Question

How do you find the antiderivative of `int sin^3xdx`?

Answer 1

`int sin^3xdx = (cos^3x)/3-cosx+C`

Explanation:

Note that:

`sin^3x = sin x*sin^2x= sin x (1-cos^2x)`

So:

`int sin^3xdx = int sin x (1-cos^2x)dx = int sin x dx -int cos^2x sinxdx`

But `-sinx dx = d cosx`

`int sin^3xdx = -int d(cosx) +int cos^2x d( cosx ) = (cos^3x)/3-cosx`