Answer
$$\int \sin \theta \sin 3 \theta d \theta =\frac{1}{4}\sin(2\theta)-\frac{1}{8}\sin (4\theta )+c$$
Work Step by Step
$$
\int \sin \theta \sin 3 \theta d \theta
$$
Since
$$
\sin m x \sin n x=\frac{1}{2}(\cos [(m-n) x]-\cos [(m+n) x])
$$
Then
\begin{align*}
\int \sin \theta \sin 3 \theta d \theta&=\frac{1}{2}\int \left( \cos(-2\theta)-\cos (4\theta )\right)d\theta \\
&=\frac{1}{2}\int \left( \cos(2\theta)-\cos (4\theta )\right)d\theta\\
&=\frac{1}{2} \left( \frac{1}{2}\sin(2\theta)-\frac{1}{4}\sin (4\theta )\right)+c\\
&=\frac{1}{4}\sin(2\theta)-\frac{1}{8}\sin (4\theta )+c
\end{align*}
