1. Integral Help

Does anyone know how to do the above integral? I'm stumped on where to even start.

2. $\int -9 \cos^3 (x) \sin^2 (x) dx =$

$-9 \int \cos (x) (1 - \sin^2 (x)) \sin^2 (x) dx =$

$-9 \int \sin^2 (x) \cos (x) - \sin^4 (x) \cos (x) dx$

Can you do it from here?

3. thanks for the help. From there you would just do u-sub with sinx, if I'm not mistaken

4. Originally Posted by cdlegendary
thanks for the help. From there you would just do u-sub with sinx, if I'm not mistaken
Yes, that will work. Then you have

$u = \sin (x)$

$du = \cos (x) dx$

$-9 \int \sin^2(x) \cos (x) - \sin^4(x) \cos(x) dx =$

$-9 \int (u^2 - u^4) du$

5. And then subbing back in after taking the integral would give you

((9sinx^5)/5)-(3sinx^3)+C

thanks for the help