integration by substitution 1

8/8/2019 Integration by Substitution 1

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Integration

SUBSTITUTION I .. f(ax + b)

Graham S McDonaldand Silvia C Dalla

A Tutorial Module for practising the integra-tion of expressions of the form f(ax + b)

q Table of contents

q Begin Tutorial

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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(ax + b)dx

where a and b are constants. We have here an unspecified function f

of a linear function of x

Letting u = ax + b thendu

dx= a , and this gives dx =

du

a

This allows us to change the integration variable from x to u

f(ax + b)dx =

f(u)

du

a

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Section 1: Theory 4

The final result is

f(ax + b)dx = 1

a

f(u) du

where u = ax + b

This is a general result for integrating functions of a linear functionof x

Each application of this result involves dividing by the coefficient

of x and then integrating

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Section 2: Exercises 5

2. Exercises

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

total).

Perform the following integrations:

Exercise 1.

(2x 1)3dx

Exercise 2.cos(3x + 5)dx

Exercise 3.e5x+2dx

q Theory q Standard integrals q Answers q Tips

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Exercise 4.

sinh3x dxExercise 5.

dx

2x

1

Exercise 6.dx

1 + (5x)2

Exercise 7.sec2(7x + 1)dx


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Exercise 8.

sin(3x 1)dxExercise 9.

cosh(1 + 2x)dx

Exercise 10.tan(9x 1)dx


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Section 3: Answers 8

3. Answers

1. 1

8

(2x

1)4 + C ,

2. 13 sin(3x + 5) + C ,

3. 15e5x+2 + C ,

4. 13 cosh3x + C ,

5.12 ln |2x 1| + C ,

6. 15 tan1 5x + C ,

7. 17 tan(7x + 1) + C ,

8. 13 cos(3x 1) + C ,9. 12 sinh(1 + 2x) + C ,

10. 19 ln | cos(9x 1)| + 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.(2x 1)3dx

Let u = 2x 1 then dudx

= 2 and dx =du

2

(2x 1)3dx = u3 du2 = 12 u3du=

1

2

1

4u4 + C

=1

8(2x

1)4 + C .

Note. The final result can also be obtained using the general pattern:f(ax + b) dx =

1

a

f(u) du

where a = 2 and

f(u)du is

u3

du. Return to Exercise 1

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Exercise 2.

cos(3

x+ 5)

dx

Let u = 3x + 5 thendu

dx= 3 and dx =

du

3

cos(3x + 5)dx =

cos u

du

3 =

1

3

cos u du

=1

3sin u + C

=1

3sin(3x + 5) + C .


1

a

f(u) du

where a = 3 and

f(u)du is

cos udu. Return to Exercise 2

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Exercise 3.

e5x+2dx


dx= 5 and dx =

du

5

e5x+2dx = eudu

5

=1

5 eudu

=1

5eu + C

=1

5e5x+2 + C .


1

a

f(u) du

where a = 5 and

f(u)du is

eu

du. Return to Exercise 3

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Exercise 4.

sinh 3x dx

Let u = 3x thendu

dx= 3 and dx =

du

3

sinh 3x dx =

sinh u

du

3 =

1

3

sinh u du

=1

3cosh u + C

=1

3cosh 3x + C .


1

a

f(u) du

where a = 3 and

f(u)du is

sinh u du. Return to Exercise 4

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Exercise 5.

dx2x 1


= 2 and dx =du

2

dx

2x 1=

du

u 1

2

=1

2

du

u

=1

2ln |u| + C

=1

2ln |2x 1| + C .


1

a

f(u) du

where a = 2 and f(u)du is duu

.

Return to Exercise 5

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Exercise 6.

dx1 + (5x)2

Let u = 5x thendu

dx= 5 and dx =

du

5

dx

1 + (5x)2

= du

1 + u2

1

5

=1

5

du

1 + u2

=1

5tan1 u + C

= =1

5tan1 5x + C .


1

a

f(u) du

where a = 5 and f(u)du is du

1+u2 du. Return to Exercise 6

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S l ti t i 18


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Exercise 7.

sec2(7x + 1)dx


dx= 7 and dx =

du

7

sec2(7x + 1)dx =

1

7 sec

2 u du

=1

7tan u + C

=1

7tan(7x + 1) + C .


1

a

f(u) du

where a = 7 and

f(u)du is

sec2 udu.

Return to Exercise 7

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S l ti t i 19


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Exercise 8.

sin(3x 1)dx


= 3 and dx =du

3

sin(3x 1)dx =1

3 sin u du

= 13

cos u + C

= 13

cos(3x 1) + C .Note. The final result can also be obtained using the general pattern:

f(ax + b) dx =1

a

f(u) du

where a = 3 and f(u)du is sin udu.Return to Exercise 8

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Exercise 9.

cosh(1 + 2x)dx

Let u = 1 + 2x thendu

dx= 2 and dx =

du

2

cosh(1 + 2x)dx = cosh u du2

=1

2sinh u + C

=1

2sinh(1 + 2x) + C .


1

a

f(u) du

where a = 2 and f(u)du is cosh u du.Return to Exercise 9

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Exercise 10.

tan(9x 1)dx


= 9 and dx =du

9

tan(9x 1) dx = tan u du

9=

1

9 tan u du

= 19

ln | cos u| + C

= 19

ln | cos(9x 1)| + C .Note. The final result can also be obtained using the general pattern:

f(ax + b) dx =1

a

f(u) du

where a = 9 and f(u)du is tan u du.Return to Exercise 10

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integration by substitution 1

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