Find $\int \sin (3x) \sin (2x) \,dx$.

Solution We use the product-to-sum identity $2\sin A\sin B=\cos(A-B)-\cos (A+B).$

Then $$\begin{eqnarray*} 2\sin (3x) \sin (2x) &=&\cos (3x-2x) -\cos (3x+2x) \\[4pt] \sin (3x) \sin (2x) &=&\dfrac{1}{2}[ \cos x-\cos (5x) ] \end{eqnarray*}$$

Then $$\begin{eqnarray*} \int \sin (3x) \sin (2x) \,dx&=&\dfrac{1}{2}\int [\cos x-\cos (5x) ] \,dx=\dfrac{1}{2}\int \cos x~dx-\dfrac{1}{2}\int \cos (5x) \,dx\\[5pt] &&\underset{\underset{\underset{\color{#0066A7}{\hbox{\(du=5~dx\)}}}{\color{#0066A7}{\hbox{\(u=5x\)}}}}{\color{#0066A7}{\uparrow }}}{=} \dfrac{1}{2}\sin x-\dfrac{1}{2}\int \cos u\,\dfrac{du}{5}=\dfrac{1}{2}\sin x-\dfrac{1}{10}\sin (5x) +C \end{eqnarray*}$$