$%\begin{array}{l} \int {{{\sin }^3}xdx} = \int {{{\sin }^2}x\sin xdx} = - \int {{{\sin }^2}xd\cos x} = \hfill \\ - \int {\left( {1 - {{\cos }^2}x} \right)d\cos x} = \int {\left( {{t^2} - 1} \right)dt} = \hfill \\ \frac{{{t^3}}}{3} - t + C = \frac{1}{3}{\cos ^3}x - \cos x + C \hfill \\ \end{array}$%

\[\frac{1}{3}{\cos ^3}x - \cos x + C\]