Question

Find the integrals of the functions $\sin 4x\sin 8x$

Answer 1

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