Tutorial to find integrals involving even powers of sin(x), using reducing power formulas, are presented. Exercises with answers are at the bottom of the page. Integrating odd powers of sine is much simpler.

Review Power Reducing Formulas

For n a positive integer, we have the following formulas in trigonometry that may be used to reduce the power of sin(x).
(a) sin^{2}x = (1/2)( 1 - cos(2x) )
(b) sin^{3}x = (1/4)( 3 sin(x) - sin(3x) )
(c) sin^{4}x = (1/8)( 3 - 4 cos(2x) + cos(4x) )
(d) sin^{5}x = (1/16)( sin(5x) - 5 sin(3x) + 10 sin(x) )
(e) sin^{6}x = (1/32)( 10 - 15 cos(2x) + 6 cos(4x) - cos(6x) )

Examples with Detailed Solutions

In what follows, C is the constant of integration.

Example 1

Evaluate the integral

sin^{2}(x) dx

Solution to Example 1:
The main idea is to use the identity sin^{2}x = (1/2)( 1 - cos(2x) ) to reduce the power and hence make the integral easily evaluated as follows:
sin^{2}(x) dx = (1/2)( 1 - cos(2x) ) dx
= (1/2)dx - (1/2) cos(2x) dx
= (1/2) x - (1/2)*(1/2) sin(2x) = x / 2 - (1/4) sin(2x) + C

Example 2

Evaluate the integral

[ sin^{2}(x) - 16 sin^{6}(x) ]dx

Solution to Example 2:
Use the power reducing formulas to rewrite the integral as follows
[ sin^{2}(x) - 16 sin^{6}(x) ] dx
= [ (1/2)( 1 - cos(2x) ) - 16 (1/32)( 10 - 15cos(2x) + 6 cos(4x) - cos(6x) ) ]dx
= (1/2) [ 1 - cos(2x) - 10 + 15cos(2x) - 6 cos(4x) + cos(6x) ] dx
= (1/2) [ - 9 + 14cos(2x) - 6 cos(4x) + cos(6x) ] dx
= (1/2) [ - 9 x + 7 sin(2x) - (3/2) sin(4x) + (1/6) sin(6x) ] + C

Exercises

Evaluate the following integrals.
(a) [ 8 sin^{6}(x) - 2 sin^{2}(x) ]dx
(b) [ 4 sin^{4}(x) + sin^{2}(x) ]dx

Answers to Above Exercises

(a) (3/2) x - (11/8) sin(2x) + (3/8) sin(4x) - (1/24) sin(6x) + C
(b) 2 x - (5/4)sin(2x) + (1/8) sin(4x) + C