Answer
$$\frac{1}{8}\sin 3x\cos x - \frac{3}{8}\sin x\cos 3x + C$$
Work Step by Step
$\begin{gathered}
\int {\sin 3x\sin x} dx \hfill \\
{\text{Use tabular integration to evaluate the integral}} \hfill \\
\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}
{\sin 3x{\text{ and its derivatives}}}&{}&{\sin x{\text{ and its integrals}}} \\
{\sin 3x}&{\mathop {\left( + \right)}\limits_ \searrow }&{\sin x} \\
{3\cos 3x}&{\mathop {\left( - \right)}\limits_ \searrow }&{ - \cos x} \\
{ - 9\sin 3x}&{\left( + \right)}&{ - \sin x}
\end{array} \hfill \\
\end{gathered} $ $$\eqalign{
& {\text{We need stop with term}}-9\sin 3x{\text{ because it is the same as the first row except }} \cr
& {\text{for constants }}\left( { - 9{\text{ on the left and }} - 1{\text{ on the right}}} \right).{\text{ Then}}{\text{:}} \cr
& {\text{We interpret the table as saying }} \cr
& \int {\sin 3x\sin x} dx = \left( {\sin 3x} \right)\left( { - \cos x} \right) - \left( {3\cos 3x} \right)\left( { - \sin x} \right) + \int {\left( { - 9\sin 3x} \right)\left( { - \sin x} \right)} dx \cr
& \int {\sin 3x\sin x} dx = - \sin 3x\cos x + 3\sin x\cos 3x + 9\int {\sin 3x\sin x} dx \cr
& \cr
& {\text{Solve for }}\int {\sin 3x\sin x} dx \cr
& \int {\sin 3x\sin x} dx = - \sin 3x\cos x + 3\sin x\cos 3x + 9\int {\sin 3x\sin x} dx \cr
& \int {\sin 3x\sin x} dx - 9\int {\sin 3x\sin x} dx = - \sin 3x\cos x + 3\sin x\cos 3x + \cr
& - 8\int {\sin 3x\sin x} dx = - \sin 3x\cos x + 3\sin x\cos 3x + C \cr
& \int {\sin 3x\sin x} dx = \frac{1}{8}\sin 3x\cos x - \frac{3}{8}\sin x\cos 3x + C \cr} $$
