2.2 Trigonometric integrals
In this section we want to look at various methods to integrate functions which are products and quotients of trigonometric functions. The starting point is the integrals of sine and cosine.
(1)
(2)

sin x dx = - cos x
 cos x dx = sin x
Next come the integrals of tangent and cotangent
Example 1.
Find
 tan x dx
Solution.
Since tan x = Ошибка!
one can substitute u = cos x with du = - sin x dx to get
Ошибка!
 tan x dx =
 Ошибка!
dx = -
 Ошибка!
du = - ln | u | = - ln | cos x | = ln
So
(3)
 tan x dx = ln | sec x |
If we are working in an interval where cos x is positive then we can just write Ошибка!
tan x dx = ln sec x .
Similarly
(4)
 cot x dx = ln | sin x |
Finally we come to the integrals of secant and cosecant. The easiest way to do these is to multiply top and bottom by sec x + tan x or csc x + cot x .
Example 2.
Find
 sec x dx
Solution.
Multiply top and bottom by sec x + tan x and then substitute u = tan x + sec x with du =
(sec 2 x + tan x sec x ) dx to get
 sec x dx = Ошибка!Ошибка!
dx = Ошибка!Ошибка!
dx = Ошибка!Ошибка!
du
= - ln | u |
= ln | tan x + sec x |
So
2.2 - 1

(5)
 sec x dx = ln | tan x + sec x |
If we are working in an interval where tan x + sec x tan x + sec x is positive then we can just write
 sec x dx = ln (tan x + sec x ).
Similarly
(6)
 csc x dx = - ln | cot x + csc x |
Now we look at some techniques for integrating products of powers of sines and cosines, i.e.

sin n x cos m x dx .
Case 1. One power is odd and positive.
The simplest case is when one of the powers is an odd positive integer. Suppose for definiteness that m is odd. In that case we split off a cos x and use it with the dx in the substitution u = sin x . We express the remaining cosine terms in terms of sine using sin 2 x + cos 2 x = 1 for the substitution. Let's look at an example.
Example 3.
Find
 Ошибка!
dx .
Solution.
x dx
 Ошибка!
dx =
 Ошибка!
cos x dx =
 Ошибка!
cos x dx =
 Ошибка!
cos
Now let u = sin x with du = cos x dx . We get
 Ошибка!
dx =
 Ошибка!
du
Now expand the top and split up the fraction.
 Ошибка!
du =
 Ошибка!
du =

(
Ошибка!
-
Ошибка!
+
Ошибка!
) du =

( u -6 – 2 u -4 + u -2 ) du
= Ошибка!
- Ошибка!
+ Ошибка!
csc 3 x - csc x
= - Ошибка!
+ Ошибка!
- Ошибка!
= - Ошибка!
csc 5 x + Ошибка!
Problem 1.
Find
 sin 3 x cos 5 x dx . Solution.
Ошибка!
-
Ошибка!
.
Case 2. Both powers are even.
In this case use the following identities to reduce the powers, substitute u = 2 x and reassess the situation.
(7) sin 2 x = 2 sin x cos x sin2 x = Ошибка!
cos2 x = Ошибка!
2.2 - 2

Example 4.
 sin 2 x cos 4 x dx .
Solution.
 sin 2 x cos 4 x dx =

(sin x cos x ) 2 cos 2 x dx =
 Ошибка!
2
Ошибка!
dx
Now let u = 2 x with Ошибка!
du = dx . The integral becomes
Ошибка!
 sin 2 u (1 + cos u ) du = Ошибка!
 sin 2 u du + Ошибка!
 sin 2 u cos u du
In the first integral use the second identity in (7). In the second, let v = sin u . The integral becomes
Ошибка!
Ошибка!
 Ошибка!
du + Ошибка!
 v 2 dv = Ошибка!
- Ошибка!
+
Ошибка!
= Ошибка!
- Ошибка!
+ Ошибка!
= Ошибка!
- Ошибка!
+
Problem 2.
Find
 sin 2 x dx and
 sin 4 x dx .
We can do similar things for integrating products of powers of tangents and secants, i.e.

tan n x sec m x dx .
Case 1. tan is to a odd positive power.
In that case we split off a tan x and use it with sec x dx in the substitution u = sec x . We express the remaining tangent terms in terms of secant using sec 2 x - tan 2 x = 1 for the substitution. Let's look at an example.
Example 3.
Find
 tan 3 x sec 3 x dx .
 tan 3 x sec 3 x dx =
 tan 2 x sec 2 x tan x sec x dx =

(sec 2 x – 1) sec 2 x tan x sec x dx
Let u = sec x with du = tan x sec x dx . The integral becomes
Ошибка!

( u 2 – 1) u 2 du =
Problem 3.
Find
 tan x sec 3 x dx .

( u 4 – u 2 ) du = Ошибка!
- Ошибка!
= Ошибка!
-
Case 2. sec is to an even positive power.
In that case we split off a sec 2 x and use it with dx in the substitution u = tan x . We express the remaining secant terms in terms of tangent using sec 2 x - tan 2 x = 1 for the substitution. Let's look at an example.
Example 3.
Find
 tan 2 x sec 4 x dx .
 tan 2 x sec 4 x dx =
 tan 2 x sec 2 x sec 2 x dx =
 tan 2 x (tan 2 x + 1) sec 2 x dx
2.2 - 3

Let u = tan x with du = sec 2 x dx . The integral becomes
Ошибка!
 u 2 ( u 2 + 1) du =
Problem 2.
Find
 tan x sec 4 x dx .

( u 4 + u 2 ) du = Ошибка!
+ Ошибка!
= Ошибка!
+
2.2 - 4