Integrate $\int \cos^4 x\,\mathrm dx.$

This is really the effect of what AWertheim is saying, and not that difficult. $$ \begin{align*} \cos^4x &= \left(\cos^2x\right)^2 \\ &=\left( \frac{1+\cos(2x)}{2} \right)^2 \\ &=\frac{1}{4}\left( 1+2\cos(2x)+\cos^2(2x) \right) \\ &= \frac{1}{4}\left( 1+2\cos(2x)+\frac{1+\cos(4x)}{2} \right) \\ &=\frac{1}{4}+\frac{\cos(2x)}{2}+\frac{1}{8}+\frac{\cos(4x)}{8} \\ &=\frac{3}{8}+\frac{\cos(2x)}{2}+\frac{\cos(4x)}{8}. \end{align*} $$

The sole advantage here being that this form is very simple to integrate.

Hint: apply the half angle formula

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

twice.