1 1

∫ 2 ( 1 + cosh 4 x ) dx + 2 ⋅ 2 sinh 2 x

x x 1 1 1 + + ⋅ sinh 4 x + sinh 2 x + c 4 4 8 8 4

=

=

=

=

2

)

2 x + 2 cosh 2 x dx

x 1 1 + ⋅ dx 4 4 2

∫

3 1 1 sinh 4 x + sinh 2 x + c x+ 8 32 4

3 4 cosh 4 x 2 cosh 2 x 1 3x 1 + + sinh 4 x + sinh 2 x + c , then y ′ = + 4 32 8 4 8 32

2 3 1 + cosh 4 x + cosh 2 x 4 8 8

=

1 1 1 2 2 1 1 1 1 1 2 =  +  + cosh 4 x + cosh 2 x = + ( 1 + cosh 4 x ) + cosh 2 x = + ⋅ ( 1 + cosh 4 x ) + cosh 2 x 4

=

4

1 1 2 + ⋅ cosh 2 2 x + cosh 2 x 4 4 4

(

8

(

1 1 + cosh 2 2 x + 2 cosh 2 x 4

=

4

4

4 2

) = 14 ( cosh 2 x + 1 )

4

2

1 =  ( cosh 2 x + 1 ) 2

2



) = cosh x 1 1 2 2 2 ∫ cosh 5x dx = ∫ (1 + sinh 5x )dx = ∫ dx + ∫ sinh 5x dx = ∫ dx + ∫ 2 ( cosh 10 x − 1 ) dx = x + 2 ∫ cosh 10 x dx = cosh 2 x

b.

4

8 8

−

2

4

1 1 1 x dx = x + ⋅ sinh 10 x − + c 2 2 10 2

∫

Check: Let y =

x sinh 10 x  1 1 +c = x1 −  + sinh 10 x + c = +

2



2

20

1 1 x sinh 10 x ⋅ cosh 10 x ⋅10 + 0 + c , then y ′ = + + 2 20 20 2

=

20

1 10 + ⋅ cosh 10 x 2 20

=

1 1 + cosh 10 x 2 2

1 1 1 1 1 = 1 −  + cosh 10 x = 1 +  cosh 10 x −  = 1 + ( cosh 10 x − 1 ) = 1 + sinh 2 5 x = cosh 2 5 x 

c.

∫ =

sinh 4 x dx =

2

2

2

∫ (sinh x ) 2

2

dx

=

∫

2

1   2 (cosh 2 x − 1 ) dx  

1 2 1 cosh 2 2 x dx − cosh 2 x dx + dx 4 4 4

∫

∫

Hamilton Education Guides

∫

2

2

=

1 4

1

=

1 4

∫ (cosh 2 x − 1 ) dx 2

1 1

=

1 4 x

∫ 2 ( 1 + cosh 4 x ) dx − 2 ⋅ 2 sinh 2 x + 4

∫ (cosh =

2

)

2 x − 2 cosh 2 x + 1 dx

1 1 ⋅ dx 4 2

∫

85

Advanced Integration

+

1.4 Integration of Hyperbolic Functions

1 1 x cosh 4 x dx − sinh 2 x + 4 4 8

∫

Check: Let y =

3 1 1 1 x x 1 1 =  +  + ⋅ sinh 4 x − sinh 2 x + c = x + sinh 4 x − sinh 2 x + c 8

4

4 8 4

8

32

3x 1 1 3 4 cosh 4 x 2 cosh 2 x + sinh 4 x − sinh 2 x + c , then y ′ = + − 8 32 4 4 32 8

4

=

2 3 1 + cosh 4 x − cosh 2 x 4 8 8

1 1 1 2 1 1 1 2 1 1 2 =  +  + cosh 4 x − cosh 2 x = + ( 1 + cosh 4 x ) − cosh 2 x = + ⋅ ( 1 + cosh 4 x ) − cosh 2 x 4

=

8 8

∫ tanh

3

4

1 1 2 + ⋅ cosh 2 2 x − cosh 2 x 4 4 4

(

)

x dx =

∫ tanh

= sinh 2 x d.

4

2

=

2

∫

∫

x dx =

∫ tanh

(

1 1 + cosh 2 2 x − 2 cosh 2 x 4

2 ∫ (1 − sec h x ) tanh x dx

x tanh x dx =

− sec h 2 x tanh x dx = − sec h 2 x ⋅ u ⋅

3

4

4

4 2

) = 14 ( cosh 2 x − 1 )

4

2

1 =  ( cosh 2 x − 1 ) 2

2



= sinh 4 x

the first integral let u = tanh x , then

∫ tanh

8

2

du d = tanh x dx dx

du 2

sec h x

x tanh x dx =

;

= − ∫ sec h 2 x tanh x dx + ∫ tanh x dx . To solve

du = sec h 2 x dx

; du = sec h 2 dx ; dx =

1 2

du

. Thus,

sec h 2 x

1 2

= − ∫ u du = − u 2 = − tan 2 x . Combining the term

2 ∫ (1 − sec h x ) tanh x dx

=

∫ ( − sec h

2

)

x tanh x + tanh x dx

1 2

1 = − ∫ sec h 2 x tanh x dx + ∫ tanh x dx = − tanh 2 x + ∫ tanh x dx = − tanh 2 x + ln cosh x + c 2

1 2

1 2

Check: Let y = − tanh 2 x + ln cosh x + c , then y ′ = − ⋅ 2 tanh x ⋅ sec h 2 x + = − tanh x sec h 2 x + e.

∫ coth

4

x dx =

∫ coth

2

sinh x cosh x

(

)

= − tanh x sec h 2 x + tanh x = tanh x 1 − sec h 2 x = tanh x tanh 2 x = tanh 3 x

x coth 2 x dx =

(

1 ⋅ sinh x + 0 cosh x

)

2 2 2 2 2 ∫ coth x ( csc h x + 1) dx = ∫ coth x csc h x dx + ∫ coth x dx

= ∫ coth 2 x csc h 2 x dx + ∫ csc h 2 x + 1 dx = ∫ coth 2 x csc h 2 x dx + ∫ csc h 2 x dx + ∫ dx . To solve the first integral let u = coth x , then ; dx = −

du 2

csc h x

du d = coth x dx dx

;

du = − csc h 2 x dx

; du = − csc h 2 x dx

. Grouping the terms together we find ∫ coth 2 x csc h 2 x dx = ∫ u 2 csc h 2 x ⋅ − 1 3

du csc h 2 x

1 3

= − ∫ u 2 du = − u 3 = − coth 3 x . Therefore,

∫ coth

4

x dx =

∫ coth

2

Hamilton Education Guides

x coth 2 x dx =

2 2 2 2 2 ∫ coth x ( csc h x + 1) dx = ∫ coth x csc h x dx + ∫ csc h x dx + ∫ dx

86

Advanced Integration

1.4 Integration of Hyperbolic Functions 1 3

1 = − coth 3 x + ∫ csc h 2 x dx + ∫ dx = − coth3 x − coth x + x + c 3

1 3

3 3

Check: Let y = − coth 3 x − coth x + x + c , then y ′ = − coth 2 x ⋅ − csc h 2 x + csc h 2 x + 1 + 0

(

)

(

)(

)

= coth 2 x csc h 2 x + csc h 2 x + 1 = csc h 2 x coth 2 x + 1 + 1 = coth 2 x − 1 coth 2 x + 1 + 1 = coth 4 x + coth 2 x − coth 2 x − 1 + 1 = coth 4 x f.

∫ tanh

4 2 4 4 2 ∫ tanh x (1 − sec h x ) dx = ∫ ( tanh x − tanh x sec h x ) dx =

∫ tanh

6

x dx =

∫ tanh

4

x dx − tanh 4 x sec h 2 x dx . In example 1.4-5, problem letter i, we found that

∫ tanh

4

∫ tanh

6

4

x tanh 2 x dx =

∫

1 x dx = − tanh 3 x − tanh x + x + c . Therefore, 3

x dx =

∫ tanh

4

x tanh 2 x dx =

4 4 2 4 2 ∫ tanh x (1 − sec h x ) dx = ∫ tanh x dx − ∫ tanh x sec h x dx

1 5

1 3

 1  =  − tanh 3 x − tanh x + x  − ∫ tanh 4 x sec h 2 x dx = − tanh5 x − tanh 3 x − tanh x + x + c  3



1 5

5 5

1 3

3 3

Check: Let y = − tanh 5 x − tanh 3 x − tanh x + x + c , then y ′ = − tanh 4 x ⋅ sec h 2 x − tanh 2 x ⋅ sec h 2 x − sec h 2 x + 1 + 0

(

(

)

= − tanh 4 x sec h 2 x − tanh 2 x sec h 2 x − sec h 2 x + 1 = − sec h 2 x tanh 4 x + tanh 2 x + 1 + 1

)(

)

(

)

= − 1 − tanh 2 x tanh 4 x + tanh 2 x + 1 + 1 = − tanh 4 x + tanh 2 x + 1 − tanh 6 x − tanh 4 x − tanh 2 x + 1 = − tanh 4 x − tanh 2 x − 1 + tanh 6 x + tanh 4 x + tanh 2 x + 1 = tanh 6 x g.

∫ coth

3

x dx =

∫ coth

2

first integral let u = coth x , then

∫ csc h ∫ coth

2

3

∫ csc h

2

x coth x dx = x dx =

∫ coth

∫ csc h 2

∫ ( csc h

x coth x dx =

2

)

x + 1 coth x dx =

du d = coth x dx dx

du

x ⋅u ⋅ −

2

csc h x

x coth x dx =

∫

2

x coth x dx + coth x dx = −

;

∫ csc h

du = − csc h 2 x dx

1 2

2

∫

x coth x dx + coth x dx . To solve the

; du = − csc h 2 dx ; dx = −

= coth x csc h 2 x +

Hamilton Education Guides

. Thus,

1 2

∫ ( csc h

2

)

x + 1 coth x dx =

1 coth 2 x + coth x dx 2

∫

∫ ( csc h

2

)

x ⋅ coth x + coth x dx

1 2

= − coth 2 x + ln sinh x + c

2

cosh x sinh x

csc h 2 x

= − ∫ u du = − u 2 = − coth 2 x . Combining the term

1 1 Check: Let y = − coth 2 x + ln sinh x + c , then y ′ = − ⋅ 2 coth x ⋅ − csc h 2 x + 2

du

(

1 ⋅ cosh x + 0 sinh x

)

= coth x csc h 2 x + coth x = coth x csc h 2 x + 1 = coth x coth 2 x = coth 3 x

87

Advanced Integration

h.

∫ sec h

3

x dx =

1.4 Integration of Hyperbolic Functions

∫ sec h

2

x sec h x dx =

2 2 ∫ (1 − tanh x ) sec h x dx = − ∫ tanh x sec h x dx + ∫ sec h x dx

= − ∫ tanh x ⋅ tanh x sec h x dx + ∫ sec h x dx . To solve the first integral let u = tanh x and dv = tanh x sec h x dx ,

∫ dv = ∫ tanh x sec h x dx

then du = sec h 2 x dx and

which implies v = − sec h x . Using the integration by

parts formula ∫ u dv = u v − ∫ v du we obtain

∫

∫

∫

− tanh 2 x sec h x dx = − tanh x ⋅ tanh x sec h x dx = tanh x sec h x − sec h x sec h 2 x dx = tanh x sec h x

∫

− sec h 3 x dx . Combining the terms we have

∫ sec h

3

x dx =

∫ tanh x ⋅ tanh x sec h x dx − ∫ sec h x dx = tanh x sec h x − ∫ sec h

∫ sec h

3

x dx term from the right hand side of the equation to the left hand side we obtain

∫ sec h

3

x dx + sec h 3 x dx = 2 sec h 3 x dx = tanh x sec h x + sec h x dx . Therefore,

∫ sec h

3

x dx =

∫

∫

(

∫

)

=

1 1 tanh x sec h x + sin −1 ( tanh x ) + c 2 2 sec h 2 x ⋅ sec h x − sec h x tanh x ⋅ tanh x 2

1 2

Check: Let y = tanh x sec h x + sin −1 (tanh x ) + c , then y ′ = +

= = i.

∫ csc h

3

sec h 2 x 2 1 − tanh 2 x

+0

=

sec h 3 x − sec h x tanh 2 x sec h 3 x − sec h x tanh 2 x sec h x sec h 2 x + = + 2 2 2 2 sec h 2 x

sec h 3 x − sec h x tanh 2 x + sec h x 2 3

∫

x dx + sec h x dx . Moving the

∫

1 tanh x sec h x + sec h x dx 2 1 2

3

3

=

(

sec h 3 x + sec h x 1 − tanh 2 x 2

) = sec h x + sec h x sec h 3

2

x

2

3

sec h x + sec h x 2 sec h x = = sec h 3 x 2 2

x dx =

∫ csc h

2

x csc h x dx =

∫ ( coth

2

)

x − 1 csc h x dx =

∫ coth

2

∫

x csc h x dx − csc h x dx

= ∫ coth x ⋅ coth x csc h x dx − ∫ csc h x dx . To solve the first integral let u = coth x and dv = coth x csc h x dx , then du = − csc h 2 x dx and

∫ dv = ∫ coth x csc h x dx which implies v = − csc h x .

Using the integration by

parts formula ∫ u dv = u v − ∫ v du we obtain

∫ coth

2

x csc h x dx =

∫ coth x ⋅ coth x csc h x dx

= coth x ⋅ − csc h x − ∫ csc h x ⋅ csc h 2 x dx = − coth x csc h x

∫

− csc h 3 x dx . Combining the terms we have

∫ csc h

3

x dx =

∫ coth x ⋅ coth x csc h x dx − ∫ csc h x dx

Hamilton Education Guides

= − coth x csc h x − ∫ csc h 3 x dx − ∫ csc h x dx . Moving 88

Advanced Integration

1.4 Integration of Hyperbolic Functions

the ∫ csc h 3 x dx term from the right hand side of the equation to the left hand side we obtain

∫

∫

∫

∫ csc h

3

x dx + csc h 3 x dx = 2 csc h 3 x dx = − coth x csc h x − csc h x dx . Therefore,

∫ csc h

3

x dx =

(

1 − coth x csc h x − csc h x dx 2

∫

1 2

1 2

−

(

x 2

4 tanh

sec h 2

x 2

4 tanh

x 2

(

=

(

2 1 1 − tanh = ⋅ 4 tanh x

x 2

1 ⋅ 4

=

2

2 cosh 2 x 2 sinh x 2 cosh x 2

. The 2nd term is simplified as follows:

cosh 2 x − sinh 2 x 2

=

cosh 2 x

1 ⋅ 4

)

2

1 = ⋅ 4

2 sinh x 2 cosh x 2

1 cosh 2 x

1 = ⋅ 4

2 sinh x 2 cosh x 2

cosh cosh 2

x 2

x x ⋅ sinh 2 2

csc h x 1 1 1 1 = = = . Therefore, x x x 2 sinh x 2 2 ⋅ sinh 2 ⋅ 2 2 ⋅ 2 cosh ⋅ sinh 2 2 2 x 2

(

=

(

)

sinh 2 x

1−

x +c 2

− − csc h 2 x ⋅ csc h x − csc h x coth x ⋅ coth x 2

then y ′ = 2 x

sec h csc h 3 x + csc h x coth 2 x − 2 4 tanh

=

1 2

sec h 2 csc h 3 x + csc h x coth 2 x +0 = − 2 4 tanh 2x

)

)

1 2

= − coth x csc h x − ln tanh

x +c, 2

Check: Let y = − coth x csc h x − ln tanh sec h 2 2x

)

x 2

)

csc h 3 x + csc h x coth 2 x − csc h x 2 3

3

=

=

csc h 3 x + csc h x coth 2 x csc h x − 2 2

(

) = csc h x + csc h x csc h

csc h 3 x + csc h x coth 2 x − 1 2

3

2

x

2

3

sec h x + sec h x 2 csc h x = = csc h 3 x 2 2

Example 1.4-7: Evaluate the following indefinite integrals: c.

e.

