Section 7.2 Trigonometric Integrals
32. Evaluate the integral. R tan2x sec xdx Solution:
SECTION 7.2 TRIGONOMETRIC INTEGRALS ¤ 637 30. 4
0 tan4 =4
0 tan2 (sec2 − 1) =4
0 tan2 sec2 −4
0 tan2
=1
0 2 [ = tan ] −4
0 (sec2 − 1) =1
331
0−
tan − 4
0
= 13 −
1 −4
− 0
= 4 − 23
31.
tan5 =
(sec2 − 1)2 tan =
sec4 tan − 2
sec2 tan +
tan
=
sec3 sec tan − 2
tan sec2 +
tan
= 14sec4 − tan2 + ln |sec | + [or 14sec4 − sec2 + ln |sec | + ] 32.
tan2 sec =
(sec2 − 1) sec =
sec3 −
sec
= 12(sec tan + ln |sec + tan |) − ln |sec + tan | + [by Example 8 and (1)]
= 12(sec tan − ln |sec + tan |) + 33. Let = = sec tan ⇒ = = sec . Then
sec tan = sec −
sec = sec − ln |sec + tan | + .
34.
sin cos3 =
sin cos · 1
cos2 =
tan sec2 =
= tan , = sec2
= 122+ = 12tan2 + Alternate solution: Let = cos to get 12sec2 + .
35. 2
6 cot2 =2
6(csc2 − 1) =
− cot − 2
6= 0 − 2
−
−√ 3 −6
=√ 3 −3
36.
csc4 cot6 =
cot6 (cot2 + 1) csc2 =
6(2+ 1) · (−) [ = cot , = − csc2 ]
=
6(2+ 1) · (−) [ = cot , = − csc2 ]
=
(−8− 6) = −199−177+ = −19cot9 −17cot7 + 37. 2
4 cot5 csc3 =2
4 cot4 csc2 csc cot =2
4(csc2 − 1)2 csc2 csc cot
=
1
√2
(2− 1)22(−) [ = csc , =− csc cot ]
=
√2 1
(6− 24+ 2) =1
77−255+133√2
1 =
8 7
√2 −85
√2 +23√ 2
−1
7 −25+13
= 120 − 168 + 70 105
√2 −15 − 42 + 35
105 = 22
105
√2 − 8 105 38. 2
4 csc4 cot4 =2
4 cot4 csc2 csc2 =2
4 cot4 (cot2 + 1) csc2
=0
1 4(2+ 1) (−)
= cot ,
=− csc2
=1
0(6+ 4)
=1
77+1551
0= 17+15 = 1235 39. =
csc =
csc (csc − cot ) csc − cot =
− csc cot + csc2
csc − cot . Let = csc − cot ⇒
= (− csc cot + csc2) . Then =
= ln || = ln |csc − cot | + .
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56. Evaluate the integral. R 1 sec θ+1dθ Solution:
SECTION 7.2 TRIGONOMETRIC INTEGRALS ¤ 693 50. 4
0
√1 − cos 4 =4 0
1 − (1 − 2 sin2(2)) =4 0
2 sin2(2) =√ 24
0
sin2(2)
=√ 24
0 |sin 2| =√ 24
0 sin 2 [since sin 2 ≥ 0 for 0 ≤ ≤ 4]
=√ 2
−12cos 24 0 = −12
√2 (0 − 1) = 12
√2
51.
sin2 =
1
2(1 − cos 2)
= 12
( − cos 2) = 12
−12
cos 2
= 121 22
− 12
1
2 sin 2 − 1
2sin 2
= , = cos 2
= , = 12sin 2
= 142− 14 sin 2 +12
−14cos 2
+ = 142−14 sin 2 −18cos 2 + 52. Let = = sec tan ⇒ = = sec . Then
sec tan = sec −
sec = sec − ln |sec + tan | + .
