`int sin^[1//3]x\ cos\ x\ dx`
Our options are to either choose u = sin x, u = sin1/3 x or u = cos x. However, only the first one of these works in this problem.
So we let
u = sin x.
Finding the differential:
du = cos x dx
Substituting these into the integral gives:
`int sin^[1//3]x\ cos\ x\ dx = int u^[1//3] du`
`=(3\ sin^[4//3]\ x)/(4)+K`
The last line is obtained by re-expressing our answer in terms of x.