Section 7.2 Trigonometric Integrals634 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION

7.2 Trigonometric Integrals

The symbolss= and

c= indicate the use of the substitutions { = sin = cos} and { = cos = − sin}, respectively.

1.

sin3 cos2 =

sin2 cos2 sin =(1− cos2 ) cos2 sin

=(1− 2)2(−) [ = cos, = − sin]

=(2 − 1)2 =

(4 − 2) = 1

55 − 1

33 + = 1

5cos5 − 1

3cos3 +

2.

sin6 cos3 =

sin6 cos2 cos =

sin6 (1− sin2 ) coss=6(1− 2)

=(6 − 8) = 1

77 − 1

99 + = 1

7sin7 − 1

9sin9 +

3. 20

sin7 cos5 = 20

sin7 cos4 cos = 20

sin7 (1− sin2 )2 cos

s= 1

07(1− 2)2 =

1

07(1− 22 + 4) =

1

0(7 − 29 + 11)

=

1

88 − 1

510 +

1

1212

10

=

1

8− 1

5+

1

12

− 0 =

15− 24 + 10

120=

1

120

4. 20

sin5 = 20

sin4 sin = 20

(1− cos2 )2 sinc= 0

1(1− 2)2(−)

=

1

0

(1− 22+

4) =

− 2

3

3+

1

5

5

10

=

1− 2

3+

1

5

− 0 =

15− 10 + 3

15=

8

15

5.

sin5(2) cos2(2) =

sin4(2) cos2(2) sin(2) =[1− cos2(2)]2 cos2(2) sin(2)

=(1− 2)2 2

− 12

[ = cos(2), = −2 sin(2) ]

= − 12

(4 − 22 + 1)2 = − 1

2

(6 − 24 + 2)

= − 12

177 − 2

55 + 1

33

+ = − 114

cos7(2) + 15

cos5(2)− 16

cos3(2) +

6. cos5(2) =

cos4(2) cos(2) =

[1− sin2(2)]2 cos(2)

=

12(1− 2)2

= sin(2), = 2 cos(2)

= 1

2

(4 − 22 + 1) = 1

2( 155 − 2

33 + ) + = 1

10sin5(2)− 1

3sin3(2) + 1

2sin(2) +

7. 20

cos2 = 20

12(1 + cos 2) [half-angle identity]

= 12

+ 1

2sin 2

20

= 12

2

+ 0− (0 + 0)

=

4

8. Let =√, so that =

1

2√ and = 2 . Then

sin3 (

√ )√

=

sin3

(2 ) = 2

sin

3 = 2

sin

2 sin = 2

(1− cos

2) sin

c= 2

(1− 2)(−) = 2

(2 − 1) = 2

133 −

+ = 2

3cos3 − 2 cos +

= 23

cos3(√ )− 2 cos

√+

9. 0

cos4(2) = 0

[cos2(2)]2 = 0

12(1 + cos(2 · 2))2 [half-angle identity]

= 14

0

[1 + 2 cos 4 + cos2(4)] = 14

0

[1 + 2 cos 4 + 12(1 + cos 8)]

= 14

0

32

+ 2cos 4 + 12

cos 8 = 1

4

32 + 1

2sin 4+ 1

16sin 8

0

= 14

32 + 0 + 0

− 0

= 38

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SECTION 7.2 TRIGONOMETRIC INTEGRALS ¤ 635

10. 0

sin2 cos4 = 14

0

(4 sin2 cos2 ) cos2 = 14

0

(2 sin cos )2 12(1 + cos 2)

= 18

0

(sin 2)2(1 + cos 2) = 18

0

(sin2 2+ sin2 2 cos 2)

= 18

0

sin2 2 + 18

0

sin2 2 cos 2 = 18

0

12(1− cos 4) + 1

8

13· 1

2sin3 2

0

= 116

− 1

4sin 4

0

+ 18(0− 0) = 1

16[( − 0)− 0] =

16

11. 20

sin2 cos2 = 20

14(4 sin2 cos2 ) =

20

14(2 sin cos)2 = 1

4

20

sin2 2

= 14

20

12(1− cos 4) = 1

8

20

(1− cos 4) = 18

− 1

4sin 4

20

= 18

2

=

16

12. 20

(2− sin )2 = 20

(4− 4 sin + sin2 ) = 20

4− 4 sin + 1

2(1− cos 2)

