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

We use the trigonometric identity: $\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]$.
Applying this to the integral:
$\int \sin 4x \sin 8x \, dx = \int \frac{1}{2} [\cos(4x-8x) - \cos(4x+8x)] \, dx$
$= \frac{1}{2} \int [\cos(-4x) - \cos(12x)] \, dx$
Since $\cos(-\theta) = \cos(\theta)$,we have:
$= \frac{1}{2} \int (\cos 4x - \cos 12x) \, dx$
Integrating term by term:
$= \frac{1}{2} [\frac{\sin 4x}{4} - \frac{\sin 12x}{12}] + C$
$= \frac{\sin 4x}{8} - \frac{\sin 12x}{24} + C$,where $C$ is an arbitrary constant.