# Integrals Involving sin(x) with Odd Power

Tutorial to find integrals involving odd powers of sin(x). Exercises with answers are at the bottom of the page.

## Examples with Detailed Solutions

In what follows, C is the constant of integration.

### Example 1

Evaluate the integral^{3}(x) dx

__Solution to Example 1:__

The main idea is to rewrite the power of sin(x) as the product of a term with power 1 and a term with an even power. Example: sin

^{3}(x) = sin

^{2}(x) sin(x). Hence the given integral may be written as follows:

sin

^{3}(x) dx = sin

^{2}(x) sin(x) dx

= (1 - cos

^{2}(x)) sin(x) dx

We now let u = cos(x), hence du/dx = -sin(x) or -du = sin(x)dx and substitute in the given integral to obtain

sin

^{3}(x) dx = - (1 - u

^{2}) du

sin

^{3}(x) dx = (1/3) u

^{3}- u + C

Substitute u by cos(x) to obtain

sin

^{3}(x) dx = (1/3)cos

^{3}(x) - cos(x) + C

### Example 2

Evaluate the integral^{5}(x) dx

__Solution to Example 2:__

Rewrite sin

^{5}(x) as follows sin

^{5}(x) = sin

^{4}(x) sin(x). Hence the given integral may be written as follows:

sin

^{5}(x) dx = sin

^{4}(x) sin(x) dx

We now use the identity sin

^{2}(x) = 1 - cos

^{2}(x) to rewrite sin

^{4}(x) in terms of power of cos(x) and rewrite the given integral as follows:

sin

^{5}(x) dx = (1 - cos

^{2}(x))

^{2}sin(x) dx

We now let u = cos(x), hence du/dx = -sin(x) or du = -sin(x)dx and substitute in the given integral to obtain

sin

^{5}(x) dx = - (1 - u

^{2})

^{2}du

Expand and calculate the integral on the right

sin

^{5}(x) dx = - (u

^{4}- 2u

^{2}+ 1) du

= -(1/5)u

^{5}+ (2/3)u

^{3}- u + C

and finally

sin

^{5}(x) dx = -(1/5)cos

^{5}(x) + (2/3)cos

^{3}(x) - cos(x) + C

## Exercises

Evaluate the following integrals.1. sin

^{7}(x)dx

2. sin

^{9}(x)dx

### Answers to Above Exercises

1. (1/7)cos^{7}(x) - (3/5)cos

^{5}(x) + cos

^{3}(x) - cos(x) + C

2. -(1/9)cos

^{9}(x) + (4/7)cos

^{7}(x) - (6/5)cos

^{5}(x) + (4/3)cos

^{3}(x) - cos(x) + C