= 20

92− 4 sin − 1

2cos 2

=

92 + 4cos − 1

4sin 2

20

=

94

+ 0− 0− (0 + 4− 0) = 9

4 − 4

13. √

cos sin3 = √

cos sin2 sin =(cos )12(1− cos2 ) sin

c=12(1− 2) (−) =

(52 − 12)

= 2772 − 2

332 + = 2

7(cos )72 − 2

3(cos )32 +

14.

sin2(1)

2=

sin

2 (−)

=

1

, = − 1

2

= −

1

2(1− cos 2) = −1

2

− 1

2sin 2

+ = − 1

2+

1

4sin

2

+

15.

cot cos

2=

cos

sin(1− sin

2)

s=

1− 2

=

1

−

= ln ||− 1

2

2+ = ln |sin|− 1

2sin

2+

16.

tan

2 cos

3=

sin2

cos2 cos

3 =

sin

2 cos

s=

2 = 1

3

3+ = 1

3sin

3 +

17.

sin2 sin 2 =

sin2 (2 sin cos) s=

23 = 124 + = 1

2sin4 +

18.

sin cos

12=

sin2 · 1

2cos

12 =

2 sin

12cos2

12

=

22 (−2 ) [ = cos12, = − 1

2sin12]

= − 433 + = − 4

3cos3

12

+

19. sin2 =

12(1− cos 2)

= 1

2

(− cos 2) = 1

2

− 1

2

cos 2

= 12

122− 1

2

12 sin 2− 1

2sin 2

= , = cos 2

= , = 12sin 2

= 1

42 − 1

4 sin 2 + 1

2

− 14

cos 2

+ = 142 − 1

4 sin 2− 1

8cos 2+

20. Let = sin . Then = cos andcos cos5(sin ) =

cos5 =

(cos2 )2 cos =

(1− sin2 )2 cos

=(1− 2 sin2 + sin4 ) cos =

[continued]

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SECTION 7.2 TRIGONOMETRIC INTEGRALS ¤ 635

10. 0

sin2 cos4 = 14

0

(4 sin2 cos2 ) cos2 = 14

0

(2 sin cos )2 12(1 + cos 2)

= 18

0

(sin 2)2(1 + cos 2) = 18

0

(sin2 2+ sin2 2 cos 2)

= 18

0

sin2 2 + 18

0

sin2 2 cos 2 = 18

0

12(1− cos 4) + 1

8

13· 1

2sin3 2

0

= 116

− 1

4sin 4

0

+ 18(0− 0) = 1

16[( − 0)− 0] =

16

11. 20

sin2 cos2 = 20

14(4 sin2 cos2 ) =

20

14(2 sin cos)2 = 1

4

20

sin2 2

= 14

20

12(1− cos 4) = 1

8

20

(1− cos 4) = 18

− 1

4sin 4

20

= 18

2

=

16

12. 20

(2− sin )2 = 20

(4− 4 sin + sin2 ) = 20

4− 4 sin + 1

2(1− cos 2)

= 20

92− 4 sin − 1

2cos 2

=

92 + 4cos − 1

4sin 2

20

=

94

+ 0− 0− (0 + 4− 0) = 9

4 − 4

13. √

cos sin3 = √

cos sin2 sin =(cos )12(1− cos2 ) sin

c=12(1− 2) (−) =

(52 − 12)

= 2772 − 2

332 + = 2

7(cos )72 − 2

3(cos )32 +

14.

sin2(1)

2=

sin

2 (−)

=

1

, = − 1

2

= −

1

2(1− cos 2) = −1

2

− 1

2sin 2

+ = − 1

2+

1

4sin

2

+

15.

cot cos

2=

cos

sin(1− sin

2)

s=

1− 2

=

1

−

= ln ||− 1

2

2+ = ln |sin|− 1

2sin

2+

16.

tan

2 cos

3=

sin2

cos2 cos

3 =

sin

2 cos

s=

2 = 1

3

3+ = 1

3sin

3 +

17.

sin2 sin 2 =

sin2 (2 sin cos) s=

23 = 124 + = 1

2sin4 +

18.

sin cos

12=

sin2 · 1

2cos

12 =

2 sin

12cos2

12

=

22 (−2 ) [ = cos12, = − 1

2sin12]

= − 433 + = − 4

3cos3

12

+

19. sin2 =

12(1− cos 2)

