Download Section 8.2: Trigonometric Integrals

Section 8.2: Trigonometric Integrals
In this section, we will make use of the following trigonometric identities:
• sin2 x + cos2 x = 1
• 1 + tan2 x = sec2 x
1
• sin2 x = (1 − cos 2x)
2
1
• cos2 x = (1 + cos 2x)
2
Consider integrals of the form
Z
sinn x cosm xdx.
There are three cases to consider:
1. If n is odd, factor out one sin x, use the identity sin2 x = 1 − cos2 x, and use the
substitution u = cos x.
2. If m is odd, factor out one cos x, use the identity cos2 x = 1 − sin2 x, and use the
substitution u = sin x.
3. If both n and m are even, use the identities
sin2 x =
Z
1 − cos(2x)
2
and
cos2 x =
1 + cos(2x)
.
2
sin2 x cos3 xdx.
Example: Evaluate
Z
2
Z
3
sin x cos xdx =
Z
=
Z
=
Z
=
sin2 x cos2 x cos xdx
sin2 x(1 − sin2 x) cos xdx
(sin2 x − sin4 x) cos xdx
(u2 − u4 )du
1 3 1 5
u − u +C
3
5
1 3
1
=
sin x − sin5 x + C.
3
5
=
Z
Example: Evaluate
Z
sin3 x cos4 xdx.
3
Z
4
sin x cos xdx =
Z
=
Z
=
Z
=
sin2 x cos4 x sin xdx
(1 − cos2 x) cos4 x sin xdx
(cos4 x − cos6 x) sin xdx
(u6 − u4 )du
1 7 1 5
u − u +C
7
5
1
1
cos7 x − cos5 x + C.
=
7
5
=
Z
Example: Evaluate
Z
√
cos5 x sin xdx.
Z
√
√
cos x sin xdx =
cos4 x sin x cos xdx
Z
√
=
(1 − sin2 x)2 sin x cos xdx
Z
√
=
(1 − u2 )2 udu
Z
√
=
(1 − 2u2 + u4 ) udu
Z
=
(u1/2 − 2u5/2 + u9/2 )du
5
2 3/2 4 7/2
2
u − u + u11/2 + C
3
7
11
2 3/2
4 7/2
2
=
sin x − sin x +
sin11/2 x + C.
3
7
11
=
π/2
Z
sin3 x cos3 xdx.
Example: Evaluate
0
π/2
Z
3
Z
3
π/2
sin3 x cos2 x cos xdx
sin x cos xdx =
0
0
Z
π/2
sin3 x(1 − sin2 x) cos xdx
=
0
Z
π/2
=
(sin3 x − sin5 x) cos xdx
Z0 1
(u3 − u5 )du
0
1
1 4 1 6 =
u − u
4
6 0
1 1
−
=
4 6
1
=
.
12
=
Z
Example: Evaluate
Z
sin2 x cos2 xdx.
2
Z
2
sin x cos xdx =
=
=
=
=
=
1
1
(1 − cos 2x) (1 + cos 2x)dx
2
2
Z
1
(1 − cos2 2x)dx
4
Z
1
sin2 2xdx
4
Z
1
(1 + cos 4x)dx
8
1
1
x + sin 4x + C
8
4
x
1
+
sin4 x + C.
8 32
Z
Example: Evaluate
Let u =
√
√
sin2 ( x)
√
dx.
x
1
x, du = √ dx. Then
2 x
√
Z
Z
sin2 ( x)
√
dx = 2 sin2 udu
x
Z
=
(1 − cos 2u)du
1
sin 2u + C
2
√
√
1
=
x − sin(2 x) + C.
2
= u−
Consider integrals of the form
Z
secn x tanm xdx.
There are three cases to consider:
1. If m is odd, factor out sec x tan x, use the identity tan2 x = sec2 x − 1, and use the
substitution u = sec x.
2. If n is even, factor out sec2 x, use the identity sec2 x = 1 + tan2 x, and use the substitution u tan x.
3. If m is even and nR is odd, use the identity tan2 x = sec2 x − 1 and use integration by
parts to evaluate secn xdx, where n is odd.
Z
Example: Evaluate
Z
sec3 x tan3 xdx.
3
Z
3
sec x tan xdx =
Z
=
Z
=
Z
=
sec2 x tan2 x sec x tan xdx
sec2 x(sec2 x − 1) sec x tan xdx
(sec4 x − sec2 x) sec x tan xdx
(u4 − u2 )du
1 5 1 3
u − u +C
5
3
1
1
5
=
sec x − sec3 x + C.
5
3
=
Z
Example: Evaluate
Z
tan2 x sec4 xdx.
2
Z
4
tan x sec xdx =
Z
=
Z
=
Z
=
tan2 x sec2 x sec2 xdx
tan2 x(1 + tan2 x) sec2 xdx
(tan2 x + tan4 x) sec2 xdx
(u2 + u4 )du
1 3 1 5
u + u +C
3
5
1
1
=
tan3 x + tan5 x + C.
3
5
=
Z
Example: Evaluate
sec3 xdx.
First,
Z
3
sec xdx =
Z
sec x sec2 xdx.
Let u = sec x and dv = sec2 xdx. Then du = sec x tan xdx and v = tan x. So
Z
Z
3
sec xdx = sec x tan x − sec x tan2 xdx
Z
= sec x tan x − sec x(sec2 x − 1)dx
Z
Z
3
= sec x tan x − sec xdx + sec xdx
Z
= sec x tan x − sec3 xdx + ln | sec x + tan x|
Z
2 sec3 xdx = sec x tan x + ln | sec x + tan x|
Z
1
1
sec3 xdx =
sec x tan x + ln | sec x + tan x| + C.
2
2