Hey all, stuck on a simple integrand..

$\displaystyle \int_0^\frac{\pi}{3} \! xcos(2x) \, \mathrm{d}x.$

I get..

$\displaystyle u = cos(2x) \; dv = xdx \; du = -2sin(2x) \; v = \frac{x^2}{x}$

$\displaystyle \int xcos(2x) dx = uv - \int vdu \Rightarrow \frac{x^2}{x}cos(2x) - \int -x^2sin(2x) $

Problem is the new integral isn't any easier to solve..