Posted by **drwls** on Tuesday, February 20, 2007 at 5:07am.

That's the same as the integral of sin^2 x dx.

Use integration by parts.

Let sin x = u and sin x dx = dv

v = -cos x

du = cos x dx

The integral is u v - integral of v du

= -sinx cosx + integral of cos^2 dx

which can be rewritten

integral of sin^2 x = -sinx cos x + integral of (1 - sin^2) dx

2 * (integral of sin^2 x dx)

= - sin x cos x + integral of dx

integral of sin^2 dx = (-1/2) sin x cos x + x/2

integral of (1-cos^2 x) dx

might be easy but i need to make sure