Prove

$\cos ^{4} 2 x=\left(\cos ^{2} 2 x\right)^{2}$

$=\left(\frac{1+\cos 4 x}{2}\right)^{2}$

$=\frac{1}{4}\left[1+\cos ^{2} 4 x+2 \cos 4 x\right]$

$=\frac{1}{4}\left[1+\left(\frac{1+\cos 8 x}{2}\right)+2 \cos 4 x\right]$

$=\frac{1}{4}\left[1+\frac{1}{2}+\frac{\cos 8 x}{2}+2 \cos 4 x\right]$

$=\frac{1}{4}\left[\frac{3}{2}+\frac{\cos 8 x}{2}+2 \cos 4 x\right]$

$\therefore \int \cos ^{4} 2 x d x=\int\left(\frac{3}{8}+\frac{\cos 8 x}{8}+\frac{\cos 4 x}{2}\right) d x$

$=\frac{3}{8} x+\frac{\sin 8 x}{64}+\frac{\sin 4 x}{8}+C$