∫ sinh x cosh x dx = 5 4 ∫ tanh x sec h x dx =

f.

∫ sinh x cosh x dx = 3 3 ∫ tanh x sec h x dx =

h.

∫ coth

i.

∫ coth

b.

d.

∫ sinh x cosh x dx = 3 2 ∫ sinh x cosh x dx =

g.

∫ coth

a.

2

2

2

x csc h 2 x dx =

2

3

5

x csc h 3 x dx =

4

2

3

x csc h 4 x dx =

Solutions: a.

∫ sinh =

2

x cosh 2 x dx =

1 1 x ⋅ sinh 4 x − + c 8 4 8

Check: Let y =

=

1 sinh 2 2 x dx 4

∫

=

1 4

1

∫ 2 ( cosh 4 x − 1 ) dx

1 8

=

1 1 cosh 4 x − 8 8

∫ ( cosh 4 x − 1) dx

=

1 1 dx cosh 4 x dx − 8 8

∫

∫

1 1 sinh 4 x − x + c 8 32

4 cosh 4 x 1 1 1 sinh 4 x − x + c , then y ′ = − +0 32 8 32 8

Hamilton Education Guides

=

=

1 ( cosh 4 x − 1) 8

89

Advanced Integration

= b.

∫ sinh

2

1.4 Integration of Hyperbolic Functions

1 1 ⋅ ( cosh 4 x − 1) 4 2

x cosh 5 x dx =

1 ⋅ sinh 2 2 x 4

=

∫ sinh

2

= sinh 2 x cosh 2 x

x cosh 4 x cosh x dx =

=

2 4 2 ∫ sinh x (1 + sinh x + 2 sinh x ) cosh x dx

=

=

∫ sinh

∫

2

∫

2 2 2 2 2 2 ∫ sinh x (cosh x ) cosh x dx = ∫ sinh x (1 + sinh x ) cos x dx

∫ ( sinh

2

)

x cosh x + sinh 6 x cosh x + 2 sinh 4 x cosh x dx

x cosh x dx + sinh 6 x cosh x dx + 2 sinh 4 x cosh x dx = 1 3

1 7

3 sinh 2 x cosh x 7 sinh 6 x cosh x + 7 3

2 5

Check: Let y = sinh 3 x + sinh 7 x + sinh 5 x + c , then y ′ = +

(

10 sinh 4 x cosh x = sinh 2 x cosh x + sinh 6 x cosh x + 2 sinh 4 x cosh x = sinh 2 x cosh x 1 + sinh 4 x + 2 sinh 2 x 5

(

= sinh 2 x cosh x 1 + sinh 2 x

c.

∫ sinh +

4

2 1 1 sinh 3 x + sinh 7 x + sinh 5 x + c 3 7 5

1 sinh 2 2 x cosh 2 x dx 8

1 1 8 6

= + ⋅ sinh 3 2 x = Check: Let y =

= −

(

2

= sinh 2 x cosh x cosh 2 x

∫ ( sinh x cosh x )

x cosh 2 x dx =

∫

)

= −

1 8

2

sinh 2 x dx =

1

)

2

= sinh 2 x cosh x cosh 4 x = sinh 2 x cosh 5 x

∫ ( 12 sinh 2 x )

1

∫ 2 ( cosh 4 x − 1 ) dx + 8 ∫ sinh

1 1 1 1 x − ⋅ sinh 4 x + sinh 3 2 x + c 48 16 16 4

=

2

2

⋅

1 ( cosh 2 x − 1 ) dx 2

2 x cosh 2 x dx

=

= −

1 sinh 2 2 x dx 8

∫

1 1 dx − cosh 4 x dx 16 16

∫

∫

1 1 1 x− sinh 3 2 x + c sinh 4 x + 48 64 16

1 4 cosh 4 x 6 sinh 2 2 x cosh 2 x 1 1 1 − + +0 x− sinh 4 x + sinh 3 2 x + c , then y ′ = 16 64 48 16 64 48

1 ( cosh 4 x − 1) + 1 sinh 2 2 x cos 2 x 8 16

1 8

1 1 8 2

= − ⋅ ( cosh 4 x − 1) + sinh 2 2 x cosh 2 x 2

1 1 1 1 = − ⋅ sinh 2 2 x + sinh 2 2 x cosh 2 x =  sinh 2 x  ⋅ ( cosh 2 x − 1) = ( sinh x cosh x )2 sinh 2 x 2

8

8



2

= sinh 2 x cosh 2 x ⋅ sinh 2 x = sinh 4 x cosh 2 x d.

∫ sinh

3

x cosh 2 x dx =

3 2 3 5 3 5 ∫ sinh x (1 + sinh x )dx = ∫ ( sinh x + sinh x ) dx = ∫ sinh x dx + ∫ sinh x dx .

Example, 1.4-5, problem letters f and g, we found ∫ sinh 5 x dx = 1 cosh 3 x − cosh x + c . Therefore, 3

∫ sinh

3

x dx =

∫ sinh

3

x cosh 2 x dx =

Hamilton Education Guides

∫ sinh

3

∫

x dx + sinh 5 x dx =

In

1 2 cosh 5 x − cosh 3 x + cosh x + c and 5 3

2 1 1  cosh 3 x − cosh x + c +  cosh 5 x − cosh 3 x + cosh x + c  3 3 5 

90

)

Advanced Integration

=

1.4 Integration of Hyperbolic Functions

1 1 2 cosh 5 x + cosh 3 x − cosh 3 x − cosh x + cosh x + c 5 3 3 1 5

1 3

Check: Let y = cosh 5 x − cosh 3 x + c , then y ′ =

1 1 cosh 5 x − cosh 3 x + c 3 5

=

1 3 ⋅ 5 cosh 4 x ⋅ sinh x − ⋅ cosh 2 x ⋅ sinh x + 0 5 3

(

)

= sinh x cosh 4 x − sinh x cosh 2 x = sinh x cosh 2 x cosh 2 x − 1 = sinh x cosh 2 x sinh 2 x = sinh 3 x cosh 2 x e.

∫ tanh

5

∫ tanh

x sec h 4 x dx =

∫

+ tanh 5 x sec h 2 x dx = −

5

x sec h 2 x sec h 2 x dx =

5 2 2 ∫ tanh x (1 − tanh x ) sec h x dx

= − ∫ tanh 7 x sec h 2 x dx

1 1 tanh8 x + tanh6 x + c 8 6

1 8

1 8

1 6

1 6

Check: Let y = − tanh 8 x + tanh 6 x + c , then y ′ = − ⋅ 8 tanh 7 x ⋅ sec h 2 x + ⋅ 6 tanh 5 x ⋅ sec h 2 x + 0

(

)

= − tanh 7 x sec h 2 x + tanh 5 x sec h 2 x = tanh 5 x sec h 2 x 1 − tanh 2 x = tanh 5 x sec h 2 x sec h 2 x = tanh 5 x sec h 4 x f.

∫ tanh

3

x sec h 3 x dx =

∫ tanh

2

2 2 ∫ (1 − sec h x )sec h x ⋅ tanh x sec h x dx

x sec h 2 x ⋅ tanh x sec h x dx =

= − ∫ sec h 4 x ⋅ tanh x sec h x dx + ∫ sec h 2 x ⋅ tanh x sec h x dx . To solve the first and the second integral let u = sec h x , then

∫ tanh

3

du d = sec h x dx dx

x sec h 3 x dx =

∫ tanh

2

∫

;

du = − sec h x tanh x dx

=

. Therefore,

∫

∫

1 5 1 3 u − u +c 3 5

du sec h x tanh x

x sec h 2 x ⋅ tanh x sec h x dx = − sec h 4 x ⋅ tanh x sec h x dx

+ sec h 2 x ⋅ tanh x sec h x dx = − u 4 ⋅ −

=

; dx = −

tanh x sec h x tanh x sec h x du + u 2 ⋅ − du tanh x sec h x tanh x sec h x

∫

=

∫u

4

∫

du − u 2 du

1 1 sec h 5 x − sec h 3 x + c 3 5 1 3

1 5

Check: Let y = sec h 5 x − sec h 3 x + c , then y ′ =

5 3 sec h 4 x ⋅ − sec h x tanh x − sec h 2 x ⋅ − sec h x tanh x 5 3

)

(

= − sec h 4 x ⋅ sec h x tanh x + sec h 2 x ⋅ sec h x tanh x = − sec h 4 x + sec h 2 x sec h x tanh x

(

)

= 1 − sec h 2 x sec h 2 x sec h x tanh x = tanh 2 x sec h 2 x sec h x tanh x = tanh 3 x sec h 3 x g. Given ∫ coth 2 x csc h 2 x dx let u = coth x , then

∫

coth 2 x csc h 2 x dx =

Hamilton Education Guides

∫

u 2 csc h 2 x

du d = coth x dx dx

;

du = − csc h 2 x dx

; dx =

−du csc h 2 x

. Thus,

1 1 du = − u 2 du = − u 3 + c = − coth 3 x + c 3 3 − csc h x 2

∫

91

Advanced Integration

1.4 Integration of Hyperbolic Functions 1 3

1 3

Check: Let y = − coth 3 x + c , then y ′ = − ⋅ 3 coth 2 x ⋅ − csc h 2 x + 0 =

h.

∫ coth

3

x csc h 3 x dx =

∫ coth x coth

2

x csc h 3 x dx =

3 ⋅ coth 2 x csc h 2 x 3

= coth 2 x csc h 2 x

2 2 4 ∫ coth x (1 + csc h x ) csc h x dx = ∫ csc h x coth x dx

1 1 + csc h 2 x coth x dx = − csc h 5 x − csc h 3 x + c 5 3

∫

3 1 1 5 Check: Let y = − csc h 5 x − csc h 3 x + c , then y ′ = − csc h 4 x ⋅ − csc h x coth x − csc h 2 x ⋅ − csc h x coth x 5

3

5

5

3

(

3

)

= csc h x coth x + csc h x coth x = csc h x coth x csc h x + 1 = csc h x coth x coth 2 x = coth 3 x csc h 3 x i.

∫ coth

3

x csc h 4 x dx =

∫ coth

3

3

x csc h 2 x csc h 2 x dx =

2

3

3 2 2 5 2 ∫ coth x ( coth x − 1) csc h x dx = ∫ coth x csc h x dx

1 1 − coth 3 x csc h 2 x dx = − coth 6 x + coth 4 x + c 6 4

∫

1 1 1 1 Check: Let y = − coth 6 x + coth 4 x + c , then y ′ = − ⋅ 6 coth 5 x ⋅ − csc h 2 x + ⋅ 4 coth 3 x ⋅ − csc h 2 x + 0 6

5

4

2

6

3

2

3

2

(

2

4

)

3

= coth x csc h x − coth x csc h x = coth x csc h x coth x − 1 = coth x csc h 2 x csc h 2 x = coth 3 x csc h 4 x Table 1.4-3 provides a summary of the basic integration formulas covered in this manual. Section 1.4 Practice Problems – Integration of Hyperbolic Functions 1. Evaluate the following integrals: a.

∫ cosh 3x dx =

b.

3x ∫ (sinh 2 x − e ) dx

d.

∫x

e.

∫3x

g.

∫ cosh

2

sec h 2 x3dx = 7

( x + 1) sinh ( x + 1) dx = h.

c.

∫ csc h 5x dx =

f.

∫x

3

∫ csc h (5x + 3) coth (5x + 3) dx =

i.

∫e

x +1

∫ coth ( x + 1) csc h ( x + 1) dx = ∫ sec h ( 3x + 2) dx = 5 ∫ coth x dx =

c.

∫ e tanh e dx = cosh (3 x + 5 ) sinh (3 x + 5) dx ∫e 6 ∫ coth x dx =

(

=

)

2 3 csc h 2 x 4 + 1 dx

=

(

)

csc h 2 x 4 + 5 dx =

sec h e x +1 dx =

2. Evaluate the following integrals: a. d. g.

∫ tanh x sec h x dx = 3 4 ∫ x sec h ( x + 1) dx = 5 ∫ tanh x dx = 5

2

Hamilton Education Guides

b. e. h.

6

2

f. i.

3x

3x

92

=

Advanced Integration

1.4 Integration of Hyperbolic Functions

Table 1.4-3: Basic Integration Formulas 1.

∫ a dx

3.

∫ a f (x ) dx = a ∫ f (x ) dx

5.

= ax + c

a≠0

x n +1 +c n +1

2.

∫x

4.

∫ [ f (x ) + g (x ) ] dx = ∫ f (x ) dx + ∫ g (x ) dx

∫ sin x dx = − cos x + c

6.

∫ cos x dx = sin x + c

7.

∫ tan x dx = ln

sec x + c

8.

∫ cot x dx = ln

9.

∫ sec x dx = ln

sec x + tan x + c

10.

∫ csc x dx = ln

12.

∫ cot x csc x dx = − csc x + c

14.

∫ cos

2

x dx =

a≠0

n

dx =

n ≠ −1

sin x + c csc x − cot x + c

11.

∫ tan x sec x dx = sec x + c

13.

∫ sin

2

x dx =

15.

∫ tan

2

x dx = tan x − x + c

16.

∫ tan

2

x dx = tan x − x + c

17.

∫ cot

2

x dx = − cot x − x + c

18.

∫ sec

2

x dx = tan x + c

19.

∫ csc

2

x dx = − cot x + c

20.

∫ x dx

21.

∫ ln xdx

22.

∫

23.

∫e

24.

∫

x

x sin 2 x − +c 2 4

= x ln x − x + c

dx = e x + c

1

1

x sin 2 x + +c 2 4

= ln x + c ax +c ln a

a x dx = 1 2

a −x

dx =

2

a 0 and a ≠ 1

1 x 1 x arc sin + c = sin −1 + c a a a a

=

1 x 1 x arc tan + c = tan −1 + c a a a a

26.

∫ a 2 − x 2 dx

=

1 x−a ln +c 2a x+a

28.

∫x

∫ sinh x dx = cosh x + c

30.

∫ cosh x dx = sinh x + c

31.

∫ tanh x dx = ln cosh x + c

32.

∫ coth x dx = ln

sinh x + c

33.

∫ sec h x dx = sin

34.

∫ csc h x dx = ln

tanh

35.

∫ tanh x sec h x dx = − sec h x + c

36.

∫ coth x csc h x dx = − csc h x + c

37.

∫ sinh

2

x dx =

38.

∫ cosh

2

x dx =

39.

∫ tanh

2

x dx = x − tanh x + c

40.

∫ coth

2

x dx = x − coth x + c

41.

∫ sec h

x dx = tanh x + c

42.

∫ csc h

25.

∫ a 2 + x 2 dx

27.

∫ x 2 − a 2 dx

29.

1

2

−1

(tanh x ) + c

sinh 2 x x − +c 4 2

Hamilton Education Guides

1

1 2

x −a

2

2

=

1 a+x ln +c 2a a−x

dx =

x x 1 1 arc sec + c = sec−1 + c a a a a

x +c 2

sinh 2 x x + +c 4 2

x dx = − coth x + c

93

Appendix – Exercise Solutions Section 1.1 Practice Problems – Integration by Parts 1. Evaluate the following integrals using the integration by parts method.

