Find the integral of the function $\cos^{4} 2x$.

We know that $\cos^{2} \theta = \frac{1+\cos 2\theta}{2}$.
$\cos^{4} 2x = (\cos^{2} 2x)^{2} = \left(\frac{1+\cos 4x}{2}\right)^{2}$
$= \frac{1}{4} (1 + 2\cos 4x + \cos^{2} 4x)$
$= \frac{1}{4} \left[1 + 2\cos 4x + \frac{1+\cos 8x}{2}\right]$
$= \frac{1}{4} \left[1 + 2\cos 4x + \frac{1}{2} + \frac{\cos 8x}{2}\right]$
$= \frac{1}{4} \left[\frac{3}{2} + 2\cos 4x + \frac{\cos 8x}{2}\right] = \frac{3}{8} + \frac{1}{2}\cos 4x + \frac{1}{8}\cos 8x$
Now,integrating with respect to $x$:
$\int \cos^{4} 2x \, dx = \int \left(\frac{3}{8} + \frac{1}{2}\cos 4x + \frac{1}{8}\cos 8x\right) \, dx$
$= \frac{3}{8}x + \frac{1}{2} \cdot \frac{\sin 4x}{4} + \frac{1}{8} \cdot \frac{\sin 8x}{8} + C$
$= \frac{3}{8}x + \frac{\sin 4x}{8} + \frac{\sin 8x}{64} + C$,where $C$ is an arbitrary constant.