Question

Question #017cf

Answer 1

`intsin^3xcos^2xdx=cos^5x/5-cos^3x/3 + C`

Explanation:

`intsin^3xcos^2xdx = intsinxsin^2xcos^2xdx`

`=intsinx(1-cos^2x)cos^2xdx`

`=intsinx(cos^2x-cos^4x)dx`

Let `u = cosx => du = -sinxdx`

`=> intsinx(cos^2x-cos^4x)dx`

`= int(cos^4x-cos^2x)(-sinx)dx`

`=int(u^4-u^2)du`

`=u^5/5-u^3/3 + C`

`=cos^5x/5-cos^3x/3 + C`