$\displaystyle
\int xcos^2(2x)~dx
$
I can tell that this needs some sort of trig identity in order to solve, but I may be wrong.
How would I set this problem up with regards to u and dv?
You cannot multiply x with the inside terms of cosine!
$\displaystyle \int x\cos^2(2x)~dx$
$\displaystyle = \int x(\frac{1+\cos{4x}}{2})$
$\displaystyle = \int \frac{x+x\cos{4x}}{2}$
Now split.
$\displaystyle = \int \frac{x}{2} + \int \frac{x\cos{4x}}{2}$
I presume you know what to do next.