1. Antideriv of trig function

∫(tan^2)x)(sec^2)x)

2. $\displaystyle \int \tan^2(x)\sec^2(x)~dx$

make $\displaystyle u =\tan(x) \implies \frac{du}{dx}= sec^2(x)$

So

$\displaystyle \int u^2~du$

What do you think about that?

3. Thanks so much! I didnt know I could use just tan(x) as u when it was squared in the problem