Integrate $\int x\sec^2(x)\tan(x)\,dx$ $$\int x\sec^2(x)\tan(x)\,dx$$
I just want to know what trigonometric function I need to use. I'm trying to integrate by parts.  My book says that the integral equals
$${x\over2\cos^2(x)}-{\sin(x)\over2\cos(x)}+C$$

Let $u = x, dv = \sec^2 x\tan xdx\implies v = \displaystyle \int\sec^2 x\tan xdx= \displaystyle \int \tan xd(\tan x)= \displaystyle \int wdw, w = \tan x$. Can you put it together ?
