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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.
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.
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]
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.
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]
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.
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.
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.
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.
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.