integration by substitution 2

8/8/2019 Integration by Substitution 2

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Integration

SUBSTITUTION II .. 1f(x)

f(x)

Graham S McDonaldand Silvia C Dalla

A Tutorial Module for practising the integra-tion of expressions of the form 1

f(x) f(x)

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/


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Table of contents

1. Theory

2. Exercises

3. Answers

4. Standard integrals

5. Tips

Full worked solutions


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Section 1: Theory 3

1. Theory

Consider an integral of the formf(x)

f(x)dx

Letting u = f(x) then

du

dx = f(x) , and this gives du = f(x)dx

So f(x)

f(x)dx =

1

f(x) f(x) dx

=

du

u

= ln

|u

|+ C

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Section 1: Theory 4

The general result is

f

(x)f(x) dx = ln |f(x)| + C

This is a useful generalisation of the standard integral

1

xdx = ln |x| + C ,

that applies straightforwardly when the top line of what we are inte-

grating is a multiple of the derivative of the bottom line

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Section 2: Exercises 5

2. Exercises

Click on EXERCISE links for full worked solutions (there are 11 ex-

ercises in total).

Perform the following integrations:

Exercise 1.

10x + 3

5x2 + 3x 1 dx

Exercise 2.

3x2

x3

6dx

Exercise 3.4x2

x3 1 dx

q Theory q Standard integrals q Answers q Tips

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Exercise 4.

2x 1

x2 x 7dx

Exercise 5.x 1

x2 2x + 7 dx

Exercise 6.8x

x2 4 dx

Exercise 7.x2

x3 + 2dx


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Exercise 8.

sin x

cos x

dx

Exercise 9.cos x

sin xdx

Exercise 10.cos x

5 + sin xdx

Exercise 11.sinh x

1 cosh x dx ,


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Section 3: Answers 8

3. Answers

1. ln |5x2 + 3x 1| + C ,2. ln |x3 6| + C ,3. 43 ln |x3 1| + C,4. ln

|x2

x

7|

+ C ,

5. 12 ln |x2 2x + 7| + C ,6. 4ln |x2 4| + C,7. 13 ln |x3 + 2| + C

8. ln | cos x| + C ,9. ln | sin x| + C,

10. ln |5 + sin x| + C,11. ln |1 cosh x| + C.

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Section 4: Standard integrals 10

f(x)

f(x) dx f(x)

f(x) dx

1

a2+x21

a tan1 x

a

1

a2x21

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

sin1xa

a2+x2 a22 sinh1 xa + xa2+x2a2

+xa2x2a2

x2a2 a22

cosh1 x

a

+ x

x2a2a2

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Section 5: Tips 11

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

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Solutions to exercises 12

Full worked solutions

Exercise 1.

10x + 3

5x2 + 3x 1 dx is of the form

f(x)

f(x)dx = ln |f(x)| + C.

Let u = 5x2 + 3x 1 then dudx

= 10x + 3, and du = (10x + 3)dx

10x + 3

5x2 + 3x 1 dx =

du

u

= ln |u| + C

= ln |5x2 + 3x 1| + C .

Return to Exercise 1

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Exercise 2.

3x2

x3 6 dx is of the form f(x)

f(x) dx = ln |f(x)| + C.

Let u = x3 6 then dudx

= 3x2 and du = 3x2dx

3x2

x3 6 dx =

1

udu

= ln |u| + C

= ln |x3 6| + C .


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Exercise 3.

4x2

x3 1 dx is of the slightly more general formk

f(x)

f(x)dx = k

f(x)

f(x)dx = k ln |f(x)| + C,

where k is a constant

i.e. the top line is a multiple of the derivative of the bottom line.

Let u = x3 1 then dudx

= 3x2 anddu

3= x2dx

4 x2

x3 1 dx = 4 1

u

du

3

=4

3ln |u| + C

=4

3

ln

|x3

1

|+ C .


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Exercise 4.

2x 1

x2 x 7 dx is of the form f(x)

f(x) dx = ln |f(x)| + C.

Let u = x2 x 7 then dudx

= 2x 1 and du = (2x 1)dx

2x 1x2 x 7 dx =

du

u

= ln |u| + C

= ln |x2 x 7| + C .


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Exercise 5.

x 1

x2 2x + 7dx is of the slightly more general form

k

f(x)

f(x)dx = k

f(x)

f(x)dx = k ln |f(x)| + C,

wherek

is a constanti.e. the top line is a multiple of the derivative of the bottom line.

Let u = x2 2x + 7 then dudx

= 2x 2 and du = (2x 2)dx

so thatdu

2= (x 1)dx

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S l i i 17


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x 1

x2

2x + 7dx =

1

u

du

2

=1

2

du

u

=1

2ln |u| + C

=1

2ln|x2

2x + 7

|+ C .


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S l ti t i 18


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Exercise 6.

8xx2 4

dx

Let u = x2 4 then du = 2x dx and du2

= xdx

8x

x2 4 dx = 8 1

u

du

2

= 4 ln |u| + C

= 4 ln |x2

4| + C .


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S l ti t i 19


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Exercise 7.

x2

x3 + 2 dx

Let u = x3 + 2 then du = 3x2dx anddu

3= x2dx

x2

x3 + 2 dx = 1

u

du

3

=1

3ln |u| + C

= 13

ln |x3 + 2| + C .


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Exercise 8.

sinxcosx dxLet u = cos x then du

dx= sin x and du(1) = sin x dx

sin x

cos x

dx = 1

u

du

(1)=

du

u

= ln |u| + C= ln | cos x| + C.


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Exercise 9.

cosxsinx dxLet u = sin x then du

dx= cos x and du = cos x dx

cos x

sin x

dx = 1

u

du

= ln |u| + C

= ln | sin x| + C.


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Exercise 10.

cos x5 + sin x

dx

Let u = 5 + sin x thendu

dx= cos x and du = cos x dx

cos x5 + sin x

dx =

duu

= ln |u| + C

= ln |5 + sin x| + C .


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integration by substitution 2

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