∫ xe

a. Given

4x

dx let u = x and dv = e 4 x dx then du = dx and

4x

1 4x e . Using the integration 4

∫

4e 4 x  1 1 e4 x e4 x 1 1 1 4x  1  = xe4 x ⋅  x −  + 1 ⋅ e 4 x + 0 = e 4 x ⋅  x −  + e 4 x = xe4 x − + e  x −  + c , then y ′ = 4 4 4  4 4 4 4 4 4  

Check: Let y = x

∫ 2 cos x dx

let u = x and dv = cos x dx then du = dx and

integration by parts formula x

∫ 2 cos x dx

∫ u dv = u v − ∫ v du

we obtain

∫

∫ ( 5 − x )e

1 1 1 1 1 1 1 1 x sin x + cos x + c , then y ′ = (1 ⋅ sin x + cos x ⋅ x ) − sin x + 0 = sin x + x cos x − sin x = x cos x 2 2 2 2 2 2 2 2 5x

dx let u = 5 − x and dv = e5 x dx then du = − dx and

the integration by parts formula 5x ∫ ( 5 − x )e dx

∫ dv = ∫ cos x dx which implies v = sin x . Using the

1 1 1 1 x ⋅ sin x − sin x dx = x sin x + cos x + c 2 2 2 2

=

Check: Let y = c. Given

dx which implies v =

1 4x 1 4x 1 1 1 1  xe − e dx = xe4 x − e 4 x + c = e 4 x  x −  + c 4 4 4 16 4 4 

dx =

b. Given

4x

∫ u dv = u v − ∫ v du we obtain

by parts formula

∫ xe

∫ dv = ∫ e

= (5 − x )

Check: Let y = e5 x −

∫ dv = ∫ e

5x

dx which implies v =

e5x . Using 5

∫ u dv = u v − ∫ v du we obtain

1 1 5x e5 x e5 x + dx = e 5 x − xe 5 x + e +c 5 5 5 25

∫

(

)

1 1 1 5 5x 1 5x 1 5x ⋅ e + 0 = 5e5 x − e5 x − xe5 x + e5 x e + c , then y ′ = 5e5 x − e5 x + 5 xe5 x + xe + 5 25 5 5 25 5

= 5e5 x − xe5 x = (5 − x )e5 x d. Given

∫ x sin 5x dx

let u = x and dv = sin 5 x dx then du = dx and

Using the integration by parts formula

∫ x sin 5x dx

∫ u dv = u v − ∫ v du

1

∫ dv = ∫ sin 5x dx which implies v = − 5 cos 5x .

we obtain

1 1 1 1 = x ⋅ − cos 5 x + cos 5 x dx = − x cos 5 x + sin 5 x + c 5 5 5 5

∫

5 1 1 1 1 1 Check: Let y = − x cos 5 x + sin 5 x + c , then y ′ = − (1 ⋅ cos 5 x − sin 5 x ⋅ 5 ⋅ x ) + cos 5 x + 0 = − cos 5 x + x sin 5 x 5 5 5 5 5 5 +

e. Given

1 5 cos 5 x = x sin 5 x = x sin 5 x 5 5

∫x

3 − x dx let u = x and dv = 3 − x dx then du = dx and

Using the integration by parts formula

∫x

3 − x dx = x ⋅ −

∫ dv = ∫

3 − x dx which implies v = −

3 2 (3 − x )2 . 3

∫ u dv = u v − ∫ v du we obtain

3 3 5 3 3 3 2 (3 − x ) 2 + 2 (3 − x ) 2 dx = − 2 x (3 − x ) 2 − 2 ⋅ 1 3 (3 − x ) 2 +1 + c = − 2 x (3 − x ) 2 − 2 ⋅ 2 (3 − x ) 2 + c 3 3 1+ 3 5 3 3 3

Hamilton Education Guides

∫

2

94

Advanced Integration

= −

Solutions

3 5 2 4 (3 − x ) 2 + c x (3 − x ) 2 − 3 15

5 3 3 3 1 4 2 4 5 2 3 2 Check: Let y = − x (3 − x ) 2 − (3 − x ) 2 + c , then y ′ = − (3 − x ) 2 + ⋅ x(3 − x ) 2 + ⋅ (3 − x ) 2 + 0 15 3 15 2 3 2 3

1 3 3 1 2 (3 − x ) 2 + x(3 − x ) 2 + 2 (3 − x ) 2 = x(3 − x ) 2 = x 3 − x 3 3

= − f. Given

∫x e

3 3x

dx let u = x3 and dv = e 3 x dx then du = 3 x 2 dx and

3x

dx which implies v =

1 3x e . Using the 3

∫ u dv = u v − ∫ v du we obtain

integration by parts formula

∫

∫ dv = ∫ e

x3e3 x 1 1 3x x3e3 x dx = x3 ⋅ e3 x − − x 2e3 x dx e ⋅ 3 x 2 dx = 3 3 3

∫

∫

In example 5.1-1, problem letter b, we showed that

∫x e

3 3x

dx =

e dx =

1 2 3x 2 3x 2 3x x e − xe + e + c . Therefore, 3 9 27

(

) (

x3e3 x 1 2 3 x 2 3 x 2 3 x 1 1 e + c , then y ′ = − x e + xe − 3 x 2 ⋅ e3 x + 3e3 x ⋅ x3 − 2 x ⋅ e3 x + 3e3 x ⋅ x 2 27 3 9 3 3 3

(

)

)

2 2 2 2 2 2 ⋅ 3e3 x + 0 = x 2e3 x + x3e3 x − xe3 x − x 2e3 x + e3 x + xe3 x − e3 x = x3e3 x 1 ⋅ e3 x + 3e3 x ⋅ x − 3 9 3 9 9 27

+

∫ cos ( ln x ) dx

let u = cos ( ln x ) and dv = dx then du =

Using the integration by parts formula

∫ cos ( ln x ) dx To integrate du =

2 3x

x3e3 x  1 2 3 x 2 3 x 2 3 x  x 3e 3 x 1 2 3 x 2 3 x 2 3x −  x e − xe + e +c = − x e + xe − e +c 3 3 9 27 3 3 9 27  

Check: Let y =

g. Given

∫x

= cos ( ln x ) ⋅ x + x ⋅

∫

− sin ( ln x ) dx and x

∫ dv = ∫ dx which implies v = x .

∫ u dv = u v − ∫ v du we obtain

sin ( ln x ) dx = x cos ( ln x ) + sin ( ln x ) dx x

(1 )

∫

∫ sin ( ln x ) dx use the integration by parts formula again, i.e., let u = sin ( ln x ) and dv = dx then

cos ( ln x ) dx and x

∫ dv = ∫ dx which implies v = x . Therefore,

∫ sin ( ln x ) dx = sin ( ln x ) ⋅ x − ∫ x ⋅

cos ( ln x ) dx = x sin ( ln x ) − cos ( ln x ) dx x

(2 )

∫

Combining equations ( 1 ) and ( 2 ) together we have

∫ cos ( ln x ) dx

= x cos ( ln x ) − sin ( ln x ) dx = x cos ( ln x ) + x sin ( ln x ) − cos ( ln x ) dx

∫

∫

Taking the integral − cos ( ln x ) dx from the right hand side of the equation to the left hand side

∫

we obtain

∫ cos ( ln x ) dx + ∫ cos ( ln x ) dx = x cos ( ln x ) + x sin ( ln x ) Therefore,

2 cos ( ln x ) dx = x cos ( ln x ) + x sin ( ln x ) + c and thus

∫

Check: Let y = = h. Given

∫ cos ( ln x ) dx

=

x x cos ( ln x ) + sin ( ln x ) + c 2 2

x x cos ( ln x ) x sin ( ln x ) sin ( ln x ) x cos ( ln x ) cos ( ln x ) + sin ( ln x ) + c , then y ′ = +0 − + + 2 2 2 2x 2 2x

cos ( ln x ) sin ( ln x ) sin ( ln x ) cos ( ln x ) cos ( ln x ) cos ( ln x ) = = cos ( ln x ) − + + + 2 2 2 2 2 2 x

∫ 3 tan

−1

x dx let u = tan −1 x and dv =

Hamilton Education Guides

1 x dx and dx then du = 3 1+ x 2

x

∫ dv = ∫ 3 dx

which implies v =

1 2 x . 6

95

Advanced Integration

Solutions

Using the integration by parts formula x

∫ 3 tan =

−1

2 x dx = tan −1 x ⋅

∫ u dv = u v − ∫ v du we obtain

1 1 x2 x 2 1 2 dx 1 1 = x 2 tan −1 x − dx = x 2 tan −1 x − − x ⋅ 2 6 6 1 + x2 6 6 6 6 1+ x

∫

∫



1



∫ 1 − 1 + x2  dx

1 2 −1 1 1 1 1 1 1 x tan x − dx + dx = x 2 tan −1 x − x + tan −1 x + c 6 6 6 6 6 1 + x2 6

∫

∫

1 1 x2 1 1 1 1 1 2 1 +0 − + ⋅ x tan −1 x − x + tan −1 x + c , then y ′ = ⋅ 2 x ⋅ tan −1 x + 6 6 6 6 6 1 + x2 6 6 1 + x2

Check: Let y =

1 1 x2 1 1 1 1 x2 + 1 1 1 1 1 1 1 −1 x x ⋅ x tan −1 x + ⋅ + ⋅ − + ⋅ − = x tan −1 x + − = x tan −1 x tan = 2 2 2 6 1+ x 3 6 1+ x 6 3 6 1+ x 6 6 6 3 3 i. Given

∫ ln x dx let u = ln x 5

integration by parts formula

∫ ln x dx = ln x 5

5

∫

⋅x− x⋅

5

and dv = dx then du =

1

⋅ 5 x 4 dx =

x5

5 dx and x

∫ dv = ∫ dx which implies v = x . Using the

∫ u dv = u v − ∫ v du we obtain

5 dx = x ln x5 − 5 dx = x ln x 5 − 5 x + c x

∫

5 x5 1   Check: Let y = x ln x5 − 5 x + c , then y ′ = 1 ⋅ ln x5 + 5 ⋅ 5 x 4 ⋅ x  − 5 + 0 = ln x5 + 5 − 5 = ln x5 + 5 − 5 = ln x5 x x   j. Given

∫x e

− ax

dx let u = x and dv = e − ax dx then du = dx and

integration by parts formula

∫x e

− ax

∫e

a a x

2

∫

(

⋅ e − ax = −

1 − ax 1 e + xe− ax + e − ax = xe− ax a a

sin 3 x dx let u = e x and dv = sin 3 x dx then du = e x dx and

sin 3 x dx = e x ⋅ −

To integrate v=

)

1 − ax 1 − ax 1 1 1 axe− ax − 2e + c , then y ′ = − 1 ⋅ e − ax − ae − ax ⋅ x − 2 ⋅ −2e − ax + 0 = − e − ax + xe a a a a a a

Using the integration by parts formula x

1 dx which implies v = − e − ax . Using the a

we obtain

∫

+

∫e

− ax

1 1 1 1 − ax 1 1 − ax dx = x ⋅ − e − ax + e dx = − x e − ax + e dx = − x e − ax − 2 e − ax + c a a a a a a

Check: Let y = −

k. Given

∫ u dv = u v − ∫ v du

∫ dv = ∫ e

∫e

x

∫ u dv = u v − ∫ v du

∫ dv = ∫ sin 3x dx which implies v = −

we obtain

1 1 x cos 3 x cos 3 x x e cos 3 x dx − − ⋅ e dx = − e x cos 3 x + 3 3 3 3

cos 3 x dx let u = e x and dv = cos 3 x dx then du = e x dx and

sin 3 x . Thus, 3

∫e

x

(1 )

∫

∫

cos 3 x dx = e x ⋅

∫ dv = ∫ cos 3x dx which implies

sin 3 x x 1 x sin 3 x 1 − ⋅ e dx = e x sin 3 x − e sin 3 x dx 3 3 3 3

∫

cos 3 x . 3

∫

(2)

Combining equations (1 ) and ( 2 ) together we obtain:

∫e

x

1 1 1 x 1 1 x sin 3 x dx = − e x cos 3 x + e sin 3 x dx e cos 3 x dx = − e x cos 3 x + e x sin 3 x − 3 9 9 3 3

Taking the −

∫

∫

1 x e sin 3 x dx from the right hand side of the equation to the left hand side we obtain 9

∫

Hamilton Education Guides

96

Advanced Integration

∫e

x

and

1 x 1 1 1 1 10 x e sin 3 x dx = − e x cos 3 x + e x sin 3 x which implies e sin 3 x dx = − e x cos 3 x + e x sin 3 x 3 9 3 9 9 9

x

sin 3 x dx =

Check: Let y = − = −

∫e

l. Given

∫

∫

sin 3 x dx +

∫e

Solutions

x

3 x 1 x 9 1 x 1  e cos 3 x + e sin 3 x  − e cos 3 x + e x sin 3 x  = − 10 10 10  3 9 

3 1 1 3 3 x 1 e cos 3 x + e x sin 3 x , then y ′ = − e x ⋅ cos 3 x + sin 3 x ⋅ 3 ⋅ e x + e x ⋅ sin 3 x + cos 3 x ⋅ 3 ⋅ e x 10 10 10 10 10 10

9 1 3 9 x 1 3 x 10 x e cos 3 x + e x sin 3 x + e x sin 3 x + e x cos 3 x = e sin 3 x + e x sin 3 x = e sin 3 x = e x sin 3 x 10 10 10 10 10 10 10

cos 5 x dx let u = e x and dv = cos 5 x dx then du = e x dx and

Using the integration by parts formula

∫e

x

∫ u dv = u v − ∫ v du

1

∫ dv = ∫ cos 5x dx which implies v = 5 sin 5x .

we obtain

e x sin 5 x 1 x 1 1 cos 5 x dx = e x ⋅ sin 5 x − − e sin 5 x dx sin 5 x ⋅ e x dx = 5 5 5 5

To integrate

∫e

Thus,

x

∫e

x

(1 )

∫

∫

1

∫ dv = ∫ sin 5x dx which implies v = − 5 cos 5x .

sin 5 x dx let u = e x and dv = sin 5 x dx then du = e x dx and

e x cos 5 x 1 1 1 sin 5 x dx = e x ⋅ − cos 5 x + + cos 5 x ⋅ e x dx cos 5 x ⋅ e x dx = − 5 5 5 5

∫

∫

(2)

Combining equations (1 ) and ( 2 ) together we obtain:

∫e −

cos 5 x dx =

 e x sin 5 x 1  e x cos 5 x 1 x e x sin 5 x e x cos 5 x e x sin 5 x 1 x − − + e cos 5 x dx  = + − e sin 5 x dx =  5 5 5 25 5 5  5 5 

∫

∫

1 1 e x cos 5 x dx . Taking the − e x cos 5 x dx from the right hand side of the equation to the left hand side we obtain 25 25

∫

∫e

∫

x

x

∫

cos 5 x dx +

e x sin 5 x e x cos 5 x e x sin 5 x e x cos 5 x 1 26 x which implies and + + e x cos 5 x dx = e cos 5 x dx = 5 25 25 5 25 25

e x cos 5 x dx =

Check: Let y = =

∫

∫

25  e x sin 5 x e x cos 5 x  5 x 1 x = + e sin 5 x + e cos 5 x  26  5 25 26 26  1 x 1 5 x 1 x 5 x 5 e sin 5 x + e cos 5 x , then y ′ = sin 5 x ⋅ 5 ⋅ e x + 0 cos 5 x ⋅ 5 ⋅ e x + e ⋅ cos 5 x − e ⋅ sin 5 x + 26 26 26 26 26 26

5 x 25 x 1 x 5 x 26 x 1 x 25 x e cos 5 x = e x cos 5 x e cos 5 x = e sin 5 x + e cos 5 x + e cos 5 x − e sin 5 x = e cos 5 x + 26 26 26 26 26 26 26

