integration by substitution 3

8/8/2019 Integration by Substitution 3

1/20

Integration

SUBSTITUTION III .. [f(x)]n f(x)

Graham S McDonaldand Silvia C Dalla

A self-contained Tutorial Module for practis-ing the integration of expressions of the form

[f(x)]n

f(x), where n = 1

q Table of contents

q Begin Tutorial

c 2004 [email protected]
http://www.cse.salford.ac.uk/http://www.cse.salford.ac.uk/mailto:[email protected]:[email protected]://www.cse.salford.ac.uk/http://www.cse.salford.ac.uk/


2/20

Table of contents

1. Theory

2. Exercises

3. Answers

4. Standard integrals

5. Tips

Full worked solutions


3/20

Section 1: Theory 3

1. Theory

Consider an integral of the form

[f(x)]nf(x)dx

Letting u = f(x) givesdu

dx= f(x) and du = f(x)dx

[f(x)]nf(x)dx =

undu =

un+1

n + 1+ C =

[f(x)]n+1

n + 1+ C

For example, when n = 1,

f(x)f(x)dx =

u du =

u2

2+ C =

[f(x)]2

2+ C

Toc Back
http://lastpage/http://prevpage/http://goback/http://goback/http://prevpage/http://lastpage/


4/20

Section 2: Exercises 4

2. Exercises

Click on EXERCISE links for full worked solutions (10 exercises in

total).

Perform the following integrations:

Exercise 1.

sin x cos x dx

Exercise 2.

sinh x cosh x dxExercise 3.

tan x sec2 x dx

q Theory q Standard integrals q Answers q Tips

Toc Back


5/20


Exercise 4.

sin2x cos2x dxExercise 5.

sinh3x cosh3x dx

Exercise 6.1

xln x dx , x > 0

Exercise 7.

sin

4

x cos x dx

Exercise 8.sinh3 x cosh x dx


Toc Back


6/20


Exercise 9.

cos3 x sin x dxExercise 10.

2x

(x2 4)2 dx


Toc Back


7/20

Section 3: Answers 7

3. Answers

1. 1

2

sin2 x + C,

2. 12 sinh2 x + C,

3. 12 tan2 x + C,

4. 14 sin2 2x + C,

5. 16 sinh2 3x + C,

6. 12 ln2 x + C

7. sin5x

5 + C,

8.

1

4 sinh

4

x + C,9. cos4 x4 + C,

10. 1x24 + C.

Toc Back


9/20

Section 4: Standard integrals 9

f(x)

f(x) dx f(x)

f(x) dx

1a2+x2

1a tan

1 xa

1a2x2

1

2a lna+xax (0 < |x|< a)

(a > 0) 1x2a2

12a ln

xax+a

(|x| > a > 0)

1a2x2 sin

1 xa

1a2+x2 ln

x+a2+x2a

(a > 0)

(a < x < a) 1x2a2 ln

x+x2a2a

(x > a > 0)

a2 x2 a22

sin1x

a

a2+x2 a22 sinh1 xa + xa2+x2a2

+xa2x2a2

x2a2 a22

cosh1 x

a

+ x

x2a2a2

Toc Back


10/20

Section 5: Tips 10

5. Tips

q

STANDARD INTEGRALS are provided. Do not forget to use thesetables when you need to

q When looking at the THEORY, STANDARD INTEGRALS, AN-SWERS or TIPS pages, use the Back button (at the bottom of thepage) to return to the exercises

q Use the solutions intelligently. For example, they can help you getstarted on an exercise, or they can allow you to check whether your

intermediate results are correct

q Try to make less use of the full solutions as you work your waythrough the Tutorial

Toc Back


11/20

Solutions to exercises 11

Full worked solutions

Exercise 1.sin x cos x dx is of the form

f(x)f(x)dx =

1

2[f(x)]

2+ C

To see this, set u = sin x, to finddu

dx

= cos x and du = cos x dx

sin x cos x dx =

u du

=

1

2 u2

+ C

=1

2sin2 x + C.

Return to Exercise 1

Toc Back


12/20


Exercise 2.

sinh

xcosh

x dxis of the form

f(

x)

f(

x)

dx= [

f(

x)]

2

+C

To see this, set u = sinh x thendu

dx= cosh x and du = cosh x dx

sinh x cosh x dx =

u du

=1

2u2 + C

= 12

sinh2 x + C.


Toc Back


13/20


Exercise 3.

tan

xsec

2 x dxis of the form

f(

x)

f(

x)

dx= [

f(

x)]

2

+C

Let u = tan x thendu

dx= sec2 x and du = sec2 x dx

tan x sec2 x dx =

u du

=1

2u2 + C

= 12

tan2 x + C.


Toc Back


14/20


Exercise 4.

sin2x cos2x dx is close to the form f(x)f(x)dx = [f(x)]2

+C

Let u = sin 2x thendu

dx= 2 cos 2x and

du

2= cos 2x dx

sin2x cos2x dx =

u

du

2

=1

2

u du

= 12

12

u2 + C

=1

4sin2 2x + C.


Toc Back


15/20


Exercise 5.

sinh 3x cosh3x dx is close to the form f(x)f(x)dx = [f(x)]2 + C

Let u = sinh 3x thendu

dx= 3 cosh 3x and

du

3= cosh 3x dx

sinh3x cosh3x dx =

u

du

3

=1

3

u du

= 13

12

u2 + C

=1

6sinh2 3x + C.


Toc Back


16/20


Exercise 6.

1x

ln x dx is of the form f(x)f(x)dx = [f(x)]2 + C

Let u = ln x thendu

dx=

1

xand du =

1

xdx

1

x ln x dx =

u du

=1

2u2 + C

=1

2

ln2 x + C .


Toc Back

S l i i 17


17/20


Exercise 7.

sin

4

x cos x dx is of the form

[f(x)]

n

f(x)dx =

[f(x)]n+1

n + 1 + C

To see this, set u = sin x to find du = cos x dx

sin4 x cos x dx =

u4 du

=u5

5+ C

= sin5

x5

+ C.


Toc Back

S l ti t i 18


18/20


Exercise 8.

sinh

3

x cosh x dx is of the form

[f(x)]

n

f(x)dx =

[f(x)]n+1

n + 1 + C

To see this, set u = sinh x then du = cosh x dx

sinh3 x cosh x dx =

u3 du

=1

4u4 + C

=1

4 sinh4

x + C.


Toc Back

S l ti t i 19


19/20


Exercise 9.

cos

3

x sin x dx is close to the form

[f(x)]

n

f(x)dx =

[f(x)]n+1

n + 1 + C

Let u = cos x then du = sin x dx

cos3 x sin x dx =

u3 (du) = u3du

= u4

4+ C

= cos4 x

4 + C.


Toc Back



20/20


Exercise 10.

2x(x2 4)2 dx is of the form [f(x)]

n

f(x)dx =

[f(x)]n+1

n + 1 + C

Let u = x2 4 then du = 2x dx

2x(x2 4)2 dx =

1

u2 du =

u2

du

= u1 + C = 1u

+ C

=

1

x2

4

+ C .


Toc Back

integration by substitution 3

Documents