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 intergral 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 intergral 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.