2. Evaluate the following integrals using the integration by parts method. a. Given

∫ x sec

2

the integration by parts formula

∫ x sec

2

∫ dv = ∫ sec

x dx let u = x and dv = sec2 x dx then du = dx and

∫ u dv = u v − ∫ v du

x dx which implies v = tan x . Using

we obtain

∫

x dx = x ⋅ tan x − tan x dx = x tan x − ln sec x + c

(

)

Check: Let y = x tan x − ln sec x + c , then y ′ = 1 ⋅ tan x + sec2 x ⋅ x − b. Given

2

∫ arc sin 3 y dy

let u = arc sin 3 y and dv = dy then du =

the integration by parts formula

Hamilton Education Guides

sec x tan x + 0 = tan x + x sec2 x − tan x = x sec2 x sec x

3dy 1− 9y

∫ u dv = u v − ∫ v du

2

and

∫ dv = ∫ dy which implies v = y . Using

we obtain

97

Advanced Integration

∫ arc sin 3 y dy

∫

3 y dy

3 y dy

3y

∫

⋅

w

(

1− 9y

2

∫

y dy

= y arc sin 3 y − 3

1− 9y

(1 )

2

use the substitution method by letting w = 1 − 9 y 2 then

1 − 9 y2

=

1 − 9 y2

3dy

∫

= arc sin 3 y ⋅ y − y ⋅

∫

To integrate

Solutions

1 1 1 = − w 2 = − 1 − 9 y2 3 3

)

1 dw = − − 18 y 6

∫

dw w

= −

dw dw . Therefore, = −18 y and dy = − dy 18 y

−1 1− 1 1 1 2 1 1 1 2 −1 1 1 w 2 dw = − ⋅ w 2 = − ⋅ 2 −1 w 2 = − ⋅ w 2 1 6 6 1 6 6 1−

∫

2

2

1 2

(2 )

Combining equations (1 ) and ( 2 ) together we obtain:

∫

arc sin 3 y dy = y arc sin 3 y −

Check: Let w = y arc sin 3 y + 3y

+

−

1 − 9 y2

∫ arc tan x dx

c. Given

∫

1 − 9 y2

(

1 1 − 9 y2 3

18 y

1 ⋅ 6

x dx

x dx

x dw

∫ 1 + x2 = ∫ w ⋅ 2x

dx

1 + x2

(

1 1 − 9 y2 3

)

1 2

1− 9y 3y

3y

−

1 − 9 y2

1 − 9 y2

dx

1 + x2

and

− 2

18 y

1 1 ⋅ ⋅ 3 2

1− 9y

2

+ 0 = arc sin 3 y

= arc sin 3 y

∫ dv = ∫ dx which implies v = x . Using the

we obtain

= x arc tan x −

x dx

(1 )

∫ 1 + x2

use the substitution method by letting w = 1 + x 2 then =

+c

3y

+ c , then w ′ = arc sin 3 y +

∫ u dv = u v − ∫ v du

∫

∫ 1 + x2

1 2

let u = arc tan x and dv = dx then du =

= arc tan x ⋅ x − x ⋅

To integrate

)

= y arc sin 3 y +

= arc sin 3 y +

1 − 9 y2

integration by parts formula

∫ arc tan x dx

3 y dy

dw dw . Therefore, = 2 x And dx = 2x dx

1 1 1 dw = ln w = ln 1 + x 2 2 2 w 2

(2)

∫

Combining equations (1 ) and ( 2 ) together we obtain:

∫ arc tan x dx

= x arc tan x −

Check: Let y = x arc tan x − d. Given

∫ sin 5x dx = ∫ sin 3

2

x dx

∫ 1 + x2

= x arc tan x −

1 ln 1 + x 2 + c 2

x x x 1 2x 1 − + 0 arc tan x + − = arc tan x ln 1 + x 2 + c , then y ′ = arc tan x + 2 2 2 2 2 1+ x 1+ x 1+ x 1 + x2

5 x ⋅ sin 5 x dx let u = sin 2 5 x and dv = sin 5 x dx then du = 10 sin 5 x cos 5 x dx and 1

∫ dv = ∫ sin 5x dx ∫ dv = ∫ sin x dx which implies v = − 5 cos 5x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ sin 5x dx = sin

To integrate

3

∫ cos

2

2

5x ⋅ −

2

∫

∫

(1 )

5 x sin 5 x dx use the integration by parts method again, i.e., let u = cos 2 5 x and dv = sin 5 x then

du = −10 sin 5 x cos 5 x dx and

∫ cos

1 cos 5 x 1 + cos 5 x ⋅ 10 sin 5 x cos 5 x dx = − sin 2 5 x cos 5 x + 2 cos 2 5 x sin 5 x dx 5 5 5

1

∫ dv = ∫ sin 5x dx which implies v = − 5 cos 5x . Therefore,

1 1 1 5 x sin 5 x dx = cos 2 5 x ⋅ − cos 5 x − cos 5 x ⋅ 10 sin 5 x cos 5 x dx = − cos3 5 x − 2 cos 2 5 x sin 5 x dx . Taking the 5 5 5

Hamilton Education Guides

∫

∫

98

Advanced Integration

Solutions

∫

integral − 2 cos 2 5 x sin 5 x dx from the right hand side of the equation to the left side we obtain

∫ cos

2

1 5 x sin 5 x dx + 2 cos 2 5 x sin 5 x dx = − cos3 5 x . Therefore, 5

∫

∫ cos

2

5 x sin 5 x dx = −

Combining equations (1 ) and ( 2 ) together we have 1

∫ sin 5x dx = − 5 sin 3

−

(

2

1 cos3 5 x 15

(2)

sin 2 5 x cos 5 x 2 1 1 − cos3 5 x 5 x cos 5 x + 2 cos 2 5 x sin 5 x dx = − sin 2 5 x cos 5 x + 2 ⋅ − cos3 5 x = − 15 5 5 15

∫

)

1 2 1 1 1 2 1  1 − cos 2 5 x cos 5 x − cos3 5 x + c = − cos 5 x +  cos3 5 x − cos3 5 x  + c = cos 3 5 x − cos 5 x + c 5 15 5 15 5 15 5 

Note that another method of solving the above problem is in the following way:

∫ sin 5x dx = ∫ sin 3

1 5

5 x ⋅ sin 5 x dx =

∫ sin 5x dx = ∫ sin 3

Therefore, = −

2

2

5 x ⋅ sin 5 x dx =

1 2 1 2 ∫ (1 − u ) du = ∫  5 u − 5  du

Check: Let y =

2 ∫ (1 − cos 5x )⋅ sin 5x dx let u = cos 5x , then

=

du du . = −5 sin 5 x and dx = − dx 5 sin 5 x

du 2 2 ∫ (1 − cos 5x )⋅ sin 5x dx = ∫ (1 − u )⋅ sin 5x ⋅ − 5 sin 5x

1 3 1 1 1 u − u+c = cos 3 5 x − cos 5 x + c 15 5 15 5

1 1 1 1 ⋅ 3 cos 2 5 x ⋅ − sin 5 x ⋅ 5 + ⋅ 5 sin 5 x + 0 = − cos 2 5 x ⋅ sin 5 x + sin 5 x cos3 5 x − cos 5 x + c , then y ′ = 15 5 5 15

(

)

= sin 5 x 1 − cos 2 5 x = sin 5 x sin 2 5 x = sin 3 5 x e. Given

∫x

2

cos x dx let u = x 2 and dv = cos x dx then du = 2 x dx and

the integration by parts formula

∫x

2

∫ u dv = u v − ∫ v du

∫ dv = ∫ cos x dx which implies v = sin x . Using

we obtain

(1 )

∫

∫

cos x dx = x 2 ⋅ sin x − sin x ⋅ 2 x dx = x 2 sin x − 2 x sin x dx

To integrate

∫ x sin x dx

use the integration by parts formula again, i.e., let u = x and dv = sin x dx then du = dx and

∫ dv = ∫ sin x dx which implies v = − cos x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain ∫ x sin x dx = x ⋅ − cos x + ∫ cos x ⋅ dx = − x cos x + sin x

(2)

Combining equations (1 ) and ( 2 ) together we have

∫x

2

cos x dx = x 2 sin x − 2 x sin x dx = x 2 sin x − 2(− x cos x + sin x ) = x 2 sin x + 2 x cos x − 2 sin x

∫

(

)

Check: Let y = x sin x + 2 x cos x − 2 sin x , then y ′ = 2 x sin x + x 2 cos x + 2(cos x − x sin x ) − 2 cos x 2

2 x sin x + x 2 cos x + 2 cos x − 2 x sin x − 2 cos x = x 2 cos x f. Given

∫e

−2 x

cos 3 x dx let u = cos 3 x and dv = e −2 x dx then du = −3 sin 3 x dx and

1 v = − e − 2 x . Using the integration by parts formula 2

∫e

−2 x

∫ u dv = u v − ∫ v du

∫ dv = ∫ e

−2 x

dx which implies

we obtain

1 3 −2 x 1 3 −2 x cos 3 x dx = cos 3 x ⋅ − e − 2 x − ⋅ sin 3 x dx = − e − 2 x cos 3 x − e e sin 3 x dx 2 2 2 2

To integrate

∫

∫e

−2 x

∫

(1 )

sin 3 x dx use the integration by parts formula again, i.e., let u = sin 3 x and dv = e −2 x dx then

du = 3 cos 3 x dx and

∫ dv = ∫ e

Hamilton Education Guides

−2 x

1 dx which implies v = − e − 2 x . Therefore, 2

99

Advanced Integration

∫e

−2 x

Solutions

1 1 −2 x 1 3 −2 x sin 3 x dx = sin 3 x ⋅ − e − 2 x + e ⋅ 3 cos 3 x dx = − e − 2 x sin 3 x + e cos 3 x dx 2 2 2 2

∫

(2)

∫

Combining equations (1 ) and ( 2 ) together we have

∫e

−2 x

cos 3 x dx = −

∫e

Taking the integral

∫e

we obtain

−2 x

1 −2 x 3 −2 x 9 −2 x 1 3 cos 3 x − sin 3 x dx = − e − 2 x cos 3 x + e − 2 x sin 3 x − e e e cos 3 x dx 2 4 4 2 2

∫

−2 x

∫

cos 3 x dx from the right hand side of the equation to the left hand side

cos 3 x dx +

1 3 9 −2 x cos 3 x dx = − e − 2 x cos 3 x + e − 2 x sin 3 x . Therefore, e 4 2 4

∫

13 − 2 x 3 1 e cos 3 x dx = − e − 2 x cos 3 x + e − 2 x sin 3 x and thus 4 4 2

∫e

∫

Check: Let y = −

(

3

Using the integration by parts formula 3

∫ x ( 5x − 1) dx = x ⋅ Check: Let y =

∫ x csc

(5 x − 1) 4 − 20

1 20

3

∫ u dv = u v − ∫ v du

∫ (5x − 1)

4

]

2

x dx let u = x and dv = csc2 x dx then du = dx and

∫ u dv = u v − ∫ v du

∫ dv = ∫ csc

2

x dx which implies v = − cot x . Using

we obtain

∫

2

∫ 3 cos

−1

∫

1 (5 x − 1)4 . 20

x dx = x ⋅ − cot x + cot x dx = − x cot x + ln sin x + c

−1

)

5 x dx let u = cos −1 5 x and dv = dx then du = −

5 x dx =

To integrate

Thus,

dx which implies v =

x (5 x − 1) 4 1 1 1 x (5 x − 1) 4 (5 x − 1) 5 + c − ⋅ (5 x − 1) 4 +1 + c = − 20 20 25 20 500

dx =

[

Using the integration by parts formula 2

3

we obtain

(

∫ 3 cos

∫ dv = ∫ (5x − 1)

dx then du = dx and

Check: Let y = − x cot x + ln sin x + c , then y ′ = − cot x − x csc2 x + i. Given

(

x (5 x − 1) 4 1 (5 x − 1) 5 + c , then y ′ = 1 (5 x − 1) 4 + 20 x (5 x − 1) 3 − 25 (5 x − 1) 4 + 0 − 20 500 500 20

the integration by parts formula 2

)

1 (5 x − 1) 4 + x (5 x − 1) 3 − 1 (5 x − 1) 4 = x (5 x − 1) 3 20 20

=

∫ x csc

2 −2 x 3 −2 x cos 3 x + sin 3 x + c e e 13 13

2e −2 x cos 3 x 3e −2 x sin 3 x 3 2 + + c , then y ′ = − − 2e − 2 x cos 3 x − 3e − 2 x sin 3 x + − 2e − 2 x sin 3 x + 3e − 2 x cos 3 x 13 13 13 13

∫ x ( 5x − 1) dx let u = x and dv = (5x − 1)

h. Given

cos 3 x dx = −

4 −2 x 6 6 9 4 −2 x 9 cos 3 x + e − 2 x cos 3 x = e −2 x cos 3 x e e cos 3 x + e − 2 x sin 3 x − e − 2 x sin 3 x + e − 2 x cos 3 x = 13 13 13 13 13 13

= g. Given

−2 x

∫

2 2 cos −1 5 x ⋅ x + x⋅ 3 3

∫

5x 1 − 25 x 2

5x 1 − 25 x 2

∫ u dv = u v − ∫ v du

dx =

Hamilton Education Guides

5 dx 1 − 25 x

2

=

cos x + 0 = − cot x + x csc2 x + cot x = x csc2 x sin x

5 1 − 25 x 2

dx and

∫ dv = ∫ dx which implies v = x .

we obtain

2 2 x cos −1 5 x + 3 3

∫

5x 1 − 25 x

2

dx

dx use the substitution method by letting w = 1 − 25 x 2 then

∫

5x w

⋅−

1 dw = − 10 50 x

∫

dw w

= −

1 10

∫

1 1 w2

dw = −

dw dw . = −50 x which implies dx = − 50 x dx

1 −1 1 1 w 2 dw = − ⋅ 2 w 2 = − 10 10

∫

1 − 25 x 2 5

100

)

Advanced Integration

and

Solutions

2 2 2 cos −15 x dx = x cos −1 5 x + 3 3 3

∫

Check: Let y =

∫ sinh

j. Given

−1

10 x

−1

3 1 − 25 x

2

3 1 − 25 x

dx

∫

∫

dw . Therefore, 2x

(

2

=

∫ u dv = u v − ∫ v du

x dx = sinh −1 x ⋅ x − x ⋅

1

10 x

= w 2 = 1 + x2

)

1 2

=

∫

1+ x

x dx 1+ x

−

1 − 25 x 2

2 ⋅ 15

−50 x 2 1 − 25 x 2

+0

1

dx and

1 + x2

∫ dv = ∫ x dx which implies v = x . Using

we obtain

= x sinh −1 x −

2

5x

2 cos −1 5 x 3

x dx let u = sinh −1 x and dv = dx then du =

To get the integral of

dx =

1 − 25 x 2

+

the integration by parts formula

∫ sinh

∫

2 1 − 25 x 2 2 x cos −1 5 x − +c 3 15

dx =

2 1 − 25 x 2 2 2 2 x cos −1 5 x − + c , then y ′ = cos −1 5 x − ⋅ 15 3 3 3

2 cos −1 5 x − 3

=

5x

∫

x dx 1+ x

(1 )

2

use the substitution method by letting w = 1 + x 2 then dw = 2 x dx which implies

2

x dx

=

1 + x2

∫

x w

⋅

1 dw = 2 2x

∫

1 w

dw =

1 2

∫

1 1 w2

dw =

−1 1− 1 1 2 12 1 1 2 = w w 2 dw = ⋅ w 2 2 2 1− 1

∫

2

(2)

1 + x2

Combining equations (1 ) and ( 2 ) together we have

∫

sinh −1 x dx = x sinh −1 x −

x dx

∫

1 + x2

(

Check: Let y = x sinh −1 x − 1 + x 2

)

1 2

(

= x sinh −1 x − 1 + x 2

)

1 2

+c x

+ c , then y ′ = sinh −1 x +

−

1+ x

∫ x sec 10 x dx let u = x and dv = sec 10 x dx then du = dx 2

2

k. Given

Using the integration by parts formula

∫ x sec 10 x dx = x ⋅ 2

Check: Let y = = l. Given

2

2x 2 1+ x

and

+ 0 = sinh −1 x

2

∫ dv = ∫ sec 10 x dx which implies v = 2

tan 10 x . 10

∫ u dv = u v − ∫ v du we obtain

1 1 tan 10 x 1 − tan 10 x dx = x tan 10 x − ln sec 10 x + c 10 10 10 100

∫

1 1 sec 10 x tan 10 x ⋅ 10 1 1 tan 10 x + x sec2 10 x − +0 x tan 10 x − ln sec 10 x + c , then y ′ = 10 100 sec 10 x 10 100

1 1 tan 10 x + x sec2 10 x − tan 10 x = x sec2 10 x 10 10 x

x

∫ 5 sinh 7 x dx let u = 5

and dv = sinh 7 x dx then du =

Using the integration by parts formula

Hamilton Education Guides

dx and 5

1

∫ dv = ∫ sinh 7 x dx which implies v = 7 cosh 7 x dx .