53.
tan2 =
(sec2 − 1) =
sec2 −
= tan −
tan − 122
= , = sec2
= , = tan
= tan − ln |sec | −122+ 54. =
sin3 . First, evaluate
sin3 =
(1 − cos2) sin =c
(1 − 2)(−) =
(2− 1)
= 133− + 1= 13cos3 − cos + 1 Now for , let = , = sin3 ⇒ = , = 13cos3 − cos , so
= 13 cos3 − cos − 1
3cos3 − cos
= 13 cos3 − cos − 13
cos3 + sin
= 13 cos3 − cos − 13(sin −13sin3) + sin + [by Example 1]
= 13 cos3 − cos + 23sin + 19sin3 +
55.
cos − 1 =
1
cos − 1· cos + 1 cos + 1 =
cos + 1 cos2 − 1 =
cos + 1
− sin2
=
− cot csc − csc2
= csc + cot +
56.
1
sec + 1 =
1
sec + 1· sec − 1 sec − 1 =
sec − 1 sec2 − 1 =
sec − 1 tan2
=
cos sin2 −
cos2 sin2 =
cos
sin2 −
1 − sin2 sin2 =s
1
2 −
csc2 +
= − 1
sin + cot + + Alternate solution:
1
sec + 1 =
cos 1 + cos =
2 cos2
2
− 1 2 cos2
2
[doubleangle identities]
=
1 −
1 2sec2
2
= − tan
2
+
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62. (a) Prove the reduction formula Z
tan2nxdx = tan2n−1x 2n − 1 −
Z
tan2n−2xdx (b) Use this formula to findR tan8xdx.
Solution:
SECTION 7.2 TRIGONOMETRIC INTEGRALS ¤ 695 62. (a)
tan2 =
tan2−2 tan2 =
tan2−2 (sec2 − 1)
=
tan2−2 sec2 −
tan2−2
=
2−2 −
tan2−2 [ = tan , = sec2 ]
= 2−1 2 − 1 −
tan2−2 = tan2−1 2 − 1 −
tan2−2
(b) Starting with = 4, repeated applications of the reduction formula in part (a) gives
tan8 = tan7
7 −
tan6 = tan7
7 −
tan5
5 −
tan4
= tan7
7 −tan5
5 +
tan3
3 −
tan2
= tan7
7 −tan5
5 +tan3
3 −
tan
1 −
1
= tan7
7 −tan5
5 +tan3
3 − tan + + 63. avg = 21
−sin2 cos3 = 21
−sin2 (1 − sin2) cos
= 21 0
0 2(1 − 2) [where = sin ] = 0 64. (a) Let = cos . Then = − sin ⇒
sin cos =
(−) = −122+ = −12cos2 + 1. (b) Let = sin . Then = cos ⇒
sin cos =
= 122+ = 12sin2 + 2. (c)
sin cos = 1
2sin 2 = −14cos 2 + 3
(d) Let = sin , = cos . Then = cos , = sin , so
sin cos = sin2 −sin cos , by Equation 7.1.2, so
sin cos = 12sin2 + 4.
Using cos2 = 1 − sin2and cos 2 = 1 − 2 sin2, we see that the answers differ only by a constant.
65. =
0(sin2 − sin3) = 0
1
2(1 − cos 2) − sin (1 − cos2)
= 0
1
2−12cos 2
+−1
1 (1 − 2)
= cos ,
=− sin
=1
2 − 14sin 2 0+ 21
0(2− 1)
=1
2 − 0
− (0 − 0) + 21
33− 1 0
= 12 + 21 3 − 1
= 12 −43
66. =4
0 (tan − tan2) =4
0 (tan − sec2 + 1)
=
ln |sec | − tan + 4 0
= ln√
2 − 1 + 4
− (ln 1 − 0 + 0)
= ln√
2 − 1 +4
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63. Find the average value of the function f (x) = sin2x cos3x on the interval [−π, π].
1
Solution:
640 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 55. ave = 21
−sin2 cos3 = 21
−sin2 (1 − sin2) cos
= 21 0
0 2(1 − 2) [where = sin ] = 0 56. (a) Let = cos . Then = − sin ⇒
sin cos =
(−) = −122+ = −12cos2 + 1. (b) Let = sin . Then = cos ⇒
sin cos =
= 122+ = 12sin2 + 2. (c)
sin cos = 1
2sin 2 = −14cos 2 + 3
(d) Let = sin , = cos . Then = cos , = sin , so
sin cos = sin2 − sin cos , by Equation 7.1.2, so
sin cos = 12sin2 + 4.
