`int sin^[1//3]x\ cos\ x\ dx`

Our options are to either choose *u* = sin *x*, *u* = sin^{1/3 }*x* or *u* = cos *x*. However, only the first one of these works in this problem.

So we let

u= sinx.

Finding the differential:

du= cosx 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*.