Download 7.4 Trigonometric Integrals

7.4. Trigonometric Integrals
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7.4 Trigonometric Integrals
1. Let u = cos x. Then du = − sin x dxx.
Z
cos4 x sin x dx = −
Z
u4 du
u5
+C
5
cos5 x
+C
=−
5
=−
2. Let u = 5ϕ. Then du = 5dϕ.
Z
sin2 5ϕdϕ =
=
=
=
=
1
sin2 u du
5
Z
1 1
(1 − cos 2u)du
5 2
1
1
u − sin 2u +C
10
2
1
1
5ϕ − sin(2 × 5ϕ) +C
10
2
1
1
ϕ − sin(10ϕ) +C
2
20
Z
3. Let u = sin 2z. Then du = 2 cos 2z dz.
Z
sin2 2z cos3 2z dz =
Z
sin2 2z cos3 2z(1 − sin2 2z)dz
Z
2
sin 2z cos 2z dz −
=
Z
sin4 2z cos 2z dz
1
1
u2 du −
u4 du
2
2
u3 u5
= − +C
6 10
sin3 2z sin5 2z
=
−
+C
6
10
Z
Z
=
4.
Z
sin x cos
Let u = cos
234
x
2
. Then du = − 12 sin
x
2
x
dx.
2
x
x
cos
cos
dx
2
2
2
Z
x x
= 2 cos2
sin
dx
2
2
Z
dx =
2 sin
x
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Chapter 7. Integration Techniques, Solution Key
Z
2 cos
2
x
2
sin
x
2
dx = −4
Z
u2 du
4u3
+C
3
x
4
= − cos3
+C
3
2
=−
5. Let u = sec x. Then du = sec x tan x dx.
Z
sec4 x tan3 xdx =
Z
tan2 x sec3 x(sec x tan x)dx
Z
(sec2 x − 1) sec3 x(sec x tan x)dx
Z
sec5 x(sec x tan x)dx −
Z
u5 du −
=
=
=
Z
Z
sec3 x(sec x tan x)dx
u3 du
u6 u4
− +C
6
4
sec6 x sec4 x
=
−
+C
6
4
=
6.
Z
tan4 x sec xdx =
=
Z
(sec2 x − 1)(sec2 x − 1) sec x dx
Z
(sec5 x − 2 sec3 x + sec x)dx
sec3 x tan x 3
sec x tan x
1
+
sec3 x dx − 2
−2×
sec x dx + ln|sec x + tan x|+C
4
4
2
2
Z
sec x tan x
sec3 x tan x 3 sec x tan x 1
+
+
sec x dx − 2
− ln|sec x + tan x|+ − ln|sec x + tan x|+C
=
4
4
2
2
2
sec3 x tan x 5 sec x tan x 3
=
−
+ ln|sec x + tan x|
4
8
8
Z
Z
=
7.
Z √
tan x sec4 x dx =
Z
Z
1
tan 2 x sec4 x dx
1
tan 2 x(tan2 x + 1) sec2 xdx
Z h
i
5
1
=
tan 2 x sec2 x + tan 2 x sec2 x dx
=
Let u = tan x. Then du = sec2 x dx.
235
7.4. Trigonometric Integrals
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Z h
Z
Z
i
5
1
5
1
tan 2 x sec2 x + tan 2 x sec2 x dx = u 2 du + u 2 du
7
=
u2
7
2
3
+
u2
3
2
7
+C
3
2 tan 2 x 2 tan 2 x
+
+C
=
7
3
8. Let u = 2x . Then du = 12 dx. If x = 0, then u = 0. If x = π2 , then u = π4 .
π
Z2
π
tan5
x
2
Z4
dx =
0
2 tan5 u du
0
π
=
2 tan4 u π
4
4
−2
Z4
0
tan3 u du
0
π
√ !4
Z2
2
2 tan4 u π4
−
=2
+ 2 tan u du
2
2
0
0
√ !2
π
2
8
−
=
− 2ln|cos u|04
16
2
π 8
=
− 1 − 2ln cos
−0
16
4
√ 2
1
= − − 2ln 2 2
4
1
= − + ln 2
2
1
= − + ln2
2
9.
236
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Chapter 7. Integration Techniques, Solution Key

π
π
Z4
Z4

V = π
cos2 x dx −
0


sin2 x dx
0
π
π
1
x 4
x 4
1
cos x sin x +
− π − cos x sin x +
2
2
2
2 0
0
1 π
1 π
+
− − +
=π
4 8
4 8
1
=π
2
π
=
2
=π
10.
a.
R
x+cot x
csc x dx = csc x csc
csc x+cot x dx
R
Let u = csc x + cot x. Then du = (− csc x cot x − csc2 x)dx.
Z
1
du
u
= −ln|u|+C
csc xdx = −
Z
= −ln|csc x + cot x|+C
b.
Z
1
dx
sin x
Z
sin x
=
dx
2
sin
x
Z
Z
csc x dx =
=
sin x
dx
(1 − cos x)(1 + cos x)
Let u = cos x. Then du = − sin x dx.
237
7.4. Trigonometric Integrals
Z
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sin x
dx = −
(1 − cos x)(1 + cos x)
du
(1 − u)(1 + u)
Z
Z
A
B
=
du +
du
1−u
1+u
=−
Z
=
Z
Z
1
2
1−u
du −
1
2
1 − cos x
Z
du −
1
2
1+u
Z
du
1
2
1 + cos x
du
1
1
= |1 − cos x|− |1 + cos x|+C
2
2
1 1 − cos x = +C
2 1 + cos x r
1 − cos x
+C
= ln
x
1+xcos
= ln tan
+C
2
OR
r
ln
1 − cos x
+C = ln
1 + cos x
s
s
(1 − cos x) (1 − cos x)
·
+C
(1 + cos x) (1 − cos x)
(1 − cos x)2
+C
1 − cos2 x
1 − cos x +C
= ln sin x 1
cos x = ln −
+C
sin x sin x = ln|csc x − cot x|+C
= ln
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