calculus - 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.

Answer

You use the identity (e.g. solving from $\cos(2x) = 2\cos^2x-1$):
$$\cos^2 x = \frac{1+ \cos(2x)}{2}$$

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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.