calculus - Integrate $int cos^4 x,mathrm dx.$

$$\int \cos^4 x\,\mathrm dx = \int (1 - \sin^2x)^2\,\mathrm dx.$$

I tried using $\cos^2(x) + \sin^2(x)$ = 1. This was on the integration by parts section of my textbook.

The integral I came out with is given me a hard time. I double checked my work and I feel I am on the right track but I feel stuck. I tried u substitution but there is no $\cos x$ to leverage with.

Answer

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.