Question:
$\sin 4 x \sin 8 x$
Solution:
It is known that,
$\sin A \cdot \sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)]$
$\therefore \int \sin 4 x \sin 8 x d x$
$=\int \frac{1}{2}\{\cos (4 x-8 x)-\cos (4 x+8 x)\} d x$
$=\frac{1}{2} \int\{\cos (-4 x)-\cos (12 x)\} d x$
$=\frac{1}{2} \int\{\cos 4 x-\cos 12 x\} d x$
$=\frac{1}{2}\left(\frac{\sin 4 x}{4}-\frac{\sin 12 x}{12}\right)+\mathrm{C}$