∫ u dv = u v − ∫ v du we obtain

101

Advanced Integration

Solutions

∫ 5 sinh 7 x dx =

1 1 x 1 dx 1 1 1 = ⋅ cosh 7 x − cosh 7 x ⋅ x cosh 7 x − cosh 7 x ⋅ dx = x cosh 7 x − sinh 7 x + c 5 7 7 35 5 35 35 245

Check: Let y =

1 1 1 1 1 1 x cosh 7 x − cosh 7 x sinh 7 x + c , then y ′ = cosh 7 x + ⋅ 7 cosh 7 x + 0 = ⋅ 7 x sinh 7 x − 35 35 245 245 35 35

x

∫

∫

1 1 1 x sinh 7 x − cosh 7 x = x sinh 7 x 5 35 5

+

Section 1.2 Solutions – Integration Using Trigonometric Substitution Evaluate the following indefinite integrals. a. Given

dx x

2

16 − x

16 − x 2 =

let x = 4 sin t , then dx = 4 cos t dt and

2

(

16 − 16 sin 2 t =

16 1 − sin 2 t

)

16 cos 2 t = 4 cos t . Substituting these values back into the original integral we obtain:

=

∫

∫

dx x 2 16 − x 2

= −

4 cos t dt

16 − x 2 4 x 4

1 1 cos t +c = − 16 16 sin t

(

16 − x 2 2

= −

16 x

x2

∫

−2 x

)

∫

− x 2 − 16 + x 2 16 x 2 16 − x 2

2 16 − x 2

16 x

=

−2 x2

⋅ x − 1 ⋅ 16 − x 2 +0 = −

2

16 16 x 2 16 − x 2

2 16 − x 2

16 − x 2 1

−

16 x

− x 2 − 16 − x 2 ⋅ 16 − x 2 16 − x 2 2

= −

2

16 x

1

=

x 2 16 − x 2

9 − x2 =

dx let x = 3 sin t , then dx = 3 cos t dt and

9 − x2

1 1 csc2 t dt = − cot t + c 16 16

=

1 4 ⋅ 16 − x 2 16 − x 2 +c ⋅ +c = − 16 4⋅ x 16 x

16 − x + c , then y ′ = − 16 x

− x 2 − 16 − x 2

b. Given

+c = −

2

Check: Let y = −

= −

1

4 cos t

∫ (4 sin t ) 2 ⋅ 4 cos t = ∫ 16 sin 2 t ⋅ 4 cos t dt = ∫ 16 sin 2 t dt

=

9 − 9 sin 2 t =

(

9 1 − sin 2 t

)

=

9 cos 2 t

= 3 cos t . Substituting these values back into the original integral we obtain:

∫ =

x2

∫

dx =

9 − x2

9 sin 2 t ⋅ 3 cos t dt = 3 cos t

∫ 9 sin

2

t dt = 9

∫

1 − cos 2t 9 dt = 2 2

∫ (1 − cos 2t ) dt

=

9 1   t − sin 2t  + c 2 2 

9 x 9 x 9 − x2 9 x x 9 9 + c = sin −1 − 9 − x2 + c t − sin t cos t + c = ⋅ sin −1 − ⋅ ⋅ 2 3 2 3 3 2 2 2 3 2 9 −1 x x 9 9 − x 2 + c , then y ′ = sin − 2 3 2 2

Check: Let y =

=

=

c. Given

∫

9 2

3 9 − x2 9

−

2 9 − x2

dx x 9 + 4x

 1  9 − x2 x2 ⋅ − − 3  2 2 9 − x2 

2

(

2 9 − 2x2 4 9 − x2

let x =

Hamilton Education Guides

)=

9 2 9 − x2

1 1−

⋅ x2 9

  − 2x 1 1 x ⋅  − 9 − x2 + 2 3 2 2 9 − x2  

(

)

     2 9 − x2 − 2x2    18 − 4 x 2 9 9 − −  =  =  4 9 − x2    4 9 − x2 2 9 − x2 2 9 − x2     −

9 − 2x2 2 9 − x2

=

3 3 tan t , then dx = sec2 t dt and 2 2

9 − 9 + 2x2 2 9 − x2

9 + 4x2 =

=

2 x2 2 9 − x2 9+4

=

(32 tan t ) 2

    

x2 9 − x2 =

9 + 4 ⋅ 94 tan 2 t

102

Advanced Integration

=

(

9 + 9 tan 2 t =

=

∫

Solutions

dx

=

x 9 + 4x2

9 1 + tan 2 t

)

3 sec 2 t 2

∫ 32 tan t ⋅ 3sec t

9 sec2 t = 3 sec t . Therefore,

= dt =

1 sec2 t 1 1 1 sec t 1 cos t 1 ⋅ sec t dt = dt = dt = ⋅ dt 3 tan t ⋅ 3 sec t 3 tan t 3 tan t 3 sin t cos t

∫

9 + 4x2

1 1 1 1 1 dt = csc t dt = ln csc t − cot t + c = ln 3 3 3 sin t 3

∫

∫

Check: Let y =

2

9 + 4x − 3

1 ln 3

2x 8x2

1 = ⋅ 3

=

2x 9 + 4x

1 ⋅ 3

1 = ⋅ 3

9+ 4 x

⋅

2

⋅

)

=

4x2 9 + 4x2

1

2

+c

2

4x2   8x2 − 2 ⋅  9 + 4 x2 − 3 9 + 4 x2   ⋅

2x 9 + 4 x2 − 3

4 x2 9 + 4 x2

2x

⋅

8 x 2 − 18 + 8 x 2 + 6 9 + 4 x 2 4x2 9 + 4x2

9 + 4x2 − 3

  6  9 + 4 x2 − 3 6  = 1 ⋅ 2x ⋅  ⋅ 3 1 − 3 4x2 9 + 4x2 4x2 9 + 4x2

2x

  ⋅ 8x − 2 ⋅  9 + 4 x2 − 3  

1 2 9+ 4 x

⋅

1 ⋅ 3

9 + 4x2 − 3

3 1 + c = ln 2x 3

2x ⋅

1 = ⋅ 3

8x2 − 2 ⋅ 9 + 4x2 + 6 9 + 4x2

∫ (49 + x2 )2 dx let x = 7 tan t , then dx = 7 sec

d. Given

−

∫

9 + 4x2 − 3

  − 2 ⋅  9 + 4 x2 − 3  

(

2x

2x

4x2

9 + 4x2 − 3

9 + 4x2

1 ⋅ 3

+ c , then y ′ =

−3

2x

2x

2

∫

∫

=

12 x

1

=

12 x 2 9 + 4 x 2

x

9 + 4x2

(

t dt and 49 + x 2 = 49 + (7 tan t ) 2 = 49 + 49 tan 2 t = 49 1 + tan 2 t

)

= 49 sec2 t . Substituting these values back into the original integral we obtain 7 sec2 t dt

7 sec2 t dt

1

∫ (49 + x2 )2 dx = ∫ ( 49 sec2 t )2 = ∫ 2401sec2 t sec2 t =



∫  ∫

7 dt 1 1 1 = ⋅ cos 2 t dt = 343 2 343 2401 sec2 t

∫

∫

 1  −1 x 1 1  1  + = = ( ) t + sin t cos t + c t t + c + sin 2    tan 686  7 686 686  2  

Check: Let y =

e.

=

1 7 1 343 + 7 x 2 − 14 x 2 = ⋅ ⋅ 2 2 686 49 + x 2 686 49 + x

+

(49 − x ) 98( 49 + x )

(

2 2

 + 5 x  dx =  x2 − 1  x2

x2 − 1

(

)

2

dx =

7

⋅

49 + x 2

(

49 + x 2

)

  1  7x  −1 x  + c +c =  tan 7 + 686 49 + x 2    

1 49 1 1 1 7 49 + x 2 − 2 x ⋅ 7 x 7x  1  −1 x = ⋅ ⋅ + ⋅ +  + c then y ′ =  tan  2 2 2 2 686 686 71 + x  686 7 49 + x  686  7 49 + x 2 49 + x  49  

+

x2

x

∫ (1 + cos 2t ) dt

1 x 2

=

∫

49 + x 2 + 49 − x 2

(

98 49 + x

x2 2

Hamilton Education Guides

=

(

1 343 − 7 x 2 1 = ⋅ 2 2 686 49 + x 98 49 + x 2

98

98 49 + x

(

)

2 2

)

=

(

1

(

)

( ) ) 686( 49 + x ) +

7 49 − x 2

2 2

=

(

)

1

98 49 + x 2

)

( 49 + x )

2 2

∫

dx + 5 x dx . In Example 5.2-1, problem letter e, the solution to the first integral was:

x −1

x2 − 1 +

)

2 2

)

+

(

1 ln x + x 2 − 1 + c . Therefore, combining the two integrals we have 2

103

Advanced Integration x2

∫

Solutions

∫

dx + 5 x dx =

2

x −1

1 x 2

Check: Let y =

x2 − 1 +

  2x × 1 +  2 x2 − 1  = f. Given

∫

∫

1 x 2

1 5 ln x + x 2 − 1 + x 2 + c 2 2

x2 − 1 +

 2x2 1 1 5 ln x + x 2 − 1 + x 2 + c , then y ′ =  x 2 − 1 + 2 2 2 2 x2 − 1 

    2  x + x2 − 1  2  5 1 2 x2 − 1 + 2x2 1 1   + 5x = 1  4x − 2 + + x ⋅ 2 + =   2 2 2  2 2 2 x2 − 1 x + x2 − 1 2 x2 − 1  2 x −1 

(

)

1 4x2 − 2 + 2 4x2 + 5x = + 5x = 2 2 x2 − 1 4 x2 − 1

x2 x2 − 1

∫ 5 tan t ⋅ 5 sec t tan t dt = ∫ 25 sec t tan

25 = tan t sec t + ln sec t + tan t 2

(

)

2

   + 5x x 2 − 1  1

+ 5x x 2 − 25 =

x 2 − 25 dx let x = 5 sec t , then dx = 5 sec t tan t dt and

x 2 − 25 dx =

  1 1 +  2 x + x2 − 1 

(

∫

25 tan 2 t = 5 tan t . Thus,

25 sec2 t − 25 =

)

∫

∫

t dt = 25 sec t sec2 t − 1 dt = 25 sec3 t dt − 25 sec t dt

25 − 25 ln sec t + tan t + c = tan t sec t − ln sec t + tan t 2

(

)+ c

=

 x 2 − 25 x 25  x − ln + ⋅  5 5 2 5 

 x 2 25 25 x 2 x + x 2 − 25 x 2 − 25  +c = x − 25 − ln x − 25 − ln x + x 2 − 25 +c =  2 2 5 2 2 5  

+

x 25 ln 5 + c = 2 2

25 25 ln x + x 2 − 25 + c Note: ln 5 is a constant which can be included in the constant c . 2 2

 1 2x2 25 x 2 x − 25 − ln x + x 2 − 25 + c , then y ′ =  x 2 − 25 + 2 2 2 2 x 2 − 25 

Check: Let y =

 × 1 +   = g. Given

∫ =

∫

x 2 − 25 −

2 2  x 2 − 25 + x 2 25 x + x 2 − 25 1  + 0 = x − 25 + x − 25 ⋅ − = ⋅  2  2 x 2 − 25 2 x 2 − 25 x + x 2 − 25 x 2 − 25 2 x 2 − 25

2x 2 x 2 − 25

(

2 x 2 − 25

)

2 x 2 − 25

=

x 2 − 25 x 2 − 25

Check: Let y =

x 2 − 25

×

x 2 − 25

x 2 − 25

2

t dt =

36 2

=

36 − x 2 − x 2

=

(36 − x ) 2

Hamilton Education Guides

2

− 25

(x

2

)

x 2 − 25

− 25

36 − x 2 =

)

=

2 x 2 − 50 2 x 2 − 25

x 2 − 25

=

36 − 36 sin 2 t =

36 cos 2 t = 6 cos t . Thus,

36 36  1  ( t + sin t cos t ) + c  t + sin 2t  + c = 2 2  2 

 36 −1 x x 36 − x 2 36 −1 x 36 x 36 − x 2  +c = + +c sin  + c = 2 sin 6 + 2 ⋅ 36 2 6 2 

36

=

36 − 2 x 2

+

2 36 − x 2

36 − x 2

36 − x 2

(x

∫ (1 + cos 2t ) dt =

36 36 −1 x x 36 − x 2 + + c , then y ′ = sin 2 2 2 6

2 36 − x 2 =

x 2 − 25

∫ 6 cos t ⋅ 6 cos t dt = ∫ 36 cos

 36  −1 x x 36 − x 2 sin + ⋅ 2  6 6 6 

+

=

36 − x 2 dx let x = 6 sin t , then dx = 6 cos t dt and

36 − x 2 dx =

  25 1 ⋅ − 2  x + x 2 − 25 

2 36 − x 2

=

1 1−

( 6x ) 2

72 − 2 x 2 2 36 − x 2

=

⋅

1 1 + 36 − x 2 + 6 2

(

2 36 − x 2 2 36 − x 2

)

=

− 2 x2 2 36 − x 2

36 − x 2 36 − x 2

=

=

36 ⋅ 2

6 6 36 − x 2

36 − x 2 36 − x 2

⋅

36 − x 2 36 − x 2

36 − x 2

104

Advanced Integration

∫

h. Given

Solutions

dx

(9 + 36 x ) ∫ (

dx

=

3

2 2

+ x2

9 36

)

let x =

3 2

3 1 1 1  tan t = tan t , then dx = sec2 t dt and 9 + 36 x 2 = 9 + 36 ⋅  tan t  2 6 2 2 

(

2

)

1 = 9 + 36 ⋅ tan 2 t = 9 + 9 tan 2 t = 9 1 + tan 2 t = 9 sec2 t . Therefore, 4

∫ =

(

dx

)

=

i. Given

∫

(

∫

=

)

3 9 sec2 2

1 1 sin t + c = 54 54

Check: Let y =

=

1 sec 2 dt 2

∫

=

3 9 + 36 x 2 2

6x 9 + 36 x

∫

9 2 sec

x 9 9 + 36 x

2

)

81 9 + 36 x 2 ⋅ 9 + 36 x 2

2

)

1 dt 1 dt 1 = = cos t dt 54 54 sec t 2 27 sec t

∫

∫

=

)

=

)

1

(9 + 36 x )

1 2 1+ 2

9 − 4x2 =

)

