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Created by T. Madas Created by T. Madas INTEGRATION by substitution

integration by substitution - MadAsMaths

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Page 1: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

INTEGRATION

by substitution

Page 2: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Question 1

Carry out the following integrations by substitution only.

1. ( ) ( ) ( )4 6 51 1

4 2 1 2 1 2 16 5

x x dx x x C− = − + − +∫

2. ( )2 1 1

2 1 ln 2 12 1 2 2

xdx x x C

x= + − + +

+∫

3. ( ) ( )1 1

2 22 24 4x x dx x C−

− = − − +∫

4. 2

2

4 1ln 6 1

6 1 3

xdx x C

x= − +

−∫

5. ( ) ( ) ( )4 6 51 1

3 1 3 1 3 154 45

x x dx x x C− = − + − +∫

6. ( ) ( )3 12 2

1

3

84 1 4 1

4 1C

xdx x x

x= + +− −

−∫

7. ( )2 1

3 2

3

2

3

22 1

2 1C

xdx x

x

= +++∫

8. ( )4 3

10ln 2 3 22

xdx x x C

x

−= + − + +

+∫

9. ( ) ( )2

24 1 12 1 2 1 ln 2 1

2 1 4 2

xdx x x x C

x= − + − + − +

−∫

10. ( )4 3 4 7

3 4 ln 3 43 4 9 9

xdx x x C

x

−= − + − +

−∫

Page 3: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Page 4: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Question 2

Carry out the following integrations by substitution only.

1. ( ) ( ) ( )3 5 42 1

6 3 1 3 1 3 115 6

x x dx x x C− = − + − +∫

2. ( )5 1 1

5 1 ln 5 15 1 5 5

xdx x x C

x= − + − +

−∫

3. ( ) ( )31

2 22 23 1 1x x dx x C+ = + +∫

4. 2

3

3

3 1ln 2 1

2 1 2

xdx x C

x= + +

+∫

5. ( ) ( ) ( )5 6 71 1

2 1 2 1 2 124 28

x x dx x x C− = − + − +∫

6. ( ) ( )3 12 2

55

3

101 2 1 2

1 2C

xdx x x

x= − +− −

−∫

7. ( )4 1

5 2

5

3

5

32 1

2 1C

xdx x

x

= +++∫

8. ( )1 3 1

ln 1 2 1 21 2 4 4

xdx x x C

x

−= + − + +

+∫

9. ( ) ( )2

26 3 9 272 3 2 3 ln 2 3

2 3 8 2 4

xdx x x x C

x= + − + + + +

+∫

10. 12

1 12 2

14 1

1

dx x C

x x

= − +

−∫

Page 5: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Page 6: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Question 3

Carry out the following integrations by substitution only.

1. ( ) ( ) ( )3 5 42 3

10 5 3 5 3 5 325 10

x x dx x x C− = − + − +∫

2. ( )12

3 2 1 3ln 2 12 1

xdx x x C

x= − + − +

−∫

3. ( ) ( )5 7

2 22 211 1

7x x dx x C− = − +∫

4. 5

6

6

5 5ln 2 7

2 7 12

xdx x C

x= + +

+∫

5. ( ) ( ) ( )4 5 62 1

2 1 5 1 5 1 5125 75

x x dx x x C− = − − + − +∫

6. 2

2

9

4

94 1

4 1C

xdx x

x

= +++∫

7. ( ) ( )3 12 2

1 1

8 8

3 14 1 4 1

4 1C

xdx x x

x= +

−− − −

−∫

8. ( )1 2 5 2

ln 1 3 1 31 3 9 9

xdx x x C

x

−= + − + +

+∫

9.

12

32

32

62ln 2 3

2 3

xdx x C

x

= + ++∫

10. 52

52

32 4

1 3151 3

xdx x C

x

= − − +

−∫

Page 7: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Page 8: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Question 4

Carry out the following integrations to the answers given, by using substitution only.

1. ( )

12

4

0

18 2 1

15x x dx− =∫

2.

3

2

3 101 ln 2

3 5 3

xdx

x= +

−∫

3. ( )1

32 2

0

11

5x x dx− =∫

4.

1

2

0

42ln 2

1

xdx

x=

+∫

5. ( )3

4

1

558082 3 1

5x x dx− =∫

6.

8

4

668

2 7

xdx

x=

−∫

7.

1

20

1

59 5

xdx

x

=−∫

8.

3

0

5 214ln 2 6

1

xdx

x

−= −

+∫

9.

125

0

10 ln 4 1

5 1 25

xdx

x

−=

+∫

10.