Using cos2 = 1 − sin2and cos 2 = 1 − 2 sin2, we see that the answers differ only by a constant.
57. =
0(sin2 − sin3) = 0
1
2(1 − cos 2) − sin (1 − cos2)
= 0
1
2 −12cos 2
+−1
1 (1 − 2)
= cos ,
=− sin
=1
2 −14sin 2 0 + 21
0(2− 1)
=1 2 − 0
− (0 − 0) + 21
33− 1 0
= 12 + 21 3− 1
= 12 −43
58. =4
0 (tan − tan2) =4
0 (tan − sec2 + 1)
=
ln |sec | − tan + 4
0 =
ln√
2 − 1 +4
− (ln 1 − 0 + 0)
= ln√
2 − 1 + 4
59. It seems from the graph that2
0 cos3 = 0, since the area below the
-axis and above the graph looks about equal to the area above the axis and below the graph. By Example 1, the integral is
sin −13sin32
0 = 0.
Note that due to symmetry, the integral of any odd power of sin or cos between limits which differ by 2 ( any integer) is 0.
60. It seems from the graph that2
0 sin 2 cos 5 = 0, since each bulge above the -axis seems to have a corresponding depression below the
-axis. To evaluate the integral, we use a trigonometric identity:
1
0 sin 2 cos 5 = 122
0[sin(2 − 5) + sin(2 + 5)]
= 122
0[sin(−3) + sin 7]
= 12 1
3cos(−3) −71 cos 72 0
= 12 1
3(1 − 1) −71(1 − 1)
= 0
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71. Find the volume obtained by rotating the region bounded by the curves about the given axis.
y = sin x, y = cos x, 0 ≤ x ≤ π
4; about y = 1 Solution:
SECTION 7.2 TRIGONOMETRIC INTEGRALS ¤ 641 61. Using disks, =
2 sin2 =
2 1
2(1 − cos 2) = 1
2 − 14sin 2
2=
2 − 0 −4 + 0
= 42
62. Using disks,
=
0 (sin2)2 = 22 0
1
2(1 − cos 2)2
= 22
0 (1 − 2 cos 2 + cos22)
= 22 0
1 − 2 cos 2 +12(1 − cos 4)
= 22 0
3
2 − 2 cos 2 −12cos 4
= 23
2 − sin 2 + 18sin 42 0
= 23
4 − 0 + 0
− (0 − 0 + 0)
= 382
63. Using washers,
=4 0
(1 − sin )2− (1 − cos )2
= 4 0
(1 − 2 sin + sin2) − (1 − 2 cos + cos2)
= 4
0 (2 cos − 2 sin + sin2 − cos2)
= 4
0 (2 cos − 2 sin − cos 2) =
2 sin + 2 cos −12sin 24 0
= √
2 +√ 2 −12
− (0 + 2 − 0)
= 2√
2 −52
64. Using washers,
=3
0
[sec − (−1)]2− [cos − (−1)]2
= 3
0 [(sec2 + 2 sec + 1) − (cos2 + 2 cos + 1)]
= 3 0
sec2 + 2 sec − 12(1 + cos 2) − 2 cos
=
tan + 2 ln |sec + tan | −12 −14sin 2 − 2 sin 3 0
= √
3 + 2 ln 2 +√
3
−6 −18
√3 −√ 3
− 0
= 2 ln 2 +√
3
− 162−18√ 3
65. = () =
0sin cos2 . Let = cos ⇒ = − sin . Then
= −1
cos
1 2 = −1
1
33cos
1 = 31 (1 − cos3).
66. (a) We want to calculate the square root of the average value of [()]2= [155 sin(120)]2 = 1552sin2(120). First, we calculate the average value itself, by integrating [()]2over one cycle (between = 0 and = 601, since there are 60cycles per second) and dividing by1
60− 0: [()]2ave= 1601 160
0 [1552sin2(120)] = 60 · 1552160 0
1
2[1 − cos(240)]
= 60 · 15521
2
−2401 sin(240)160
0 = 60 · 15521
2
1
60 − 0
− (0 − 0)
= 15522 The RMS value is just the square root of this quantity, which is 155√
2 ≈ 110 V.
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