Question

How do you find the antiderivative of `Cos(2x)Sin(x)dx`?

Answer 1

`intcos2xsinxdx = cosx-2/3cos^3x + "constant"`

Explanation:

`cos2x -= 2cos^2x-1`

`therefore intcos2xsinxdx = int(2cos^2x -1)sinxdx = 2intcos^2xsinxdx -intsinxdx`

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

Then `cosx +2intcos^2xsinxdx = cosx-2intu^2du = cosx-2/3u^3=cosx-2/3cos^3x +"constant"`