Example 6. Find the integral $\int \sin 7 x \sin 3 x d x$.
Solution. Since

$$
\sin 3 x \sin 7 x=\frac{1}{2}[\cos (7-3) x-\cos (7+3) x]=
$$

$$
=\frac{1}{2}(\cos 4 x-\cos 10 x)
$$

then

$$
\begin{gathered}
\int \sin 7 x \sin 3 x d x=\frac{1}{2} \int[\cos 4 x-\cos 10 x] d x= \\
=\frac{1}{2} \int \cos 4 x d x-\frac{1}{2} \int \cos 10 x d x=\frac{1}{8} \sin 4 x-\frac{1}{20} \sin 10 x+C
\end{gathered}
$$