(

9 9 + 36 x 2 −324 x 2

⋅x

2 9 + 36 x 2

=

1 9 + 36 x 2 2

3 3 sin t , then dx = cos t dt and 2 2

dx let x =

(

(

72 x

81 9 + 36 x 2

(9 + 36 x )⋅ (

dt

∫ 2 729 sec t

∫

+c

1

=

1 2

=

93 sec3 t

+ c , then y ′ =

81

x

t

∫2

1 sec 2 dt 2

1 ⋅ 9 9 + 36 x 2 − 9 ⋅

9 9 + 36 x 2

9 − 4x2

2

=

+c =

2

x

(

1 sec 2 dt 2 3 2× 3

(

9 + 36 x 2

81 9 + 36 x 2 =

9−4

81+ 324 x 2 −324 x 2

=

)

(

9 + 36 x 2

81 9 + 36 x 2

)

1

(9 + 36 x )

3

2 2

(32 sin t ) 2

9 − 4 ⋅ 94 sin 2 t

=

9 cos 2 t = 3 cos t . Therefore,

9 − 9 sin 2 t =

9 1 − sin 2 t

9 − 4x2

3 cos t 3 1 6 3 cos 2 t 1 − sin 2 t 1 sin 2 t ⋅ cos t dt = = = dt ⋅ dt − 3 dt = 3 dt 3 3 dt 3 sin t 2 3 2 sin t sin t sin t sin t sin t 2

x

dx =

∫

∫

=

∫

∫

∫

∫

3 − 2x

∫

− 3 sin t dt = 3 csc t dt − 3 sin t dt = 3 ln csc t − cot t + 3 cos t + c = 3 ln

= 3 ln

9 − 4x2

3−

+

2x

Check: Let y = 3 ln

3−

−

×

9 − 4x2

 8x2 − 2 ⋅  3 −  ⋅

6x

4x2

9 − 4x2

3−

(

8x2 − 6 9 − 4x2 + 2 9 − 4x2

)−

4x2 9 − 4x2   3  − 6 9 − 4 x 2 + 18   

= 3− −

−

9 − 4x2 ⋅ 2x 9 − 4x2 4x

9 − 4x2

=

9 − 4x2 x 9 − 4x2

Hamilton Education Guides

9 − 4x2

9 − 4x2

+ 3⋅

2x

3

+c

9 − 4x

 9 − 4x2  ⋅ 9 − 4x2  − 2 9 − 4x

4x 9 − 4x2

4x

  18  3 − 9 − 4 x 2   

9 − 4 x2 x 9 − 4 x2

×

 2 2  3 − 9 − 4x  ⋅ 2x 9 − 4x   9 − 4 x2 9 − 4 x2

=

 9 − 4x2  

4x2

9 − 4x2

=

2

2

  6 x  8 x 2 − 6 9 − 4 x 2 + 18 − 8 x 2   − = 2 2 2 3 − 9 − 4x ⋅ 4x 9 − 4x

4x

9 − 4x2

=

2 9− 4 x

⋅

 ⋅ 2x − 2 ⋅  3 − 

1

8x ⋅

2x

9 − 4 x 2 + c , then y ′ = 3 ⋅

+

3−

=

2 9 − 4x2

∫

9 − 4x2 + c

2x

8x

∫

(9 − 4 x ) 9 − 4 x x (9 − 4 x ) 2

2

6x

=

9 − 4x2

3−

4x 9 − 4x2 4x

−

9 − 4x2 2

=

=

9 x 9 − 4x2

9 − 4x2 x

105

Advanced Integration

Solutions

Section 1.3 Solutions – Integration by Partial Fractions a. Evaluate the integral

dx

∫ x2 + 5x + 6 .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second - Factor the denominator x 2 + 5 x + 6 into (x + 2 ) (x + 3) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way: 2

1

=

x + 5x + 6

1

( x + 2 ) ( x + 3)

B A + x+2 x+3

=

Fourth - Solve for the constants A and B by equating coefficients of the like powers.

1

2

x + 5x + 6

=

A (x + 3) + B (x + 2 ) (x + 2) (x + 3)

1 = A (x + 3) + B (x + 2 ) = Ax + 3 A + Bx + 2 B 1 = ( A + B ) x + (3 A + 2 B ) therefore, A+ B = 0

3 A + 2B = 1

which result in having A = 1 and B = −1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

dx

A

1

B

1

∫ x2 + 5x + 6 = ∫ x + 2 dx + ∫ x + 3 dx = ∫ x + 2 dx − ∫ x + 3 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. 1

1

∫ x + 2 dx − ∫ x + 3 dx = ln

x + 2 − ln x + 3 + c

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = ln x + 2 − ln x + 3 + c , then y ′ = b. Evaluate the integral

1 ( x + 3) − ( x + 2 ) = x + 3 − x − 2 = 1 1 ⋅1 − ⋅1 + 0 = (x + 2) (x + 3) x 2 + 3x + 2 x + 6 x 2 + 5 x + 6 x+2 x+3

x 2 +1

∫ x 3 − 4 x dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x3 − 4 x into x x 2 − 4 = x(x − 2 ) (x + 2 ) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way:

x2 + 1

=

3

x − 4x

x2 + 1 A B C = + + x x−2 x+2 x (x − 2 ) (x + 2 )

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.

x2 + 1

(

x3 − 4 x

=

A (x − 2 ) (x + 2 ) + Bx (x + 2 ) + Cx (x − 2 ) x (x − 2 ) (x + 2 )

) (

) (

)

x 2 + 1 = A x 2 + 2 x − 2 x − 4 + B x 2 + 2 x + C x 2 − 2 x = Ax 2 − 4 A + Bx 2 + 2 Bx + Cx 2 − 2Cx x 2 + 1 = ( A + B + C )x 2 + (2 B − 2C )x − 4 A therefore, A+ B+C =1

Hamilton Education Guides

2 B − 2C = 0

−4 A = 1

106

Advanced Integration

Solutions

which result in having A = −

5 1 5 , B = , and C = 8 4 8

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

x 2 +1

A

B

1 1

C

5

1

5

1

∫ x 3 − 4 x dx = ∫ x dx + ∫ x − 2 dx + ∫ x + 2 dx = − 4 ∫ x dx + 8 ∫ x − 2 dx + 8 ∫ x + 2 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. −

1 1 5 dx + 4 x 8

∫

1

5

1

1

∫ x − 2 dx + 8 ∫ x + 2 dx = − 4 ln

x +

5 5 ln x − 2 + ln x + 2 + c 8 8

Seventh - Check the answer by differentiating the solution. The result should match the integrand.

5 5 1 −2(x − 2 )(x + 2 ) + 5 x(x + 2 ) + 5 x(x − 2 ) 1 5 5 + + = Let y = − ln x + ln x − 2 + ln x + 2 + c , then y ′ = − 4 8 8 8 x(x − 2 )(x + 2 ) 4 x 8(x − 2 ) 8(x + 2 )

=

− 2 x 2 + 8 + 5 x 2 + 10 x + 5 x 2 − 10 x

(

8x x2 − 4

c. Evaluate the integral

)

=

8x2 + 8

(

8x x2 − 4

=

)

( ) 8 x (x − 4 ) 8 x2 + 1 2

x2 + 1

=

x3 − 4 x

1

∫ 36 − x 2 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second - Factor the denominator 36 − x 2 into (6 − x )(6 + x ) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way:

1 36 − x

2

=

1

(6 − x ) (6 + x )

=

A B + 6−x 6+ x

Fourth - Solve for the constants A and B by equating coefficients of the like powers.

1 36 − x

2

=

A (6 + x ) + B (6 − x ) (6 − x ) (6 + x )

1 = A (6 + x ) + B (6 − x ) = 6 A + Ax + 6 B − Bx 1 = ( A − B ) x + (6 A + 6 B ) therefore, 6 A + 6B = 1 which result in having A =

A− B = 0

1 1 , and B = 12 12

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

1

A

B

1

1

1

1

∫ 36 − x2 dx = ∫ 6 − x dx + ∫ 6 + x dx = 12 ∫ 6 − x dx + 12 ∫ 6 + x dx Sixth - Integrate each integral individually using integration methods learned in previous sections. 1 1 1 1 1 1 dx + dx = ln 6 − x + ln 6 + x + c 12 6 − x 12 6 + x 12 12

∫

∫

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y =

1 6+ x+6− x 1 1 1 1 1 1 = ln 6 − x + ln 6 + x + c , then y ′ = ⋅ + ⋅ +0 = 12 12 12(6 − x ) (6 + x ) 12 6 − x 12 6 + x 36 − x 2

d. Evaluate the integral

x+5

∫ x3 + 2 x2 + x dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

Hamilton Education Guides

107

Advanced Integration

Solutions

(

)

Second - Factor the denominator x3 + 2 x 2 + x into x x 2 + 2 x + 1 = x(x + 1)2 . Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand can be represented in the following way:

x+5

x3 + 2 x 2 + x

=

(

x+5

=

)

2

x x + 2x + 1

x+5

=

x(x + 1)2

A B C + + x x + 1 (x + 1)2

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.

x+5

x3 + 2 x 2 + x

(

A (x + 1)2 + Bx (x + 1) + Cx

=

) (

x(x + 1)(x + 1)2

)

x + 5 = A x 2 + 2 x + 1 + B x 2 + x + Cx = Ax 2 + 2 Ax + A + Bx 2 + Bx + Cx x + 5 = ( A + B )x 2 + (2 A + B + C )x + A therefore, 2A + B + C = 1

A+ B = 0

A=5

which result in having A = 5 , B = −5 , and C = −4 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

x+5

A

1

C

B

1

1

∫ x3 + 2 x2 + x dx = ∫ x dx + ∫ x + 1 dx + ∫ (x + 1) 2 dx = 5∫ x dx − 5∫ x + 1 dx − 4∫ (x + 1) 2 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. 5

1

1

1

∫ x dx − 5∫ x + 1 dx − 4∫ (x + 1) 2 dx = 5 ln

x − 5 ln x + 1 +

4 +c x +1

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = 5 ln x − 5 ln x + 1 + =

5(x + 1) 2 − 5 x (x + 1) − 4 x 4 1 4 1 + c , then y ′ = 5 ⋅ − 5 ⋅ +0 = − 2 x +1 x x + 1 (x + 1) x (x + 1)2

5 x 2 + 10 x + 5 − 5 x 2 − 5 x − 4 x 3

2

x + 2x + x

e. Evaluate the integral

=

x+5

x + 2x2 + x 3

1

∫ x3 − 2 x2 + x dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second – Factor the denominator x3 − 2 x 2 + x into x(x − 1)2 . Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand can be represented in the following way: 1

x(x − 1)2

=

A B C + + x x − 1 (x − 1)2

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.

1

x3 − 2 x 2 + x

(

) (

A (x − 1)2 + Bx (x − 1) + Cx

=

)

x(x − 1)2

1 = A x 2 − 2 x + 1 + B x 2 − x + Cx = Ax 2 − 2 Ax + A + Bx 2 − Bx + Cx

1 = ( A + B )x 2 + (− 2 A − B + C )x + A therefore, A+ B = 0

−2 A − B + C = 0

A =1

which result in having A = 1 , B = −1 , and C = 1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

Hamilton Education Guides

108

Advanced Integration dx

Solutions A

B

1

C

1

1

∫ x3 − 2 x2 + x = ∫ x dx + ∫ x − 1 dx + ∫ (x − 1)2 dx = ∫ x dx − ∫ x − 1 dx + ∫ (x − 1) 2 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. 1

1

1

∫ x dx − ∫ x − 1 dx + ∫ (x − 1) 2 dx

1 +c x −1

= ln x − ln x − 1 −

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = ln x − ln x − 1 − =

x2 − x2 − 2x + 2x + 1 3

=

2

x − 2x + x

f. Evaluate the integral

(x − 1)2 − x (x − 1) + x = x 2 − 2 x + 1 − x 2 + x + x 1 1 1 1 + c , then y ′ = − + +0 = 2 x −1 x x − 1 (x − 1) x3 − 2 x 2 + x x(x − 1)2 3

1

x − 2x2 + x

x2 + 3

∫ x2 − 1 dx . x2 + 3

∫ x2 − 1 dx = ∫

First – Rewrite the integral in the following form:

(x − 1)+ 4 dx = 2

2

x −1

 x2 − 1

4





4



∫  x2 − 1 + x2 − 1  dx = ∫ 1 + x 2 − 1  dx .

Then, check to see if the integrand of the second integral is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second - Factor the denominator x 2 − 1 into (x − 1) (x + 1) .

Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand can be represented in the following way:

4 x 2 −1

=

4

=

(x − 1) (x + 1)

B A + x −1 x +1

Fourth - Solve for the constants A and B by equating coefficients of the like powers.

4 x 2 −1

=

A (x + 1) + B (x − 1) (x − 1) (x + 1)

4 = A (x + 1) + B (x − 1) = Ax + A + Bx − B 4 = ( A + B )x + ( A − B ) therefore,

A− B = 4

A+ B = 0

which result in having A = 2 and B = −2 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

x2 + 3



4

4



A

B

1

1

∫ x2 − 1 dx = ∫ 1 + x2 − 1  dx = ∫ dx + ∫ x2 − 1 dx = ∫ dx + ∫ x − 1 dx + ∫ x + 1 dx = ∫ dx + 2∫ x − 1 dx − 2∫ x + 1 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. 1

1

∫ dx + 2∫ x − 1 dx − 2∫ x + 1 dx =

x + 2 ln x − 1 − 2 ln x + 1 + c

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = x + 2 ln x − 1 − 2 ln x + 1 + c , then y ′ = 1 + 2 ⋅ =

(x

2

)

+ x − x − 1 + 2x + 2 − 2x + 2

(x

2

)

+ x − x −1

g. Evaluate the integral

=

x2 − 1 + 4 2

x −1

=

(x − 1) (x + 1) + 2(x + 1) − 2(x − 1) 1 1 − 2⋅ +0 = (x − 1) (x + 1) x −1 x +1

x2 + 3 x2 − 1

1

∫ x3 − 1 dx .

Hamilton Education Guides

109

Advanced Integration

Solutions

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x3 − 8 into (x − 1) x 2 + x + 1 . Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way:

1

3

x −1

1

=

=

(x − 1) (x 2 + x + 1)

A Bx + C + 2 x −1 x + x +1

Fourth - Solve for the constants A , B and C by equating coefficients of the like powers.

