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8-30-2016
Trigonometric Integrals
For trig integrals involving powers of sines and cosines, there are two important cases:
1. The integral contains an odd power of sine or cosine.
2. The integral contains only even powers of sines and cosines.
I will look at the odd power case first. It turns out that the same idea can be used to integrate some
powers of secants and tangents, so I’ll digress to do some examples of those as well.
Example. Compute
Z
Z
(sin 5x)3 1 + 4(cos 5x)2 dx.
(sin 5x)3 1 + 4(cos 5x)2 dx =
Z
1 − (cos 5x)2
u = cos 5x,
Z
(sin 5x)2 1 + 4(cos 5x)2 sin 5x dx =
1 + 4(cos 5x)2 sin 5x dx =
du = −5 sin 5x dx,
du
dx = −
5 sin 5x
Z
Z
4
1
1
1
(1 − u2 )(1 + 4u2 ) du = −
(1 + 3u2 − 4u4 ) du = − (u + u3 − u5 ) + C =
5
5
5
5
4
1
1
4
1
cos 5x + (cos 5x)3 − (cos 5x)5 + C = − cos 5x − (cos 5x)3 + (cos 5x)5 + C.
−
5
5
5
5
25
−
In this example, the key point was in the second line. I obtained an integral with lots of cos 5x’s and a
single sin 5x. This allowed me to make the substitution u = cos 5x, because the sin 5x was available to make
du.
I got the sin 5x by “pulling it off” the odd power of sin 5x. Then I converted the rest of the stuff to
cos 5x’s using the identity (sin θ)2 + (cos θ)2 = 1. This is the generic procedure when you have at least one
odd power of sine or cosine.
Example. Compute
Z Z 5(sin x)2/3 + 1 (cos x)3 dx.
Z 5(sin x)2/3 + 1 1 − (sin x)2 (cos x) dx =
5(sin x)2/3 + 1 (cos x)3 dx =
Z 5u2/3 + 1
1 − u2 du =
u = sin x,
Z du = cos x dx,
du
dx =
cos x
15
1
5u2/3 + 1 − 5u8/3 − u2 du = 3u5/3 + u − u11/3 − u3 + C =
11
3
3(sin x)5/3 + sin x −
1
15
(sin x)11/3 − (sin x)3 + C.
11
3
1
If you have an integral involving sines and cosines in which all the powers are even, the method I just
described usually won’t work. Instead, it is better to apply the following double angle formulas:
1
(1 − cos 2θ)
2
1
(cos θ)2 = (1 + cos 2θ)
2
(sin θ)2 =
Any even power of sin x or cos x can be expressed as a power of (sin x)2 or (cos x)2 . Use the identities
above to substitute for (sin x)2 or (cos x)2 , and multiply out the result. The net effect is to reduce the
powers that occur in the integral, while at the same time increasing the arguments (x → 2x).
Example. Compute
Z
Z
(cos 5x)2 (sin 5x)2 dx.
Z
1
1
1
1 − (cos 10x)2 dx =
(1 + cos 10x)
(1 − cos 10x) dx =
(cos 5x) (sin 5x) dx =
2
2
4
Z
Z
1
1
1
1
1
(1 − cos 20x) dx = (x −
sin 20x) + C.
(sin 10x)2 dx =
4
4
2
8
20
2
2
Example. Compute
Z
Z (sin 5x)4 dx.
I’ll use the double angle formula (twice):
Z
2
Z
1
1 1 − 2 cos 10x + (cos 10x)2 dx =
(1 − cos 10x) dx =
2
4
Z 1
1
1
1
1
1
x − sin 10x + (x +
1 − 2 cos 10x + (1 + cos 20x) dx =
sin 20x) + c =
4
2
4
5
2
20
(sin 5x)4 dx =
Z
[(sin 5x)2 ]2 dx =
Z 1
1
3
x−
sin 10x +
sin 20x + c.
