# Integral of sin^6(x)dx

by shft600
Tags: integral, sin6xdx
 Mentor P: 21,397 Use the identity $$sin^2(x) = \frac{1 - cos(2x)}{2}$$ The right side above is easy to integrate. $$sin^4(x) = \left(\frac{1 - cos(2x)}{2}\right)^2 = (1/4)(1 - 2cos(2x) + cos^2(2x)$$ The first two terms on the far right side are easy to integrate, but the last one requires a little more work, this time using the identity cos2(2x) = (1/2)(1 + cos(4x)). Do you see how this idea could be extended?