Evaluate:

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

$\Rightarrow \cos ^{2} x=\frac{1+\cos 2 x}{2}$

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

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

$\Rightarrow \cos ^{2} 4 x=\frac{1+\cos 8 x}{2}$

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

Now the question becomes

$\Rightarrow \frac{1}{4} \int d x+\frac{1}{2} \int \cos 4 x d x+\frac{1}{8} \int d x+\frac{1}{8} \int \cos 8 x d x$

We know $\int \cos a x d x=\frac{1}{a} \sin a x+c$

$\Rightarrow \frac{x}{4}+\frac{1}{8} \sin 4 x+\frac{x}{8}+\frac{\sin 8 x}{64}+c$

$\Rightarrow \frac{24 x+8 \sin 4 x+\sin 8 x}{64}+c$