8
20
160
Example. Why would it be a bad idea to use the double angle formulas to compute
Z
(cos x)3 dx?
Suppose I try to apply the double angle formula for cosine:
Z
Z
Z
1
3
2
(1 + cos 2x) cos x dx.
(cos x) dx = (cos x) cos x dx =
2
The integral can be done in this form, but you either need to apply one of the angle addition formulas to
cos 2x cos x or use integration by parts. The problem is that having trig functions with different arguments
in the same integral makes the integral a bit harder to do.
It would have been better to do the integral by using the “odd power” technique:
Z
Z
Z
3
2
1 − (sin x)2 cos x dx =
(cos x) dx = (cos x) cos x dx =
u = sin x,
du = cos x dx,
2
dx =
du
cos x
Z
1
1
(1 − u2 ) du = u − u3 + C = sin x − (sin x)3 + C.
3
3
To integrate some powers of secants and tangents, here are two useful approaches:
1. Use (sec θ)2 = 1 + (tan θ)2 to convert the integrand to something with lots of tan θ’s and a single
(sec θ)2 . Then substitute u = tan θ.
2. Use (tan θ)2 = (sec θ)2 − 1 to convert the integrand to something with lots of sec θ’s and a single
sec θ tan θ. Then substitute u = sec θ.
Example. Compute
Z
Z
(sec 3x)4 dx =
(sec 3x)4 dx =
Z
Z
(sec 3x)2 (sec 3x)2 dx.
(sec 3x)2 (sec 3x)2 dx =
u = tan 3x,
1
3
Z
(1 + u2 ) du =
Z
2
du = 3(sec 3x) dx,
1 + (tan 3x)2 (sec 3x)2 dx =
du
dx =
3(sec 3x)2
1
1
1
1
(u + u3 ) + C = tan 3x + (tan 3x)3 + C.
3
3
3
9
In this example, I pulled off a (sec 3x)2 , then converted the rest of the stuff to tan 3x’s using (sec θ)2 =
1 + (tan θ)2 . The (sec 3x)2 was exactly what I needed to make du for the substitution u = tan 3x.
Notice that the argument 3x did not play an important role in the problem.
Example. Compute
Z
Z
(sec 6x)(tan 6x)3 dx.
3
(sec 6x)(tan 6x) dx =
Z
2
(tan 6x) (sec 6x tan 6x dx) =
Z
(sec 6x)2 − 1 (sec 6x tan 6x dx) =
u = sec 6x,
du
du = 6 sec 6x tan 6x dx, dx =
6 sec 6x tan 6x
Z
Z
du
1 3
2
2
(u − 1)(sec 6x tan 6x) ·
= 6 (u − 1) du = 6
u − u + C = 2(sec 6x)3 − 6 sec 6x + C.
6 sec 6x tan 6x
3
In this example, I pulled off a sec 6x tan 6x, then converted the rest of the stuff to sec 6x’s using (tan θ)2 =
(sec θ)2 − 1. The sec 6x tan 6x was exactly what I nneded to make du for the substitution u = sec 6x.
Example. Compute
Z
Z
Z
(tan θ)3 dθ.
3
(tan θ) dθ =
2
Z
(sec θ) tan θ dθ −
2
(tan θ) tan θ dθ =
Z
tan θ dθ =
Z
3
Z
(sec θ)2 − 1 tan θ dθ =
sec θ(sec θ tan θ dθ) −
Z
sin θ
dθ.
cos θ
I can do the first integral using u = sec θ, so du = sec θ tan θ dθ and dθ =
Z
sec θ(sec θ tan θ dθ) =
Z
u du =
1 2
1
u + C = (sec θ)2 + C.
2
2
I can do the second integral using w = cos θ, so dw = − sin θ dθ and dθ =
Z
Therefore,
Example. Compute
sin θ
dθ =
cos θ
Z
Z
Z
du
:
sec θ tan θ
dw
:
− sin θ
1
dw = ln |w| + C = ln | cos θ| + C.
w
(tan θ)3 dθ =
1
(sec θ)2 + ln | cos θ| + C.
