Math 166
September 21-22, 2015
Trigonometric integrals
In this section we learn how to handle various integrals involving trigonometric functions.
0.1
Powers of sine and cosine
The following summarizes the general idea of computing integrals involving
products of powers of sine and cosine.
R
In order to compute sinm x cosn x dx when either m or n is odd we
do the following. If m is odd, then m = 2k + 1 for some k, and we write
Z
Z
m
n
sin x cos x dx = sin x sin2k x cosn x dx
Z
= sin x(1 − cos2 x)k cosn x dx,
and then use the substitution u = cos x.
If n is odd, then n = 2j + 1 for some j, and we write
Z
Z
sinm x cosn x dx = sinm x cos x cos2j x dx
Z
= sinm x cos x(1 − sin2 x)j dx,
and then use the substitution u = sin x.
If m and n are both even use the trigonometric identities
sin2 x =
1 − cos 2x
,
2
cos2 x =
1
1 + cos 2x
.
2
Example 1
Z
sin3 x cos2 x dx
The idea will be to use the trigonometric identity sin2 x+cos2 x = 1 together
with a u-substitution. Observe that
Z
Z
sin3 x cos2 x dx = sin x sin2 x cos2 x dx
Z
= sin x(1 − cos2 x) cos2 x) dx.
Now, using u = cos x, we get
Z
Z
sin x(1 − cos2 x) cos2 x) dx = − (1 − u2 )u2 du
Z
= − (u2 − u4 ) du
1 5 1 3
u − u +C
5
3
1
1
5
= cos x − cos3 x + C
5
3
=
Example 2
cos5 x dx
We’ll use the same trick as above. That is, we’ll first use our favorite trigonometric identity, followed by a u-substitution. Observe that,
Z
Z
5
cos x dx = cos x cos4 x dx
Z
= cos x(cos2 x)2 dx
Z
= cos x(1 − sin2 x)2 dx.
Now, using u = sin x, we get
Z
Z
cos x(1 − sin2 x)2 dx = (1 − u2 )2 du
Z
= (1 − 2u2 + u4 ) du
1
2
= u − u3 + u5 + C
3
5
2
1
= sin x − sin3 x + sin5 x + C.
3
5
2
Example 3
Z
sin2 x cos4 x dx
The previous technique no longer works here. Instead we will use the following trigonometric identities:
sin2 x =
1 − cos 2x
2
and
cos2 x =
1 + cos 2x
.
2
Observe that
2
Z
Z
1 − cos 2x 1 + cos 2x
2
4
sin x cos x dx =
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
1
1
2
3
=
x + sin 2x − cos 2x dx − cos 2x dx .
8
2
For the first remaining integral above, we may apply the same trick:
Z
Z
1 + cos 4x
dx
cos2 2x dx =
2
1
1
=
x + sin 4x + C.
2
4
For the second integral, we apply the technique used in Examples 1 and 2:
let u = sin 2x and observe that
Z
Z
3
cos 2x dx = cos 2x(1 − sin2 2x) dx
Z
1
=
(1 − u2 ) du
2
1
1
= u − u3 + C
2
6
1
1
= sin 2x − sin3 2x + C
2
6
3
0.2
Eliminating square roots
For integrals involving square roots of functions of sine and cosine, the following
identities are sometimes useful.
sin2 x =
1 − cos 2x
,
2
cos2 x =
1 + cos 2x
2
Example
Z
π/4
√
1 + cos 4x dx
0
Note that using the second identity above, we have that
1 + cos 4x = 2 cos2 2x.
Using this in our integral we get
Z
π/4
√
Z
π/4
1 + cos 4x dx =
0
√
2 cos 2x dx
0
√
π/4
2
sin 2x
2
0
√
2
=
.
2
=
0.3
Powers of secant and tangent
The main tools for integrals involving products of powers of secant and tangent
are the following:
• tan2 x + 1 = sec2 x
•
d
dx
[tan x] = sec2 x
•
d
dx
[sec x] = sec x tan x
The following rules of thumb might be useful in guiding you through these
problems.
4
Z
Rules of thumb for computing integrals of the form
secn x tanm x dx:
• If n is even, the integral will involve the substitution u = tan x.
• If n is odd and m is odd, the integral will involve the substitution u = sec x.
• If n is odd and m is even, the integral will require integration by parts.
Example 4
Z
tan4 x dx
In a similar fashion, we apply our favorite identity. Observe that
Z
Z
tan4 x dx = tan2 x tan2 x dx
Z
= tan2 x(sec2 x − 1) dx
Z
Z
= tan2 x sec2 x dx − tan2 x dx
Z
Z
2
2
= tan x sec x dx − (sec2 x − 1) dx
For the first integral, we use the substitution u = tan x. Indeed,
Z
Z
1
tan2 x sec2 x dx = u2 du = tan3 x + C.
