Originally Posted by

**x3bnm** Sorry ILikeSerena. In my last post I accidentally typed $\displaystyle u = \tan^2(x)$ instead of $\displaystyle u = \tan(x)$. I edited the last post just before your posting. Sorry for any confusion. Here's the complete proof.

Let $\displaystyle u = \tan(x)$ and $\displaystyle dv = \sec^2(x)\,\,dx$

Now $\displaystyle du = \sec^2(x)\,\,dx$ and $\displaystyle v = \tan(x)$

So

$\displaystyle \begin{align*}\int \tan(x) \sec^2(x)\,\, dx =& \tan^2(x) - \int \tan(x) \sec^2(x)\,\,dx \\ =& \tan^2(x) - \frac{\tan^2(x)}{2} + C\\ =& \frac{\tan^2(x)}{2} + C \end{align*}$