Find the integral of the function $\sin ^{4} x$.

Consider $\sin ^{4} x = (\sin ^{2} x)^{2}$.
Using the identity $\sin ^{2} x = \frac{1 - \cos 2x}{2}$,we have:
$\sin ^{4} x = \left(\frac{1 - \cos 2x}{2}\right)^{2} = \frac{1}{4}(1 - 2\cos 2x + \cos ^{2} 2x)$.
Using the identity $\cos ^{2} 2x = \frac{1 + \cos 4x}{2}$,we get:
$\sin ^{4} x = \frac{1}{4} \left(1 - 2\cos 2x + \frac{1 + \cos 4x}{2}\right) = \frac{1}{4} \left(\frac{3}{2} - 2\cos 2x + \frac{1}{2}\cos 4x\right) = \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$.
Now,integrate with respect to $x$:
$\int \sin ^{4} x \, dx = \int \left(\frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x\right) \, dx$.
$= \frac{3}{8}x - \frac{1}{2} \left(\frac{\sin 2x}{2}\right) + \frac{1}{8} \left(\frac{\sin 4x}{4}\right) + C$.
$= \frac{3}{8}x - \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C$,where $C$ is the constant of integration.