
Stuck on simple integral
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..

You nearly always choose the polynomial function to be $\displaystyle \displaystyle u$, because differentiating reduces its power, and eventually becomes a constant...

Do you know "ILATE" rule? According to that u should be x and dv should be cos(2x). Now try.

Forgot about LATE. Thanks.