Ok, I'm having a bit of trouble with this one:

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

I'm guessing that I would want to use this trig identity:

$\displaystyle sec^2(x) = 1 + tan^2(x)$

which means...

$\displaystyle tans^2(x) = sec^2(x) - 1$

So, the integral with the identity would now look like this:

$\displaystyle \int sec(x)(sec^2(x) - 1)dx$

Then it would be:

$\displaystyle \int sec^3(x) - sec(x)dx$

Now, with u sub, I would let $\displaystyle u = sec(x)$ so this means $\displaystyle du = sec(x)tan(x)dx$

So $\displaystyle dx = \frac {du}{sec(x)tan(x)}$

Now:

$\displaystyle \int u^3 - u (\frac {du}{sec(x)tan(x)})$

Now I'm kind of stuck here, I know how to take the antiderivative of the first part, but not the $\displaystyle \frac {du}{sec(x)tan(x)}$ part.

Any help?