12

32

5 2 5 21 3ln

2 5 2 4 4

xdx

x

− = +

− ∫

Page 9: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Page 10: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Question 5

Carry out the following integrations to the answers given, by using substitution only.

1. ( )

32

4

0

2432 2 3

20x x dx− =∫

2.

2

0

4 12 ln3

4 1 2

xdx

x= −

+∫

3. ( )1

92 3 2

0

21

33x x dx− =∫

4. ( )4

2

0

53

1212ln

9

xdx

x=

+∫

5. ( )2

4

1

35692 3 1

5x x dx− =∫

6.

6

2

6 272

93 2

xdx

x=

−∫

7.

1

20

1

716 7

xdx

x

=−∫

8.

6

5

1 22 7ln 2

4

xdx

x

−= − −

−∫

9.

123

0

9 ln 4 1

3 1 6

xdx

x

−=

+∫

10. ( )

32

0

2 3 31 ln 4

2 3 2

xdx

x

−= −

+∫

Page 11: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Page 12: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Question 6

Carry out the following integrations.

1. ( ) ( ) ( )3 12 2

2 21 2 1 2 1

3 31

xdx x x C x x C

x= + − + + = − + +

+∫

2. ( )

( ) ( )( )

1 2

3 2

2 1 1 4 12 1 2 1

2 42 1 4 2 1

x xdx x x C C

x x

− − += − + + + + = − +

+ +∫

3. ln 1 1 ln 11

xdx x x C x x C

x= − + + = + − + +

+∫

4. ( ) ( ) ( )3 12 2

2 21 2 1 2 1

3 31

xdx x x C x x C

x= − + − + = + − +

−∫

5. 4 1 11

2 ln 2 52 5 2

xdx x x C

x

+= + − +

−∫

6. ( ) ( )2 2

21 1 1 1 1 1

2 1 2 1 ln 2 1 ln 2 116 4 8 4 4 82 1

x x x C x x x Cx

dxx

− + − + − + = + + − +=−∫

7. ( )2 1 1

2 1 ln 2 1 ln 2 12 1 2

xdx x x C x x C

x

+= − + − + = + − +

−∫

8. ( ) ( ) ( )3 12 2

62 3 9 2 3 2 3 2 3

2 3

xdx x x C x x C

x= + − + + = − + +

+∫

9. ( )3 1 3 11 3 11

2 3 ln 2 3 ln 2 32 3 4 4 2 4

xdx x x C x x C

x

−= − − + + = − + +

+∫

10. ( ) ( )2

2 28 11 2 2 1 2 ln 1 2 2 2 ln 1 2

1 2 2

xdx x x x C x x x C

x= − − + − − − + = − − − − +

−∫

Page 13: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Page 14: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Question 7

Carry out the following integrations using the substitutions given.

1. ( ) ( )5 32 2

2 21 1 1

5 3x x dx x x C− = − − − +∫ Use 1u x= − , or 1u x= −

2. ( ) ( )3 12 2

62 1 3 2 1

2 1

xdx x x C

x= + − + +

+∫ Use 2 1u x= + , or 2 1u x= +

3. 3 31cos sin sin

3x dx x x C= − +∫ Use sinu x=

4. 4 31sec tan tan

3x dx x x C= + +∫ Use tanu x=

5. ( )

1 1 2ln

24 2

xdx C

x x x

−= +

− +∫ Use u x=

6. 2 2

2

2

9 3 9 39 ln

2 9 3

x xdx x C

x x

+ + −= + + +

+ +∫ Use 2 9u x= +

7. ln1 cos

cos 1sin

Cx

dx xx

= ++

−∫ Use cosu x=

8. ( )21

2 2ln 1 21 2

Cdx x xx

= +− + + −+ −∫ Use 2u x= −

9. ( )( )3

2 22

sec tan 1 tan 3tan 2 1 tan15

x x x dx x x C+ = − + +∫ Use 1 tanu x= +

10. ( )

3ln9 3 1

9 1 3 1C

xdx

x x x= +

− +∫ Use u x=

Page 15: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Page 16: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Question 8

Carry out the following integrations.

