How to integrate $\cos^2x$? It seems like I am stuck on such a simple problem:
How to I find the antiderivative of $\cos^2x$?
I have tried partial integration, it doesn't seem to work (for me). Some help on how to integrate it would be nice.
 A: You use the identity (e.g. solving from $\cos(2x) = 2\cos^2x-1$):
$$\cos^2 x = \frac{1+ \cos(2x)}{2}$$

Addendum: the previous hint will give you the easiest solution, but you mentioned an attempt with integration by parts - that would work too:
$$\begin{array}{rl}
\displaystyle  \color{blue}{\int \cos^2x \, \mbox{d}x} & \displaystyle
 = \int \cos x \, \mbox{d} \sin x \\[6pt]
& \displaystyle 
= \cos x \sin x - \int \sin x \, \mbox{d} \cos x \\[6pt]
& \displaystyle 
= \cos x \sin x + \int \sin^2 x \, \mbox{d}x \\[6pt]
& \displaystyle 
= \cos x \sin x + \int 1-\cos^2 x \, \mbox{d}x \\[6pt]
& \displaystyle 
= \cos x \sin x + x -  \color{blue}{\int \cos^2x \, \mbox{d}x} \\[6pt]
\Rightarrow \displaystyle 2 \int \cos^2x \, \mbox{d}x & \displaystyle
= \cos x \sin x + x  + C 
\end{array}$$
Then you divide by 2.
A: $$\int (\cos x)^2\, dx=\int \frac{1+\cos(2x)}{2}\, dx$$
$$=\frac{1}{2}\left(\int \, dx+\frac{1}{2}\int \cos(2x)\, d(2x)\right)$$
$$=\frac{1}{2}\left(x+\frac{1}{2}\sin(2x)\right)+C=\frac{x}{2}+\frac{\sin(2x)}{4}+C$$