3
The second integral is straightforward:
Z
(sec2 x − 1) dx = tan x − x + C.
We conclude that
Z
tan4 x dx =
1
tan3 x − tan x + x + C.
3
Example 5
Z
sec3 x tan x dx
Here, we start by isolating sec x tan x with a view toward using the substi-
5
tution u = sec x, du = sec x tan x dx. Observe that
Z
Z
sec3 x tan x dx = sec2 x sec x tan x dx
Z
= u2 du
=
1
sec3 x + C
3
Example 6
Z
sec3 x dx
Here, the previous techniques fail. Instead, we use integration by parts. Let
u = sec x,
dv = sec2 x dx.
Then
du = sec x tan x dx,
v = tan x.
Therefore,
Z
sec3 x dx = sec x tan x −
Z
sec x tan2 x dx
Z
sec x(sec2 x − 1) dx
Z
Z
= sec x tan x + sec x dx − sec3 x dx.
= sec x tan x −
R
Now, solving the above algebraic equation for sec3 x dx, we get that
Z
Z
1
1
sec x dx.
sec3 x dx = sec x tan x +
2
2
R
Note that we can compute sec x dx by multiplying and dividing the integrand
by sec x + tan x and using the substitution u = sec x + tan x. Note then that
du = (sec x tan x + sec2 x) dx. Thus
Z
Z
sec x + tan x
sec x dx = sec x
dx
sec x + tan x
Z
sec2 x + sec x tan x
=
dx
sec x + tan x
Z
du
=
u
= ln | sec x + tan x| + C
We conclude that
Z
1
1
sec3 x dx = sec x tan x + ln | sec x + tan x| + C.
2
2
6
0.4
Problems
The following problems were given as in class exercises on September 22.
Problem 1
Z
sec4 x dx
We will use the substitution u = tan x, du = sec2 x dx. Observe that
Z
Z
sec4 x dx = sec2 x sec2 x dx
Z
= (tan2 x + 1) sec2 x dx
Z
= (u2 + 1) du
1
tan3 x + tan x + C.
3
=
Problem 2
Z
tan3 x dx
Observe that
Z
tan3 x dx =
Z
tan2 x tan x dx
Z
(sec2 x − 1) tan x dx
Z
Z
= sec2 x tan x dx − tan x dx
=
The first integral above can be done again using the substitution u = tan x.
Indeed, observe that
Z
Z
2
sec x tan x dx = u du
=
1
tan2 x + C
2
The second integral can be done using the substituion u = cos x. Indeed, observe
7
that
Z
−
Z
sin x
dx
cos x
Z
du
=
u
= ln | cos x| + C
tan x dx =
We conclude that
Z
Problem 3
tan3 x dx =
Z
1
tan2 x + ln | cos x| + C.
2
sec3 x tan3 x dx
Here, our plan will be to isolate sec x tan x dx in the integral above and
use the substitution u = sec x. In order to make this work, will need to turn
any remaining powers of tangent into terms involving secant by using our main
identity. Observe that
Z
Z
sec3 x tan3 x dx = sec2 x tan2 x sec x tan x dx
Z
= sec2 x(sec2 x − 1) sec x tan x dx
Z
= u2 (u2 − 1) du
Z
= (u4 − u2 ) du
1 5 1 3
u − u +C
5
3
1
1
5
= sec x − sec3 x + C.
5
3
=
Problem 4
Z
sec x tan2 x dx
Here, we start by using our identity:
Z
Z
Z
Z
2
2
3
sec x tan x dx = sec x(sec x − 1) dx = sec x dx − sec x dx.
8
We have seen that the first integral was done in Example 6, while the second
integral was computed through the course of doing Example 6. Using the work
done there, we conclude
Z
1
3
sec x tan2 x dx = sec x tan x + ln | sec x + tan x| + C
2
2
Problem 5
Z
√
sin2 x
dx
1 − cos x
The main√hint to doing this problem is to first multiply and divide the
integrand by 1 + cos x. Doing this gives
√
Z
Z
sin2 x
1 + cos x
sin2 x
√
√
√
dx =
dx
1 − cos x
1 − cos x 1 + cos x
√
Z
sin2 x 1 + cos x
√
dx
=
1 − cos2 x
√
Z
sin2 x 1 + cos x
√
=
dx
sin2 x
√
Z
sin2 x 1 + cos x
=
dx
sin x
Z
√
= sin x 1 + cos x dx.
At this point we can use the substitution u = 1 + cos x. Then
Z
Z
√
sin2 x
√
dx = sin x 1 + cos x dx
1 − cos x
Z
√
=−
u du
2
= − (1 + cos x)3/2 + C.
3
9