2
1
dx.
(sec x)2 tan x
In this problem, I’ll use the identity
(sec x)2 − (tan x)2 = 1.
Applying this to the top of the fraction, I get
Z
Z Z
(sec x)2 − (tan x)2
(sec x)2
(tan x)2
1
dx
=
dx
=
−
dx =
(sec x)2 tan x
(sec x)2 tan x
(sec x)2 tan x (sec x)2 tan x
Z Z
Z
Z
1
sin x
tan x
2
dx
=
cot
x
dx
−
−
·
(cos
x)
dx
=
ln
|
sin
x|
−
sin x cos x dx =
tan x (sec x)2
cos x
1
ln | sin x| − (sin x)2 + C.
2
I used the formula
Z
cot x dx = ln | sin x| + C.
cos x
If you didn’t know this, you could derive it by writing cot x =
. Then substitute u = sin x.
sin x
Z
I used u = sin x to do sin x cos x dx.
Example. Compute
Z
sec x dx.
This integral uses a trick:
Z
Z
Z
sec x + tan x
(sec x)2 + sec x tan x
sec x dx = sec x ·
dx =
dx =
sec x + tan x
sec x + tan x
du
u = sec x + tan x, du = sec x tan x + (sec x)2 dx, dx =
sec x tan x + (sec x)2
Z
Z
(sec x)2 + sec x tan x
du
du
=
·
= ln |u| + C = ln | sec x + tan x| + C.
2
u
sec x tan x + (sec x)
u
4
Example. Compute
I can compute
Z
Z
Z
(sec x)3 dx.
(sec x)3 dx using parts:
(sec x)3 dx =
Z
(sec x)(sec x)2 dx = sec x tan x −
Z
d
dx
sec x tan x −
sec x tan x +
Z
Z
dx
sec x
−
sec x tan x
ց
→
tan x
sec x (sec x) − 1 dx = sec x tan x −
sec x dx −
Z
sec x(tan x)2 dx =
(sec x)2
+
2
Z
Z
(sec x)3 − sec x dx =
(sec x)3 dx = sec x tan x + ln | sec x + tan x| −
Z
(sec x)3 dx.
Thus,
2
Z
(sec x)3 dx = sec x tan x + ln | sec x + tan x| −
Z
(sec x)3 dx = sec x tan x + ln | sec x + tan x|
Z
(sec x)3 dx =
Z
(sec x)3 dx
1
1
sec x tan x + ln | sec x + tan x| + C
2
2
This integral also comes up a lot, so you should make a note of it.
Remark. Using the methods of the last two examples, you can show:
Z
csc x dx = − ln | csc x + cot x| + C.
Z
1
1
(csc x)3 dx = − csc x cot x − ln | csc x + cot x| + C.
2
2
In general, integrals involving powers of cosecant and cotangent use the same ideas as integrals involving
powers of secant and tangent.
Z
Example. Compute (cot 5x)2 dx.
Remember the trig identity
(csc x)2 = 1 + (cot x)2 .
So
Z
(cot 5x)2 dx =
Z
1
(csc 5x)2 − 1 dx = − cot 5x − x + C.
5
The examples show that certain patterns that arise in trig integrals are good, in the sense that they
allow you to do a substitution which makes the integral easy. Here is a review of some of the “good patterns”:
• Lots of cos x’s and a single sin x.
5
• Lots of sin x’s and a single cos x.
• Lots of tan x’s and a single (sec x)2 .
• Lots of sec x’s and a single sec x tan x.
• Lots of cot x’s and a single (csc x)2 .
• Lots of csc x’s and a single csc x cot x.
You should aim for these patterns whenever possible.
c 2016 by Bruce Ikenaga
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