# 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
sin3(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: sin3(x) = sin2(x) sin(x). Hence the given integral may be written as follows:
sin3(x) dx = sin2(x) sin(x) dx
=
(1 - cos2(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
sin3(x) dx = - (1 - u2) du
sin3(x) dx = (1/3) u3 - u + C
Substitute u by cos(x) to obtain
sin3(x) dx = (1/3)cos3(x) - cos(x) + C

### Example 2

Evaluate the integral
sin5(x) dx

Solution to Example 2:
Rewrite sin5(x) as follows sin5(x) = sin4(x) sin(x). Hence the given integral may be written as follows:
sin5(x) dx = sin4(x) sin(x) dx
We now use the identity sin2(x) = 1 - cos2(x) to rewrite sin4(x) in terms of power of cos(x) and rewrite the given integral as follows:
sin5(x) dx = (1 - cos2(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
sin5(x) dx = - (1 - u2)2 du
Expand and calculate the integral on the right
sin5(x) dx = - (u4 - 2u2 + 1) du
= -(1/5)u
5 + (2/3)u3 - u + C
and finally
sin5(x) dx = -(1/5)cos5(x) + (2/3)cos3(x) - cos(x) + C

## Exercises

Evaluate the following integrals.
1.
sin7(x)dx
2.
sin9(x)dx

### Answers to Above Exercises

1. (1/7)cos7(x) - (3/5)cos5(x) + cos3(x) - cos(x) + C
2. -(1/9)cos
9(x) + (4/7)cos7(x) - (6/5)cos5(x) + (4/3)cos3(x) - cos(x) + C

## More References and links

integrals and their applications in calculus.