How to solve $\int {\cos ^4 x} \ dx$ using $\cos 2x$?

Question

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?

Answer 1

Do it twice.

$$\begin{align}\cos^4 (x) = [\cos^2(x)]^2 &= \left(\dfrac{\cos(2x)+1}{2}\right)^2 \\ &=\frac{1}{4}\left(\color{blue}{\cos^2(2x)}+2\cos(2x)+1\right) \\ &= \frac{1}{4}\left(\color{blue}{\dfrac{\cos(4x)+1}{2}}+2\cos(2x)+1\right)\end{align}$$

Answer 2

Another way:

$$ \cos^4 x = \left( \frac{e^{ix}+e^{-ix}}{2} \right)^4=\frac{1}{16} \left( e^{4ix} + 4 e^{2ix} + 6 + 4 e^{-2ix} + e^{-4ix} \right)$$

$$\begin{aligned} \int \cos^4 x \, dx & = \frac{1}{16} \left(-\frac{i}{4} e^{4ix} -2i e^{2ix} + 6 x+ 2i e^{-2ix} +\frac{i}{4} e^{-4ix} \right)+\text{const.}\\ &= \frac{3}{8}x + \frac{1}{4} \sin 2 x +\frac{1}{32} \sin 4x + \text{const.} \end{aligned}$$