Answer
\[ = \frac{2}{3}{\sin ^3}x + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\sin x\,\sin \,2x\,dx} \hfill \\
\hfill \\
set\,\,\,\,\sin \,2x\, = 2\,\sin x\cos x \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\int_{}^{} {\sin x\,\sin \,2x\,dx} = \int_{}^{} {\sin \,x\,\,\left( {2\,\sin x\,\,\cos x} \right)} \hfill \\
\hfill \\
multiply \hfill \\
\hfill \\
= 2\int_{}^{} {{{\sin }^2}x\,\,\,\cos x\,dx} \hfill \\
\hfill \\
set\,\,\sin \,x\, = t\,\, \to \,\cos x\,dx = dt \hfill \\
\hfill \\
2\int_{}^{} {{{\sin }^2}x\,\,\,\cos x\,dx} = 2\int_{}^{} {{t^2}} dt\,\, \hfill \\
\hfill \\
in\,tegrate \hfill \\
\hfill \\
= 2\left( {\frac{{{t^3}}}{3}} \right) + C \hfill \\
\hfill \\
substituting\,back\,\,t = \,\sin x \hfill \\
\hfill \\
= \frac{2}{3}{\sin ^3}x + C \hfill \\
\end{gathered} \]
