Question

Solve $\int \sin 4x\sin 8xdx$

Answer 1

Consider the given integral.

$I=\int \sin 4x\sin 8xdx$

$I=\int \sin 8x\sin 4xdx$

We know that

$2\sin A\sin B=\cos (A-B)-\cos (A+B)$

Therefore,

$I=\frac{1}{2}\int 2\sin 8x\sin 4xdx$

$I=\frac{1}{2}\int (\cos (8x-4x)-\cos (8x+4x))dx$

$I=\frac{1}{2}\int (\cos 4x-\cos 12x)dx$

$I=\frac{1}{2}[\frac{\sin 4x}{4}-\frac{\sin 12x}{12}]+C$

Hence, this is the answer.