# 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 1Evaluate 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 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 2Evaluate 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 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
## ExercisesEvaluate the following integrals.1. sin ^{7}(x)dx
2. sin ^{9}(x)dx
## Answers to Above Exercises1. (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
## More References and linksintegrals and their applications in calculus. |