# Stuck on simple integral

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• February 25th 2011, 09:40 PM
Oiler
Stuck on simple integral
Hey all, stuck on a simple integrand..

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

I get..

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

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

Problem is the new integral isn't any easier to solve..
• February 25th 2011, 10:00 PM
Prove It
You nearly always choose the polynomial function to be $\displaystyle u$, because differentiating reduces its power, and eventually becomes a constant...
• February 25th 2011, 10:02 PM
sa-ri-ga-ma
Do you know "ILATE" rule? According to that u should be x and dv should be cos(2x). Now try.
• February 25th 2011, 10:23 PM
Oiler
Forgot about LATE. Thanks.