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posted by on .

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

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