Download Section 7.3, Some Trigonometric Integrals

Section 7.3, Some Trigonometric Integrals
Homework: 7.3 #1–31 odds
We will look at five commonly encountered types of trigonometric integrals:
R
R
1. sinn x dx and cosn x dx
R
2. sinm x cosn x dx
R
R
R
3. sin mx cos nx dx, sin mx sin nx dx, and cos mx cos nx dx
R
R
4. tann x dx and cotn x dx
R
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5. tanm x secn dx and cotm x cscn x dx
We will demonstrate how to calculate these by example. Throughout this section, we will be using
many trigonometric identities, including:
sin2 x + cos2 x = 1
1 + tan2 x = sec2 x
1 + cot2 x = csc2 x
1 − cos 2x
sin2 x =
2
1
+
cos
2x
cos2 x =
2
1
sin mx cos nx =
sin(m + n)x + sin(m − n)x
2
1
sin mx sin nx = − cos(m + n)x − cos(m − n)x
2
1
cos mx cos nx =
cos(m + n)x + cos(m − n)x
2
1
Integrals of the form
R
sinn x dx and
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cosn x dx
We will look at examples when n is odd and when n is even. When n is odd, we will use sin2 x +
2x
2x
cos2 x = 1. When n is even, we will use either sin2 x = 1−cos
or cos2 x = 1+cos
.
2
2
Examples
R
1. Find cos5 x dx.
We will use the identity cos2 x = 1 − sin2 x, so we will substitute cos4 x = (1 − sin2 x)2 .
Z
Z
cos5 x dx = (1 − sin2 x)2 cos x dx
Z
=
1 − 2 sin2 x + sin4 x cos x dx
Z
=
cos x − 2 sin2 x cos x + sin4 x cos x dx
= sin x −
2
1
sin3 x + sin5 x + C
3
5
R
2. Find sin4 x dx
We will start by using sin2 x =
Z
Z 4
sin x dx =
1−cos 2x
.
2
1 − cos 2x
2
2
dx
Z
=
=
=
=
=
1
1 − 2 cos 2x + cos2 2x dx
4
Z
Z
Z
1
1
1
dx −
cos 2x dx +
cos2 2x dx
4
2
4
Z
x 1
1
1 + cos 4x
− sin 2x +
dx
4 4
4
2
x
1
x 1
− sin 2x + +
sin 4x + C
4 4
8 32
3x 1
1
− sin 2x +
sin 4x + C,
8
4
32
where we also used that cos2 x =
2
Integrals of the form
1+cos 2x
2
R
in the third-to-last line.
sinm x cosn x dx
If either m or n is an odd positive integer, we will use the identity sin2 x + cos2 x = 1. If both m
and n are even and positive, we will use the half-angle identities.
Examples
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1. Find cos5 x sin−4 x dx
Since the exponent for cosine is odd, we will replace cos4 x by (1−sin2 x)2 = 1−2 sin2 x+sin4 x:
Z
Z
−4
5
cos x sin x dx = cos x(1 − 2 sin2 x + sin4 x) sin−4 x dx
Z
Z
Z
= cos x sin−4 x dx − 2 cos x sin−2 x dx + cos x dx
1
= − (sin x)−3 + 2(sin x)−1 + sin x + C
3
(Be careful with notation, since sin−1 x refers to the inverse sine function, not 1/(sin x).)