1. 1

2 4ln 22

dx x x Cx

= − + ++∫

2. ( ) ( ) ( )2

21 1 1 1 11 ln 1 2 1 2 1 2 ln 1 2

1 2 4 8 16 4 8

xdx x x x x x x C

x= − + − − = − − + − − − +

−∫

3. ( ) ( ) ( )22

2

3 3164 128 32

354 1 4 1 ln 4 1

6424 12 35ln 4 1

3 2

4 1x x xx x x C

xdx

x+ − + + +− + + + =

+=

+∫

4. 4 3 3 11

ln 2 12 1 2 4

xdx x x C

x

−= − + + +

+∫

5. 1

6ln 55

xdx x x C

x

+= + − +

−∫

6. 2

212 4ln 2

2 2

xdx x x x C

x= + + − +

−∫

7. 1

2 1 4ln 1 22 1

dx x x Cx

= − − − + ++ −∫

8. 4

8ln 44

xdx x x C

x

+= + − +

−∫

Page 17: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Question 9

Carry out the following integrations.

1. ( ) ( )32

1 3 2 115

x x dx x x C+ = + + +∫

2. ( )2

23

3 2

1 32 3

42 3

xdx x x C

x x

+= + + +

+ +∫

3. 3

2 2

2

3 5 3ln 1

1 2

x xdx x x C

x

+= + + +

+∫

4. 2 1 2 5

ln 3 13 1 3 9

xdx x x C

x

+= + − +

−∫

5. ( ) 2

1 1

11 1

xdx C

xx x

+= − +

−− −∫ , use 1

1xu

− =

6. 3 4

4 4 4

4

4 12 1 2ln 1 1

1 1

x xdx x x x C

x

+= − + + + + +

+ +∫

Page 18: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Question 10

Carry out the following integrations to the answers given.

1. ( )

12

2

0

1 3ln

3 42

xdx

x

= +

−∫

2. ( )

2

2

1

2 ln 27

122 1

xdx

x

+=

−∫

3.

2

0

2 17

64 1

xdx

x

+=

+∫

4. ( )

36

0

1ln16

2dx

x x

=+∫

5.

3

2

6

9

24 2

xdx

x−

= −−∫

6. ( )

32

5

1

12

2 1

xdx

x

+=

−∫

7.

1

2

3

1 1ln 3

2 1 6 4

xdx

x= +

−∫

8.

7 2

1

652

152

xdx

x−

=+∫

9.

5

2

1

4 20

32 1

xdx

x=

−∫

10. ( )

1

2

0

1ln 2

21

xdx

x

= − ++∫

Page 19: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Page 20: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Question 11

Carry out the following integrations to the answers given.

1.

3

0

1161

15x x dx+ =∫

2. ( )2 3

20

64 1 5

1

xdx

x

= ++∫

3.

0 2

1

1ln 2

1 2

xdx

x−

= − +−∫

4.

100

0

140ln 2 20

20dx

x= −

−∫

5. 0

14 12 1 4

30x x dx− =∫

6. ( )5

0

2 107sin cos 1 sin

14x x x dx

π

+ =∫

7. 5 2

2

356

151

xdx

x=

−∫

Page 21: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

Question 12

Carry out the following integrations to the answers given.

1.

2 2

20

124

xdx

x

π= −

−∫ , use 2sinx θ=

2. ( )2

2 21

1 13 1

44dx

x x

= −

−∫ , use 2cosx θ=

3.

( )( )

1

22

0

1 12

81

dx

x

π= +

+∫ , use tanx θ=

Page 22: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

4. ( )2

2 22

1 13 2

21dx

x x

= −

−∫ , use secx θ=

5.

34

20

1

63 4dx

x

π=

−∫ , use 3

sin2

x θ=

6.

( )32

1

2

0

1 1

21 3

dx

x

=

+∫ , use 1

tan3

x θ=

Page 23: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

7.

1

20

1

42dx

x

π=

−∫ , use 2 sinx θ=

8.

12

20

1 3

364 3dx

x

π=

+∫ , use 3

tan2

x θ=

9.

( )32

1

2

0

1 3

124

dx

x

=

−∫ , use 2sinx θ=

Page 24: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

10.

22

2

13 1

12

xdx

x

π−= − −∫ , use cosecx θ=

11.

1

20

1 3

94 3dx

x

π=

−∫ , use 2

sin3

x θ=

12. 3 2

21

3 1121

xdx

x

π= − −

+∫ , use tanx θ=

Page 25: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

13. ( )2

2

0

116 4 6 3

3x dx π− = +∫ , use 4sinx θ=

14.

( )

2

20

32

1 1

83 4

dx

x

=

+∫ , use 2

tan3

x θ=

15. 2

2

0

8 316 3 2

9x dx

π− = +∫ , use

4sin

3x θ=

Page 26: integration by substitution - MadAsMaths

Created by T. Madas

Created by T. Madas

16.

( )

3

22

0

27 1

8 49

dx

x

π= +

+∫ , use 3tanx θ=