= 1

2

(− cos 2) = 1

2

− 1

2

cos 2

= 12

122− 1

2

12 sin 2− 1

2sin 2

= , = cos 2

= , = 12sin 2

= 1

42 − 1

4 sin 2 + 1

2

− 14

cos 2

+ = 142 − 1

4 sin 2− 1

8cos 2+

20. Let = sin . Then = cos andcos cos5(sin ) =

cos5 =

(cos2 )2 cos =

(1− sin2 )2 cos

=(1− 2 sin2 + sin4 ) cos =

[continued]

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636 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION

Now let = sin. Then = cos and

=(1− 22 + 4) = − 2

33 + 1

55 + = sin− 2

3sin3 + 1

5sin5 +

= sin(sin )− 23

sin3(sin ) + 15

sin5(sin ) +

21.

tan sec3 =

tan sec sec2 =2 [ = sec, = sec tan ]

= 133 + = 1

3sec3 +

22.

tan2 sec4 =

tan2 sec2 sec2 =

tan2 (tan2 + 1) sec2

=2(2 + 1) [ = tan , = sec2 ]

=(4 + 2) = 1

55 + 1

33 + = 1

5tan5 + 1

3tan3 +

23.

tan2 =(sec2 − 1) = tan− +

24.(tan2 + tan4 ) =

tan2 (1 + tan2 ) =

tan2 sec2 =

2 [ = tan, = sec2 ]

= 133 + = 1

3tan3 +

25. Let = tan. Then = sec2, sotan4 sec6=

tan4 sec4 (sec2) =

tan4(1 + tan2)2 (sec2)

=4(1 + 2)2 =

(8 + 26 + 4)

= 199 + 2

77 + 1

55 + = 1

9tan9+ 2

7tan7+ 1

5tan5+

26. 40

sec6 tan6 = 40

tan6 sec4 sec2 = 40

tan6 (1 + tan2 )2 sec2

= 1

06(1 + 2)2

= tan , = sec2

= 1

06(4 + 22 + 1) =

1

0(10 + 28 + 6)

=

11111 + 2

99 + 1

7710

= 111

+ 29

+ 17

= 63 +154 +99693

= 316693

27.

tan3 sec =

tan2 sec tan =(sec2 − 1) sec tan

=(2 − 1) [ = sec, = sec tan] = 1

33 − + = 1

3sec3 − sec+

28. Let = sec, so = sec tan. Thus,tan5 sec3=

tan4 sec2 (sec tan) =

(sec2− 1)2 sec2 (sec tan)

=(2 − 1)22 =

(6 − 24 + 2)

= 177 − 2

55 + 1

33 + = 1

7sec7 − 2

5sec5 + 1

3sec3 +

29.

tan3 sec6 =

tan3 sec4 sec2 =

tan3 (1 + tan2 )2 sec2

=3(1 + 2)2

= tan, = sec2

=3(4 + 22 + 1) =

(7 + 25 + 3)

= 188 + 1

36 + 1

44 + = 1

8tan8 + 1

3tan6 + 1

4tan4 +

c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

636 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION

Now let = sin. Then = cos and

=(1− 22 + 4) = − 2

33 + 1

55 + = sin− 2

3sin3 + 1

5sin5 +

= sin(sin )− 23

sin3(sin ) + 15

sin5(sin ) +

21.

tan sec3 =

tan sec sec2 =2 [ = sec, = sec tan ]

= 133 + = 1

3sec3 +

22.

tan2 sec4 =

tan2 sec2 sec2 =

tan2 (tan2 + 1) sec2

=2(2 + 1) [ = tan , = sec2 ]

=(4 + 2) = 1

55 + 1

33 + = 1

5tan5 + 1

3tan3 +

23.

tan2 =(sec2 − 1) = tan− +

24.(tan2 + tan4 ) =

tan2 (1 + tan2 ) =

tan2 sec2 =

2 [ = tan, = sec2 ]

= 133 + = 1

3tan3 +

25. Let = tan. Then = sec2, sotan4 sec6=

tan4 sec4 (sec2) =

tan4(1 + tan2)2 (sec2)

=4(1 + 2)2 =

(8 + 26 + 4)

= 199 + 2

77 + 1

55 + = 1

9tan9+ 2

7tan7+ 1

5tan5+

26. 40

sec6 tan6 = 40

tan6 sec4 sec2 = 40

tan6 (1 + tan2 )2 sec2

= 1

06(1 + 2)2

= tan , = sec2

= 1

06(4 + 22 + 1) =

1

0(10 + 28 + 6)

