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

In what follows, C is the constant of integration.

Example 1: Evaluate the integral

sin^{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 intergral 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

sin^{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 intergral 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