1

=

3

x −1

(

(

) (x − 1) (x

A x 2 + x + 1 + (Bx + C ) (x − 1)

)

2

)

+ x +1

1 = A x 2 + x + 1 + (Bx + C ) (x − 1) = Ax 2 + Ax + A + Bx 2 − Bx + Cx − C 1 = ( A + B ) x 2 + ( A − B + C )x + ( A − C ) therefore, A+ B = 0 which result in having A =

A− B+C = 0

A−C =1

1 2 1 , B = − , and C = − 3 3 3

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

1

∫ x3 − 1

dx =

∫

A Bx + C 1 1 dx + 2 dx = dx + x −1 3 x −1 x + x +1

∫

∫

− 13 x −

2 3

1

1

1

x+2

∫ x2 + x + 1 dx = 3 ∫ x − 1 dx − 3 ∫ x2 + x + 1 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. To solve the second integral let u = x 2 + x + 1 , then x+2 =

1 (2 x + 1) + 3 . Therefore, 2 2

1 1 1 x+2 1 1 1 dx − dx = dx − 3 x −1 3 x2 + x + 1 3 x −1 3

∫

=

=

∫

∫

(2 x + 1) + 32 x2 + x + 1

dx =

1 1 1 dx − 3 x −1 6

∫

2x + 1

∫ x2 + x + 1

1 1 1 2 x + 1 du 1 1 1 1 1 1 2x + 1 ⋅ − dx − dx − dx − dx = 2 2 3 x −1 6 2x + 1 2 u 2 x + x +1 3 x −1 6 x + x +1

∫

∫

∫

∫

1 1 1 1 ln x − 1 − ln u − ⋅ tan −1 3 6 2 3 2

=

∫

1 2

du du . Also, x + 2 can be rewritten as = 2 x + 1 and dx = 2x + 1 dx

2 x +1 2 3 2

+c =

∫

dx −

3

1 2 dx 3 x2 + x + 1

∫

1

∫ (x + 1 ) 2 + 3 dx 2

4

1 1 1 2 2(2 x + 1) +c ln x − 1 − ln x 2 + x + 1 − ⋅ tan −1 3 6 2 3 2 3

1 1 1 2x + 1 1 1 3 2x + 1 ln x − 1 − ln x 2 + x + 1 − tan −1 tan −1 +c + c = ln x − 1 − ln x 2 + x + 1 − 3 6 3 6 3 3 3 3

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = y′ =

−

1 1 3 2x + 1 ln x − 1 − ln x 2 + x + 1 − tan −1 + c , then 3 6 3 3

3 1 1 1 1 − ⋅ ⋅ (2 x + 1) − ⋅ ⋅ 3 3 x − 1 6 x2 + x + 1

1 1 +  2 x +1   3 

2

⋅

( 2 ⋅ 3 )− 0 ⋅ (2 x + 1) + 0 = 1 − 2 x + 1 3(x − 1) 6(x + x + 1) ( 3) 2

2

1 2x + 1 2 3 3 2 3 6 x 2 + 6 x + 6 − (2 x + 1) ⋅ (3 x − 3) 2 = = − − ⋅ ⋅ − 2 2 2 3(x − 1) 6 x 2 + x + 1 3 3 + (2 x + 1) 3 3 + 4x + 4x + 1 3 + (2 x + 1) 18(x − 1) x 2 + x + 1

Hamilton Education Guides

(

)

(

)

110

Advanced Integration

=

Solutions

6 x 2 + 6 x + 6 − 6 x 2 + 6 x − 3x + 3

) (x + 1) (x + x + 1) − (x − 1) = = 2(x − 1)(x + x + 1) (

−

3

18 x − 1

2

3

3

=

9x + 9

(

3

)

18 x − 1

x3 + x 2 + x + x 2 + x + 1 − x3 + 1

(x − 1)(2 x 3

2

h. Evaluate the integral

2

4x2 + 4x + 4

2

+ 2x + 2

)

−

1

(

=

)

2

4 x + x +1

(

x +1 3

(x − 1)(2 x 3

2

)

2 x −1

2x2 + 2x + 2

=

−

+ 2x + 2

)

=

(

1 2

)

2 x + x +1

1

3

x −1

1

∫ x4 − 1 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)(

)

(

)

Second - Factor the denominator x 4 − 1 into x 2 − 1 x 2 + 1 = (x − 1) (x + 1) x 2 + 1 . Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way:

1

x4 − 1

=

1

(x − 1) (x + 1) (x

2

=

)

+1

A B Cx + D + + x − 1 x + 1 x2 + 1

Fourth - Solve for the constants A , B , C and D by equating coefficients of the like powers.

1

4

x −1

=

(

(

)

)

A(x + 1) x 2 + 1 + B (x − 1) x 2 + 1 + (x − 1) (x + 1) (Cx + D )

(

(x − 1) (x + 1) (x 2 + 1)

(

)

)

1 = A(x + 1) x 2 + 1 + B(x − 1) x 2 + 1 + (x − 1)(x + 1)(Cx + D ) 1 = Ax3 + Ax 2 + Ax + A + Bx3 − Bx 2 + Bx − B + Cx3 + Dx 2 − Cx − D 1 = ( A + B + C )x3 + ( A − B + D )x 2 + ( A + B − C )x + ( A − B − D ) therefore, A+ B+C = 0

A− B+ D = 0

A− B− D =1

A+ B−C = 0

1 1 1 which result in having A = , B = − , C = 0 , and D = − 4 2 4

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

1

A

B

∫ x4 − 1 dx = ∫ x − 1 dx + ∫ x + 1 dx + ∫

Cx + D 2

x +1

dx =

1 1 1 1 1 1 dx − dx − dx 4 x −1 4 x +1 2 x2 + 1

∫

∫

∫

Sixth - Integrate each integral individually using integration methods learned in previous sections.

1 1 1 1 1 1 1 1 1 dx − dx − dx = ln x − 1 − ln x + 1 − tan −1 x + c 4 x −1 4 x +1 2 x2 + 1 4 4 2

∫

∫

∫

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = =

1 1 1 1 1 1 1 1 1 ⋅1 − ⋅ ⋅1 − ⋅ 2 ⋅1 + 0 ln x − 1 − ln x + 1 − tan −1 x + c , then y ′ = ⋅ 4 4 2 4 x −1 4 x +1 2 x +1

1 1 1 − − 4(x − 1) 4(x + 1) 2 x 2 + 1

(

2

− 2x − 2x + 2x + 2

(

2

)(

4 x −1 4 + x

2

)

i. Evaluate the integral

=

=

)

(

4 4

)

4 x −1

=

(x + 1) (x 2 + 1) − (x − 1) (x 2 + 1) − 2(x − 1) (x + 1) 4(x − 1) (x + 1) (x 2 + 1)

=

x3 + x + x 2 + 1 − x3 − x + x 2 + 1

(

)(

4 x2 − 1 4 + x2

)

1

x4 − 1

1

∫ x3 + 64 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is a rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x3 + 64 into (x + 4 ) x 2 − 4 x + 16 .

Hamilton Education Guides

111

Advanced Integration

Solutions

Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way:

1

x3 + 64

=

(x + 4) (x

1 2

− 4 x + 16

=

)

A Bx + C + x + 4 x 2 − 4 x + 16

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. 3

1

x + 64

(

(

)

A x 2 − 4 x + 16 + (Bx + C ) (x + 4 )

=

)

(x + 4) (x 2 − 4 x + 16)

1 = A x 2 − 4 x + 16 + (Bx + C )(x + 4 ) = Ax 2 − 4 Ax + 16 A + Bx 2 + 4 Bx + Cx + 4C 1 = ( A + B )x 2 + (− 4 A + 4 B + C )x + (16 A + 4C ) therefore, −4 A + 4 B + C = 0

A+ B = 0 which result in having A =

16 A + 4C = 1

1 1 1 , B=− , and C = 48 48 6

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

1

∫ x3 + 64

dx =

∫

A Bx + C 1 1 dx + 2 dx = dx + x+4 48 x + 4 x − 4 x + 16

∫

∫

−

1 48

x+

1 6

1

1

1

x −8

∫ x2 − 4 x + 16 dx = 48 ∫ x + 4 dx − 48 ∫ x2 − 4 x + 16 dx

Sixth - Integrate each integral individually using integration methods learned in previous du du . Also, x − 8 can be = 2 x − 4 and dx = 2x − 4 dx

sections. To solve the second integral let u = x 2 − 4 x + 16 , then rewritten as x − 8 = (x − 2 ) − 6 =

1 (2 x − 4) − 6 . Therefore, 2

(x − 2) − 6 dx = 1 1 dx − 1 1 1 1 1 1 1 x −8 dx − dx − dx = 48 x + 4 48 x + 4 48 x 2 − 4 x + 16 48 x 2 − 4 x + 16 48 x + 4 48

∫

∫

∫

∫

(2 x − 4) − 6

∫ x2 − 4 x + 16 dx

=

1 1 1 dx − 48 x + 4 96

=

1 48

=

1 1 12 x−2 1 1 12 x−2 ln x + 4 − ln x 2 − 4 x + 16 + tan −1 +c = ln x + 4 − ln x 2 − 4 x + 16 − tan −1 +c 48 96 8 ⋅ 12 48 96 96 12 12

∫

1

1

2x − 4

∫

1 2

1

6

1

1

1

∫ x2 − 4 x + 16 dx + 48 ∫ x2 − 4 x + 16 dx = 48 ∫ x + 4 dx − 96 ∫ 1

1

1

∫ x + 4 dx − 96 ∫ u du + 8 ∫ (x − 2)2 + 12 dx =

6 2 x − 4 du + ⋅ u 2 x − 4 48

1

∫ (x − 2)2 + 12 dx

1 1 1 x−2 ln x + 4 − ln u + tan −1 +c 48 96 8 12 12

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y =

y′ =

x−2 1 12 1 + c , then tan −1 ln x + 4 − ln x 2 − 4 x + 16 + 96 48 96 12

1 1 1 1 12 ⋅ − ⋅ ⋅ (2 x − 4 ) + ⋅ 48 x + 4 96 x 2 − 4 x + 16 96

1 1 +  x − 2   12 

2

⋅

(1 ⋅ 12 )− 0 ⋅ (x − 2) + 0 = 1 − 48(x + 4 ) 48(x ( 12 ) 2

+

x−2 1 6 12 12 12 = − + ⋅ ⋅ 2 48(x + 4 ) 48 x 2 − 4 x + 16 96 12 + (x − 2 )2 12 48 x − 4 x + 16

=

1 −x + 8 + 48(x + 4 ) 48 x 2 − 4 x + 16

=

(

16 + 32

(

48(x + 4 ) x 2 − 4 x + 16

)

Hamilton Education Guides

)

=

=

(x

(

2

) 48(x + 4 ) (x

)

(

− 4 x + 16 + (− x + 8)(x + 4 )

(

2

− 4 x + 16

48

48(x + 4 ) x 2 − 4 x + 16

)

=

)

=

3

2

−x + 2 + 6 1 + 48(x + 4 ) 48 x 2 + 4 x + 16

=

)

x−2 2

(

− 4 x + 16

)

)

x 2 − 4 x + 16 − x 2 − 4 x + 8 x + 32

(

48(x + 4 ) x 2 − 4 x + 16

1

2

x − 4 x + 16 x + 4 x − 16 x + 64

=

)

3

1

x + 64

112

Advanced Integration

Solutions

Section 1.4 Practice Problems – Integration of Hyperbolic Functions 1. Evaluate the following integrals: a. Given

∫ cosh 3x dx let u = 3x , then

∫ cosh 3x dx = ∫ cosh u ⋅

1 du 1 1 = cosh u du = sinh u + c = sinh 3 x + c 3 3 3 3

∫

1 d 1 1 1 d d sinh 3 x + c , then y ′ = ⋅ sinh 3 x + c = ⋅ cosh 3 x ⋅ 3 x + 0 = ⋅ cosh 3 x ⋅ 3 = cosh 3 x 3 3 dx 3 3 dx dx

Check: Let y = b. Given

du d du . Therefore, = 3 x = 3 which implies dx = dx dx 3

3x 3x ∫ (sinh 2 x − e ) dx = ∫ sinh 2 x dx + ∫ e dx

let:

a. u = 2 x , then

du du d du and = 2 ; du = 2dx ; dx = = 2x ; dx dx dx 2

b. v = 3 x , then

dv dv d dv . = 3x ; = 3 ; dv = 3dx ; dx = dx dx 3 dx

∫ sinh 2 x dx + ∫ e

Therefore, =

3x

dx =

∫ sinh u ⋅

1 1 v 1 1 du dv = sinh u du + + e3 x ⋅ e dv = cosh u + c1 + ev + c2 2 3 2 3 2 3

∫

∫

∫

1 1 1 1 cosh 2 x + e3 x + c1 + c2 = cosh 2 x + e 3 x + c 3 2 3 2 1 1 1 d 1 d d 1 1 d d cosh 2 x + e3 x + c then y ′ = ⋅ cosh 2 x + ⋅ e3 x + c = ⋅ sinh 2 x ⋅ 2 x + ⋅ e3 x ⋅ 3 x + 0 2 3 2 dx 3 dx dx 3 2 dx dx

Check: Let y = = c. Given

1 1 2 3 ⋅ sinh 2 x ⋅ 2 + ⋅ e3 x ⋅ 3 = ⋅ sinh 2 x + ⋅ e3 x = sinh 2 x + e3 x 2 3 2 3

∫ csc h 5x dx let u = 5x , then

∫ csc h 5x dx = ∫ csc h u ⋅

du d du . Therefore, 5 x = 5 which implies du = 5dx ; dx = = 5 dx dx

1 du 1 5x 1 u = csc h u du = ln tanh +c + c = ln tanh 5 5 5 2 5 2

∫

1 1 1 5x d 5x 1 d ln tanh + c , then y ′ = ⋅ ln tanh c = ⋅ + 5 2 dx 2 5 dx 5 tanh

Check: Let y =

1 1 = ⋅ 5 tanh

5x 2

1 5x 5   ⋅ ⋅  sec h 2 ⋅ +0 = 10 2 2 

5 sec h 2 52x tanh 52x

1 = ⋅ 2

sec h 2 52x tanh 52x

1 = 2

5x 2

⋅

d  5x  +0  tanh dx  2 

1 − tanh 2 52x ⋅ tanh 52x

1 = ⋅ 2

1−

sinh 2 5 x 2

cosh 2 5 x

sinh 5 x

2

2

cosh 5 x 2

cosh 2 5 x − sinh 2 5 x 2

2

cosh 2 5 x

=

2⋅

d. Given

∫x

2

∫x

2

2 sinh 5 x 2 cosh 5 x 2

Check: Let y = 2

2⋅

=

2 sinh 5 x 2 cosh 5 x 2

∫3x

3

∫x

2

sec h 2u ⋅

du 3x

2

=

5x

cosh 2 1 1 1 1 = = = = csc h5 x ⋅ 5 x 5 x 5 x sinh 5x 2 cosh 2 5 x ⋅ sinh 5 x 2 cosh 2 ⋅ sinh 2 sinh 2 ⋅ 2 2

2

sec h 2 x3dx let u = x3 , then

sec h 2 x3dx =

e. Given

=

1 cosh 2 5 x

du du d 3 du = x ; = 3x 2 ; du = 3 x 2 dx ; dx = 2 . Therefore, dx dx dx 3x

1 1 1 sec h 2u du = tanh u + c = tanh x 3 + c 3 3 3

∫

1 1 d 3 1 1 d d tanh x3 + c , then y ′ = ⋅ tanh x3 + c = ⋅ sec h 2 x3 ⋅ x + 0 = ⋅ sec h 2 x3 ⋅ 3 x 2 = x 2 sec h 2 x3 3 3 dx dx 3 3 dx

(

)

csc h 2 x 4 + 1 dx let u = x 4 + 1 , then

Hamilton Education Guides

(

)

du du d 4 du = x +1 ; = 4x3 ; du = 4 x3dx ; dx = 3 . Therefore, dx dx dx 4x

113

Advanced Integration 2

∫3x

3

(

Solutions

)

(

)

(

) (

2 3 du 1 1 1 x csc h 2u ⋅ 3 = csc h 2u du = − coth u + c = − coth x 4 + 1 + c 3 6 6 6 4x

∫

csc h 2 x 4 + 1 dx =

(

∫

)

(

)

)

1 1 d d 1 d 4 Check: Let y = − coth x 4 + 1 + c , then y ′ = − ⋅ coth x 4 + 1 + c = − ⋅ − csc h 2 x 4 + 1 ⋅ x +1 + 0 6 dx dx 6 dx 6

(

f. Given

∫

(

)

)

(

)

4 x3 1 2 ⋅ csc h 2 x 4 + 1 = x3 csc h 2 x 4 + 1 ⋅ csc h 2 x 4 + 1 ⋅ 4 x3 = 6 3 6

=

∫x

3

(

)

csc h 2 x 4 + 5 dx let u = 2 x 4 + 5 , then

(

)

x3 csc h 2 x 4 + 5 dx =

du

=

8 x3

1 8

∫

)

