`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`

`= (3u^[4//3])/(4)+K`

`=(3\ sin^[4//3]\ x)/(4)+K`

The last line is obtained by re-expressing our answer in terms of x.