<s>[INST] How to solve $\int {\cos ^4 x} \ dx$ using $\cos 2x$? How would I go about solving $$\int \cos{^4} x \ dx $$ given that $$\cos 2x = \cos^2 x − \sin^2 x = 1 − 2 \sin ^2 x = 2 \cos^2 x − 1?$$ I know that I could break it down to $$(\cos{^2}x){^2}$$ but how do I proceed from here?
 [/INST] Another way:
$$ \cos^4 x = \left( \frac{e^{ix}+e^{-ix}}{2} \right)^4=\frac{1}{16}
\left( e^{4ix} + 4 e^{2ix} + 6 + 4 e^{-2ix} + e^{-4ix} \right)$$
$$\begin{aligned}
\int \cos^4 x \, dx & = \frac{1}{16} \left(-\frac{i}{4} e^{4ix} -2i e^{2ix} + 6 x+ 2i e^{-2ix} +\frac{i}{4} e^{-4ix} \right)+\text{const.}\\
&= \frac{3}{8}x + \frac{1}{4} \sin 2 x +\frac{1}{32} \sin 4x + \text{const.}
\end{aligned}$$
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