=

11111 + 2

99 + 1

7710

= 111

+ 29

+ 17

= 63 +154 +99693

= 316693

27.

tan3 sec =

tan2 sec tan =(sec2 − 1) sec tan

=(2 − 1) [ = sec, = sec tan] = 1

33 − + = 1

3sec3 − sec+

28. Let = sec, so = sec tan. Thus,tan5 sec3=

tan4 sec2 (sec tan) =

(sec2− 1)2 sec2 (sec tan)

=(2 − 1)22 =

(6 − 24 + 2)

= 177 − 2

55 + 1

33 + = 1

7sec7 − 2

5sec5 + 1

3sec3 +

29.

tan3 sec6 =

tan3 sec4 sec2 =

tan3 (1 + tan2 )2 sec2

=3(1 + 2)2

= tan, = sec2

=3(4 + 22 + 1) =

(7 + 25 + 3)

= 188 + 1

36 + 1

44 + = 1

8tan8 + 1

3tan6 + 1

4tan4 +

c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

638 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION

40. Let = csc, = csc2 . Then = − csc cot, = − cot ⇒csc3 = − csc cot− csc cot2 = − csc cot− csc (csc2 − 1)

= − csc cot +

csc− csc3

Solving for

csc3 and using Exercise 39, we getcsc3 = − 1

2csc cot+ 1

2

csc = − 1

2csc cot+ 1

2ln |csc− cot|+ . Thus, 3

6csc3 =

− 1

2csc cot+ 1

2ln |csc− cot|

36

= − 12· 2√

3· 1√

3+ 1

2ln 2√

3− 1√

3

+ 12· 2 ·√3− 1

2ln2−√3

= − 1

3+√

3 + 12

ln 1√3− 1

2ln2−√3

≈ 17825

41.

sin 8 cos 52a=

12[sin(8− 5) + sin(8+ 5)] = 1

2

(sin 3+ sin 13)

= 12(− 1

3cos 3− 1

13cos 13) + = − 1

6cos 3− 1

26cos 13+

42.

sin 2 sin 6 2b=

12[cos(2 − 6)− cos(2 + 6)]

= 12

[cos(−4)− cos 8] = 1

2

(cos 4 − cos 8)

= 12

14

sin 4 − 18

sin 8

+ = 18

sin 4 − 116

sin 8 +

43. 20

cos 5 cos 10 2c= 20

12[cos(5− 10) + cos(5+ 10)]

= 12

20

[cos(−5) + cos 15] = 12

20

(cos 5+ cos 15)

= 12

15

sin 5 + 115

sin 1520

= 12

15− 1

15

= 1

15

44.

sin sec5=

sin

cos5

c=

1

5(−) =

1

44+ =

1

4 cos4 + = 1

4sec

4+

45. 60

√1 + cos 2=

60

1 + (2 cos2 − 1) =

60

√2 cos2 =

√2 60

√cos2

=√

2 60

|cos| =√

2 60

cos [since cos 0 for 0 ≤ ≤ 6]

=√

2sin

60

=√

2

12− 0

= 12

√2

46.

cos + sin

sin 2 =

1

2

cos+ sin

sin cos = 1

2

(csc + sec)

= 12

ln |csc− cot|+ ln |sec+ tan|

+ [by Exercise 39 and (1)]

47.

1− tan2

sec2 =

cos

2− sin

2 =

cos 2 = 1

2sin 2+

48.

cos− 1=

1

cos− 1· cos+ 1

cos+ 1 =

cos+ 1

cos2 − 1 =

cos+ 1

− sin2

= − cot csc− csc2

= csc + cot+

49. tan2 =

(sec2 − 1) =

sec2 −

= tan− tan− 122

= , = sec2

= , = tan

= tan− ln |sec|− 1

22 +

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638 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION

40. Let = csc, = csc2 . Then = − csc cot, = − cot ⇒csc3 = − csc cot− csc cot2 = − csc cot− csc (csc2 − 1)

= − csc cot +

csc− csc3

Solving for

csc3 and using Exercise 39, we getcsc3 = − 1

2csc cot+ 1

2

csc = − 1

2csc cot+ 1

2ln |csc− cot|+ . Thus, 3

6csc3 =

− 1

2csc cot+ 1

2ln |csc− cot|

36

= − 12· 2√

3· 1√

3+ 1

2ln 2√

3− 1√

3

+ 12· 2 ·√3− 1

2ln2−√3

= − 1

3+√

3 + 12

ln 1√3− 1

2ln2−√3

≈ 17825

41.

