Question

Find the integral of the function
$\sin 4x\sin 8x$

Answer 1

Consider the given function.

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

$I=\int \sin 4x\sin 2\left(4x\right)dx$

$I=\int \sin 4x\left(2\sin 4x\cos 4x\right)dx$

$I=2\int {\sin }^{2}4x\cos 4xdx$

Let $t=\sin 4x$

$\frac{dt}{dx}=4\cos 4x$

$\frac{dt}{4}=\cos 4xdx$

Therefore,

$I=\frac{2}{4}\int t^{2}dt$

$I=\frac{1}{2}\left[\frac{t^3}{3}\right]+C$

$I=\left[\frac{t^3}{6}\right]+C$

On putting the value of $t$, we get

$I=\frac{{\sin }^{3}4x}{6}+C$

Hence, this is the answer.