Answer
See the work step by step section for the method used to obtain the answer\[ = - \cos x + \frac{{{{\cos }^3}x}}{3} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{{\sin }^3}xdx} \hfill \\
\hfill \\
{\text{first}}\,\,{\text{rewrite}}\,\,{\text{the}}\,\,{\text{integrand}} \hfill \\
\hfill \\
{\sin ^3}x = \,\left( {\sin x} \right)\,\left( {{{\sin }^2}x} \right) \hfill \\
\hfill \\
now\,use \hfill \\
\hfill \\
{\sin ^2}x = 1 - {\cos ^2}x \hfill \\
\hfill \\
{\sin ^3}x = \,\left( {\sin x} \right)\,\left( {1 - {{\cos }^2}x} \right) \hfill \\
\hfill \\
Multiply \hfill \\
\hfill \\
{\sin ^3}x = \sin x - \sin x{\cos ^2}x \hfill \\
\hfill \\
integrate\,\, \hfill \\
\hfill \\
= \int_{}^{} {\sin xdx + \int_{}^{} {{{\cos }^2}x\,\left( { - \sin x} \right)dx} } \hfill \\
\hfill \\
then \hfill \\
\hfill \\
= - \cos x + \frac{{{{\cos }^3}x}}{3} + C \hfill \\
\end{gathered} \]