sin 8 cos 52a=

12[sin(8− 5) + sin(8+ 5)] = 1

2

(sin 3+ sin 13)

= 12(− 1

3cos 3− 1

13cos 13) + = − 1

6cos 3− 1

26cos 13+

42.

sin 2 sin 6 2b=

12[cos(2 − 6)− cos(2 + 6)]

= 12

[cos(−4)− cos 8] = 1

2

(cos 4 − cos 8)

= 12

14

sin 4 − 18

sin 8

+ = 18

sin 4 − 116

sin 8 +

43. 20

cos 5 cos 10 2c= 20

12[cos(5− 10) + cos(5+ 10)]

= 12

20

[cos(−5) + cos 15] = 12

20

(cos 5+ cos 15)

= 12

15

sin 5 + 115

sin 1520

= 12

15− 1

15

= 1

15

44.

sin sec5=

sin

cos5

c=

1

5(−) =

1

44+ =

1

4 cos4 + = 1

4sec

4+

45. 60

√1 + cos 2=

60

1 + (2 cos2 − 1) =

60

√2 cos2 =

√2 60

√cos2

=√

2 60

|cos| =√

2 60

cos [since cos 0 for 0 ≤ ≤ 6]

=√

2sin

60

=√

2

12− 0

= 12

√2

46.

cos + sin

sin 2 =

1

2

cos+ sin

sin cos = 1

2

(csc + sec)

= 12

ln |csc− cot|+ ln |sec+ tan|

+ [by Exercise 39 and (1)]

47.

1− tan2

sec2 =

cos

2− sin

2 =

cos 2 = 1

2sin 2+

48.

cos− 1=

1

cos− 1· cos+ 1

cos+ 1 =

cos+ 1

cos2 − 1 =

cos+ 1

− sin2

= − cot csc− csc2

= csc + cot+

49. tan2 =

(sec2 − 1) =

sec2 −

= tan− tan− 122

= , = sec2

= , = tan

= tan− ln |sec|− 1

22 +

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642 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION

(b) 220 =

[()]

2

ave ⇒

2202 = [()]2

ave = 1160

160

02 sin2(120) = 602

160

0

12[1− cos(240)]

= 302− 1

240sin(240)

1600

= 302

160− 0− (0− 0)

= 1

22

Thus, 2202 = 122 ⇒ = 220

√2 ≈ 311 V.

67. Just note that the integrand is odd [(−) = −()].

Or: If 6= , calculate

−sin cos =

−12[sin(− )+ sin(+ )] = 1

2

−cos(− )

− − cos(+ )

+

−

= 0

If = , then the first term in each set of brackets is zero.

68. − sin sin =

−

12[cos(− )− cos(+ )] .

If 6= , this is equal to1

2

sin(− )

− − sin(+ )

+

−

= 0.

If = , we get −

12[1− cos(+ )] =

12− −

sin(+ )

2(+ )

−

= − 0 = .

69. − cos cos =

−

12[cos(− )+ cos(+ )] .

If 6= , this is equal to1

2

sin(− )

− +

sin(+ )

+

−

= 0.

If = , we get −

12[1 + cos(+ )] =

12− +

sin(+ )

2(+ )

−

= + 0 = .

70.1

−() sin =

1

−

=1

sin

sin

=

=1

−sin sin. By Exercise 68, every

term is zero except theth one, and that term is

· = .

7.3 Trigonometric Substitution

1. Let = 3 sec , where 0 ≤ 2or ≤ 3

2. Then

= 3 sec tan and√2 − 9 =

√9 sec2 − 9 =

9(sec2 − 1) =

√9 tan2

= 3 |tan | = 3 tan for the relevant values of .1

2√2 − 9

=

1

9 sec2 · 3 tan 3 sec tan = 1

9

cos = 1

9sin + =

1

9

√2 − 9

+

Note that− sec( + ) = sec , so the figure is sufficient for the case ≤ 32.

2. Let = 3 sin , where−2≤ ≤

2. Then = 3cos and

√9− 2 =

9− 9 sin2 =

9(1− sin2 ) =

√9 cos2

= 3 |cos | = 3cos for the relevant values of .

c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

1