How do you find the integral of sin^3(x) cos^2(x) dx?

1 Answer
Oct 12, 2015

I = 1/5cos^5x-1/3cos^3x+C

Explanation:

I = int sin^3xcos^2xdx = int sin^2xcos^2xsinxdx
I = int (1-cos^2x)cos^2xsinxdx
cosx=t => -sinxdx=dt => sinxdx=-dt

I = int (1-t^2)t^2(-dt) = int (t^4-t^2)dt = t^5/5-t^3/3+C

I = 1/5cos^5x-1/3cos^3x+C