Example 2 (to $2^{\circ}, 3^{\circ}$ ). Find $\int \cos ^{6} 2 x d x$.
Solution. We have:

$$
\begin{aligned}
& \cos ^{6} 2 x=\left(\frac{1+\cos 4 x}{2}\right)^{3}=\frac{1}{8}\left(1+3 \cos 4 x+3 \cos ^{2} 4 x+\cos ^{3} 4 x\right)= \\
& =\frac{1}{8}\left(1+3 \cos 4 x+\frac{3}{2}(1+\cos 8 x)+\left(1-\sin ^{2} 4 x\right) \cos 4 x\right)= \\
& =\frac{1}{8}\left(\frac{5}{2}+4 \cos 4 x+\frac{3}{2} \cos 8 x-\sin ^{2} 4 x \cos 4 x\right)
\end{aligned}
$$

Therefore,

$$
\int \cos ^{6} 2 x d x=\frac{1}{8}\left(\frac{5}{2} x+\sin 4 x+\frac{3}{16} \sin 8 x-\frac{1}{12} \sin ^{3} 4 x\right)+C
$$