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.
In what follows, C is the constant of integration.
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
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.