R
2. Find cos2 x sin4 x dx.
2
4
1+cos 2x
1−cos 2x
2
We will substitute cos x =
and sin x =
. Then,
2
2
Z
cos2 x sin4 x dx =
=
=
=
=
=
=
2
1 + cos 2x 1 − cos 2x
dx
2
2
Z
1
(1 + cos 2x)(1 − 2 cos 2x + cos2 2x) dx
8
Z
1
(1 − cos 2x − cos2 2x + cos3 2x) dx
8
Z
Z
Z
Z
1
1
1
1
2
dx −
cos 2x dx −
cos 2x dx +
cos3 2x dx
8
8
8
8
Z
Z
x
1
1
1 + cos 4x
1
−
sin 2x −
dx +
cos 2x(1 − sin2 2x) dx
8 16
8
2
8
x
1
x
1
1
1
−
sin 2x −
−
sin 4x +
sin 2x −
sin3 2x + C
8 16
16 64
16
48
x
1
1
−
sin 4x −
sin3 2x + C
16 64
48
Z
3
RIntegrals of the form
cos mx cos nx dx
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sin mx cos nx dx,
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sin mx sin nx dx, and
1
Here,
we
will
use
the
identities
sin
mx
cos
nx
=
sin(m
+
n)x
+
sin(m
−
n)x
, sin mx sin nx =
2
1
1
− 2 cos(m + n)x − cos(m − n)x , and cos mx cos nx = 2 cos(m + n)x + cos(m − n)x .
Examples
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1. Find sin 4x cos 5x dx Since sin mx cos nx = 21 sin(m + n)x + sin(m − n)x , we will use this with m = 4 and n = 5:
Z
Z
1
sin 9x + sin(−x) dx
sin 4x cos 5x dx =
2
1
1
= − cos 9x + cos x + C,
18
2
where we used that cos(−x) = cos x.
Rπ
2. Find −π sin mx sin nx dx, where m and n are positive integers.
First, consider m 6= n. Then,
Z π
Z
1 π sin mx sin nx dx = −
cos(m + n)x − cos(m − n)x dx
2
−π
−π
π
1
1
1
sin(m + n)x −
sin(m − n) =−
2 m+n
m−n
−π
=0
If m = n, then
Z π
Z
1 π sin mx sin nx dx = −
cos(2m)x − 1 dx
2 −π
−π
π
1 1
=−
sin(2m)x − x 2 2m
−π
=π
4
Integrals of the form
R
tann x dx and
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cotn x dx
In the tangent case, we will use tan2 x = sec2 x − 1. In the cotangent case, we will use cot2 x =
csc2 x − 1. Here, we will only replace tan2 x or cot2 x, distribute, integrate what we can, then repeat
as necessary.
Examples
1. Find
R
Z
tan4 x dx
tan4 x dx =
Z
tan2 x(sec2 x − 1) dx
Z
Z
2
2
= tan x sec x dx − tan2 x dx
Z
1
= tan3 x − (sec2 x − 1) dx
3
1
= tan3 x − tan x + x + C
3
2. Find
R
Z
5
cot5 x dx
cot5 x dx =
Z
cot3 x(csc2 x − 1) dx
Z
Z
3
2
= cot x csc x − cot3 x dx
Z
1
= − cot4 x − cot x(csc2 x − 1) dx
4
Z
Z
1
cos x
= − cot4 x − cot x csc2 x dx +
dx
4
sin x
1
1
= − cot4 x + cot2 x + ln | sin x| + C
4
2
Integrals of the form
R
tanm x secn dx and
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cotm x cscn x dx
If n is even, use either sec2 x = tan2 x + 1 or csc2 x = cot2 x + 1 to replace all but 2 powers of sec x
or csc x, then you can use a u-substitution to integrate.
If m is odd, we will use that the derivative of sec x is sec x tan x (or the derivative of csc x is
− csc x cot x), and replace m − 1 powers of either tangent or cotangent using a Pythagorean identity.
Examples
1. Find
R
Z
tan1/2 x sec4 x dx
tan1/2 x sec4 x dx =
tan1/2 x(tan2 +1) sec2 x dx
Z
Z
= tan5/2 x sec2 x dx + tan1/2 x sec2 x dx
=
2. Find
R
Z
Z
2
2
tan7/2 x + tan3/2 x + C
7
3
cot3 x csc3/2 x dx
cot3 x csc3/2 x dx =
Z
cot x(csc2 x − 1) csc3/2 x dx
Z
Z
= cot x csc x csc5/2 x dx − cot x csc x csc1/2 x dx
2
2
= − csc7/2 x + csc3/2 +C
7
3