Math 101 – SOLUTIONS TO WORKSHEET 13
TRIGONOMETRIC INTEGRALS
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0
0
Formulas to memorize: (sin x) = cos x, (cos x) = sin x, (tan x) = sec2 x,
cos(2x) = 2 cos2 x − 1
sin(2x) = 2 sin x cos x
cos2 x =
1 + cos(2x)
2
sin2 x =
1 − sin(2x)
x
(1) Evaluate
the integrals
´
(a) sin4 x cos3 x dx
Solution: The power of cosine is odd, while the power of sine is even, so let u = sin x,
du = cos x dx. Then sin4 x = u4 , cos2 x = 1 − u2 and
ˆ
ˆ
sin4 x cos3 x dx
u7
u5
+
+C
u4 − u6 du =
5
7
1
1
sin5 x + sin7 x + C .
5
7
=
(b)
ˆ
u4 (1 − u2 ) du =
=
´
sin5 x cos4 x dx
Solution: This time let u = cos x, du = − sin x. Then
ˆ
ˆ
5
4
sin x cos x dx
sin2 x
2
cos4 x sin x dx
ˆ
2
1 − u2 u4 du =
1 − 2u2 + u4 u4 du
=
ˆ
= −
ˆ
2
1
1
=
2u6 − u4 − u8 du = u7 − u5 − u9 + C
7
5
9
1
1
2
cos7 x − cos5 x − cos9 x + C
=
7
5
9
(c)
´
sin4 x cos4 x dx
Solution: We have sin x cos x =
ˆ
sin4 x cos4 x dx
=
=
=
=
=
1
16
1
2
sin(2x) so
ˆ
sin4 (2x) dx
2
ˆ 1
1 − cos(4x)
dx
16
2
ˆ 1
1
1
2
1 − cos(4x) + cos (4x) dx
16
2
4
ˆ
1
1
1
1 + cos(8x)
x − sin(4x) +
dx
16
8
64
2
1
1
1
1
x−
sin(4x) +
x+
sin(8x) + C .
16
128
128
64 · 8
Date: 1/2/2016, Worksheet by Lior Silberman. This instructional material is excluded from the terms of UBC Policy 81.
1
Solution: Use sin2 x = 1−cos(2x)
, cos2 x = 1+cos(2x)
to get
2
2
ˆ
ˆ
1
2
2
sin4 x cos4 x dx =
(1 − cos(2x)) (1 + cos(2x))
16
ˆ
1
2
((1 − cos(2x))(1 + cos(2x))) dx
=
16
ˆ
2
1
=
1 − cos2 (2x) dx
16
ˆ
1
=
1 − 2 cos2 (2x) + cos(4x) dx
2 cos2 θ − 1 = cos(2θ)
16
ˆ
1
=
(cos(4x) − cos(8x)) dx
16
1
1
sin(4x) −
sin(8x) + C .
=
64
128
(2) Powers of tangent and secant
´ π/4
(a) Evaluate 0 tan x dx
´ π/4
´ π/4 sin x
Solution:
tan x = 0 cos
x dx. We note that sin x dx is roughly d(cos x) = − sin x dx
0
so letting u = cos x we have
ˆ π/4
ˆ x=π/4
− du
u=cos(π/4)
= − [log |u|]u=1
tan x =
u
0
x=0
1
1
1
= − log( √ ) − log 1 = − log √ =
log 2 .
2
2
2
´ +π/4
(b) Evaluate −π/4 tan x dx
Solution: Since tan(−x) = tan(x) and the domain is symmetric about x = 0, the integral
vanishes. In detail, let u = −x. Then du = − dx abd
ˆ x=+π/4
ˆ u=−π/4
tan x dx =
tan(−u)(− du)
x=−π/4
u=π/4
ˆ
u=−π/4
=
tan(u) du
tan(−u) = − tan(u)
u=π/4
ˆ
ˆ
u=π/4
= −
tan(u) du
u=−π/4
ˆ
−
ˆ
a
b
=
b
a
x=+π/4
= −
tan(x) dx
x=−π/4
2
substitute u = x.