(

)

csc h u du =

2 x4 + 5 1 1 u +c ln tanh + c = ln tanh 2 8 8 2

d 1 1 2x4 + 5 1 1 d 2x4 + 5 d 2x4 + 5 ⋅ tanh +0 ln tanh + c = ⋅ + c , then y ′ = ⋅ ln tanh 4 8 tanh 2 x + 5 dx 2 8 8 dx 2 dx 2 2

Check: Let y =

=

∫

x3 csc h u ⋅

(

du du du d 2x4 + 5 ; = 2 x 4 + 5 ; du = 8 x3dx ; dx = 3 . Therefore, = dx dx dx 8x

1 2 x 4 +5 2

8 tanh

sec h 2 2x4 + 5 d 2x4 + 5 ⋅ sec h 2 ⋅ +0 = dx 2 2 8 tanh

2 x 4 +5 2 2 x 4 +5 2

4

2 2 x +5 8 x3 x3 1 − tanh x3 2 ⋅ ⋅ ⋅ = = 4 2 2 2 tanh 2 x 2 + 5

1−

4 sinh 2 2 x + 5 2

4 cosh 2 2 x + 5 2

4 sinh 2 x + 5 2

4 cosh 2 x + 5 2

x3 ⋅ = 2

4 4 cosh 2 2 x + 5 − sinh 2 2 x + 5 2 2 2 2 x 4 +5

cosh

sinh

2 2 x 4 +5 2

x3 = ⋅ 2

4 cosh 2 x + 5 2

=

sinh 2 ⋅

∫ cosh

g. Given

∫ cosh

x

7

7

2 x 4 +5 2

=

(

=

2

4

cosh 2 x 2 + 5 x3 x3 ⋅ = = 4 4 2 cosh 2 2 x 4 + 5 ⋅ sinh 2 x 4 + 5 2 cosh 2 x 2 + 5 ⋅ sinh 2 x 2 + 5 2 2

2

x

(

3 4

sinh 2 x + 5

( x + 1) sinh ( x + 1)dx

sinh

2 2 x 4 +5

4 cosh 2 x + 5

( x + 1) sinh ( x + 1)dx

Check: Let y = h. Given

3

1 4 cosh 2 2 x + 5

∫u

7

= x3 csc h 2 x 4 + 5

)

)

let u = cosh (x + 1) , then sinh (x + 1) ⋅

du = sinh (x + 1)

du du d du . Thus, = cosh (x + 1) ; = sinh (x + 1) ; dx = sinh (x + 1) dx dx dx

∫ u du 7

=

1 1 8 u + c = cosh 8 ( x + 1) + c 8 8

1 1 cosh8 (x + 1) + c , then y ′ = ⋅ 8 cosh 7 (x + 1) ⋅ sinh (x + 1) + 0 = cosh 7 (x + 1) sinh (x + 1) 8 8

∫ csc h (5x + 3) coth (5x + 3) dx let u = 5x + 3 , then

∫ csc h (5x + 3) coth (5x + 3) dx = ∫ csc h u coth u

du d ( 5 x + 3) ; du = 5 ; du = 5dx ; dx = du . Therefore, = dx dx 5 dx

1 du 1 1 = csc h u coth u du = − csc h u + c = − csc h (5 x + 3 ) + c 5 5 5 5

∫

1 d 1 csc h (5 x + 3) coth (5 x + 3) Check: Let y = − csc h (5 x + 3) + c , then y ′ = − ⋅ − csc h (5 x + 3) coth (5 x + 3) ⋅ (5 x + 3) + 0 = ⋅5 5 5 5 dx

i. Given

∫e

=

5 csc h (5 x + 3) coth (5 x + 3) = csc h (5 x + 3) coth (5 x + 3) 5

∫e

x +1 sec h e x +1 dx let u = e x +1 , then du = d e x +1 ; du = e x +1 ; du = e x +1 ⋅ dx ; dx = du . Therefore, dx dx dx e x +1

x +1 sec h e x +1 dx =

(

∫e

x +1

sec h u ⋅

)

du e

x +1

=

Check: Let y = sin −1 tanh e x +1 + c , then y ′ =

Hamilton Education Guides

∫ sec h u du = sin 1 1 − tanh 2 e x +1

−1

⋅

(tanh u ) + c

(

)

= sin −1 tanh e x +1 + c

d tanh e x +1 + 0 = dx

sec h 2e x +1 sec h 2e x +1

⋅

d x +1 e dx

114

Advanced Integration

Solutions

sec h 2e x +1

=

sec h e x +1

⋅ e x +1 = e x +1 sec h e x +1

2. Evaluate the following integrals: a.

∫ tanh ∫ tanh

5

x sec h 2 x dx let u = tanh x , then

5

x sec h 2 x dx =

Check: Let y = b. Given

∫ coth

6

∫u

5

du d du du . Thus, = sec h 2 x ; du = sec h 2 x dx ; dx = tanh x ; = dx dx dx sec h 2 x

du

⋅ sec h 2 x ⋅

sec h 2 x

1

∫ u du = 5 + 1 u 5

=

5 +1

+c =

1 1 6 u + c = tanh 6 x + c 6 6

1 1 tanh 6 x + c then y ′ = ⋅ 6 (tanh x ) 6 −1 ⋅ sec h 2 x + 0 = (tanh x ) 5 sec h 2 x = tanh 5 x sec h 2 x 6 6

(x + 1) csc h 2 ( x + 1) dx

; du = − csc h 2 (x + 1) dx ; dx = −

let u = coth (x + 1) , then du

csc h 2 (x + 1)

. Thus,

∫ coth

6

d du du = coth (x + 1) ; = − csc h 2 (x + 1) c dx dx dx

(x + 1) csc h 2 ( x + 1) dx

=

∫u

6

⋅ csc h 2 (x + 1) ⋅

− du

csc h 2 (x + 1)

1 1 = − u 6 du = − u 7 + c = − coth7 ( x + 1) + c 7 7

∫

1 1 Check: Let y = − coth 7 (x + 1) + c then y ′ = − ⋅ 7[ coth (x + 1) ] 7 −1 ⋅ − csc h 2 (x + 1) + 0 = coth 6 (x + 1) csc h 2 (x + 1) 7 7

c. Given

∫e

3x

∫e

3x

tanh e3 x dx let u = e3 x , then

Check: Let y = d. Given

∫x

3

∫e

tanh e3 x dx =

∫x

3

3x

(

)

)

∫x

(

(

3

⋅ sec h u ⋅

1 1 1 tanh u du = ln cosh u + c = ln cosh e 3 x + c 3 3 3

∫

)

[ (

du 4x

3

=

( (

)

[

)] sec h (x + 1) 1 d ⋅ tanh (x + 1) + 0 = dx 1 − tanh (x + 1) 4 sec h (x + 1) (

1 1 1 sec h u ⋅ du = sin −1 (tanh u ) + c = sin −1 tanh x 4 + 1 + c 4 4 4

∫

)]

2

4

4

) )

2

4

2

4

4

sec h 2 x 4 + 1 4 x3 d 4 ⋅ 4 x3 = ⋅ sec h x 4 + 1 = x3 sec h x 4 + 1 x +1 = 4 4 dx 4 sec h x + 1

(

)

(

∫ sec h ( 3x + 2) dx let u = 3x + 2 , then

Check: Let y =

×

(

du du d 4 du x +1 ; = = 4x3 ; du = 4 x3dx ; dx = 3 . Thus, dx dx dx 4x

1 −1 sin tanh x 4 + 1 + c then y ′ = 4

∫ sec h ( 3x + 2) dx = ∫ sec h u ⋅

f. Given

=

sec h x 4 + 1 dx let u = x 4 + 1 , then

Check: Let y =

e. Given

du

3e3 x

3 1 1 1 ln cosh e3 x + c , then y ′ = ⋅ ⋅ sinh e3 x ⋅ 3e3 x + 0 = e3 x tanh e3 x = e3 x tanh e3 x 3 3 3 cosh e3 x

sec h x 4 + 1 dx =

×

tanh u ⋅

du d 3 x du du = 3e3 x ; du = 3e3 x dx ; dx = 3 x . Therefore, e ; = dx dx dx 3e

)

(

)

du d ( 3x + 2) ; du = 3 ; du = 3dx ; dx = du . Thus, = dx 3 dx dx

1 1 du 1 = sec h u du = sin −1 (tanh u ) + c = sin −1 [ tanh (3 x + 2 ) ] + c 3 3 3 3

∫

1 −1 sin [ tanh (3 x + 2 ) ] + c , then y ′ = 3

1 3 1 − tanh 2 (3 x + 2 )

⋅

d tanh (3 x + 2 ) + 0 = dx

sec h 2 (3 x + 2 ) 3 sec h 2 (3 x + 2 )

2 2 d (3x + 2) = sec h (3x + 2) ⋅ 3 = 3 ⋅ sec h (3x + 2) = sec h (3x + 2) dx 3 sec h (3 x + 2 ) 3 sec h (3 x + 2 )

∫e

cosh (3 x + 5 )

sinh (3 x + 5) dx let u = cosh (3 x + 5) , then

Hamilton Education Guides

du d du d = sinh (3 x + 5) ⋅ (3 x + 5) = cosh (3 x + 5) ; dx dx dx dx

115

Advanced Integration

;

Solutions

du du . Therefore, = sinh (3 x + 5) ⋅ 3 ; dx = dx 3 sinh (3 x + 5)

∫e

cosh (3 x + 5 )

sinh (3 x + 5) dx =

Check: Let y =

∫e

u

sinh (3 x + 5) ⋅

du 1 u 1 1 = e du = eu + c = e cosh (3 x +5 ) + c 3 3 3 sinh (3 x + 5) 3

∫

d 1 3 1 cosh (3 x +5 ) e + c , then y ′ = ⋅ ecosh (3 x + 5 ) ⋅ sinh (3 x + 5) ⋅ (3 x + 5) + 0 = ⋅ ecosh (3 x + 5 ) ⋅ sinh (3 x + 5) dx 3 3 3

= ecosh (3 x + 5 ) sinh (3 x + 5) g.

∫ tanh

5

x dx =

∫ tanh

3

3 2 3 2 3 ∫ tanh x (1 − sec h x )dx = − ∫ tanh x sec h x dx + ∫ tanh x dx . To solve the first

x tanh 2 x dx =

du d du du . Therefore, = sec h 2 x ; du = sec h 2 x dx ; dx = = tanh x ; dx dx dx sec h 2 x

integral let u = tanh x , then

du

∫

∫

− tanh 3 x sec h 2 x dx = − u 3 sec h 2 x ⋅ letter d, we found that

∫ tanh

∫ tanh

x tanh 2 x dx =

= −

5

x dx =

∫ tanh

3

3

sec h 2 x

x dx = −

1 1 = − u 3du = − u 4 + c = − tanh 4 x + c . In Example 5.4-6, problem 4 4

∫

1 tanh 2 x + ln cosh x + c . Therefore, 2

3 2 3 3 2 ∫ tanh x (1 − sec h x ) dx = − ∫ tanh x sec h x dx + ∫ tanh x dx

1 1 1 1   1 tanh 4 x + tanh 3 x dx = − tanh 4 x +  − tanh 2 x + ln cosh x + c  = − tanh 4 x − tanh 2 x + ln cosh x + c 4 4 2 4   2

∫

4 2 sinh x 1 1 Check: Let y = − tanh 4 x − tanh 2 x + ln cosh x + c , then y ′ = − tanh 3 x ⋅ sec h 2 x − tanh x ⋅ sec h 2 x + +0 4 2 4 2 cosh x

(

(

)

)(

= − tanh 3 x sec h 2 x − tanh x sec h 2 x + tanh x = − sec h 2 x tanh 3 x + tanh x + tanh x = − 1 − tanh 2 x tanh 3 x + tanh x

(

)

)

+ tanh x = − tanh 3 x + tanh x − tanh 5 x − tanh 3 + tanh x = − tanh 3 x − tanh x + tanh 5 x + tanh 3 + tanh x = tanh 5 x h.

∫ coth

5

x dx =

∫ coth

3

x coth 2 x dx =

du d du du . Thus, = − csc h 2 x ; du = − csc h 2 x dx ; dx = − = coth x ; dx dx dx csc h 2 x

integral let u = coth x , then

∫ coth

3 x csc h 2 x dx =

∫u

3 2 3 3 2 ∫ coth x (1 + csc h x ) dx = ∫ coth x csc h x dx + ∫ coth x dx . To solve the first

3

csc h 2 x ⋅ −

letter g, we found that

∫ coth

x dx = −

∫ coth

x coth 2 x dx =

= −

5

x dx =

∫ coth

3

3

du csc h 2 x

1 1 = − u 3du = − u 4 + c = − coth 4 x + c . In example 5.4-6, problem 4 4

∫

1 coth 2 x + ln sinh x + c . Grouping the terms together we have 2

1 3 2 3 3 2 4 3 ∫ coth x (1 + csc h x )dx = ∫ coth x csc h x dx + ∫ coth x dx = − 4 coth x + ∫ coth x dx

1 1 1  1  coth 4 x +  − cot 2 x + ln sinh x + c  = − coth4 x − coth 2 x + ln sinh x + c 4 2 4 2  

− 4 ⋅ coth 3 x ⋅ − csc h 2 x 2 ⋅ coth x ⋅ − csc h 2 x cosh x 1 1 Check: Let y = − coth 4 x − coth 2 x + ln sinh x + c , then y ′ = − + +0 4 2 sinh x 4 2

(

(

)

)(

)

= coth 3 x csc h 2 x + coth x csc h 2 x + coth x = csc h 2 x coth 3 x + coth x + coth x = coth 2 x − 1 coth 3 x + coth x + coth x = coth 5 x + coth 3 x − coth 3 x − coth x + coth x = coth 5 x i.

∫ coth

6

x dx =

∫ coth

4

x coth 2 x dx =

∫

4 2 4 2 4 4 2 ∫ coth x (1 + csc h x ) dx = ∫ ( coth x csc h x + coth x ) dx = ∫ coth x csc h x dx .

+ coth x dx . In example 5.4-6, problem letter e, we found that

Hamilton Education Guides

1

∫ coth x dx = − 3 coth 4

3

x − coth x + x + c . Therefore,

116

Advanced Integration

∫ coth

6

x dx =

Solutions

∫ coth

4

x coth 2 x dx =

4 2 4 2 4 4 2 ∫ coth x (1 + csc h x ) dx = ∫ coth x csc h x dx + ∫ coth x dx = ∫ coth x csc h x dx

1 1  1  +  − coth 3 x − coth x + x + c  = − coth5 x − coth 3 x − coth x + x + c 3 3 5  

5 ⋅ coth 4 x ⋅ − csc h 2 x 3 ⋅ coth 2 x ⋅ − csc h 2 x 1 1 Check: Let y = − coth 5 x − coth 3 x − coth x + x + c , then y ′ = − − 5 3 3 5

(

)

+ csc h 2 x + 1 + 0 = coth 4 x ⋅ csc h 2 x + coth 2 x ⋅ csc h 2 x + csc h 2 x + 1 = csc h 2 x coth 4 x + coth 2 x + 1 + 1

(

)(

)

= coth 2 x − 1 coth 4 x + coth 2 x + 1 + 1 = coth 6 x + coth 4 x + coth 2 x − coth 4 x − coth 2 x − 1 + 1 = coth 6 x

Hamilton Education Guides

117

About the Author Dan Hamilton received his B.S. degree in Electrical Engineering from Oklahoma State University and Master's degree, also in Electrical Engineering from the University of Texas at Austin. He has taught a number of math and engineering courses as a visiting lecturer at the University of Oklahoma, Department of Mathematics, and as a faculty member at Rose State College, Department of Engineering Technology, at Midwest City, Oklahoma. He is currently working in the field of aerospace technology and has published several books and numerous technical papers.

Hamilton Education Guides

118