# Math Help - Trig Integral

1. ## Trig Integral

Find an expression for

$\cos 2x + \cos 4x + \cos 6x + \dots + \cos (2nx)$

Hence prove that

$\int_0^{\pi/2} \frac{\sin(2n+1)x}{\sin x} \ dx = \frac{\pi}{2}$

Attempt: I found that the sum is the real part of:
$(e^{2ix})^{1}(e^{2ix})^{2}+(e^{2ix})^{3}+\dots+(e^ {2ix})^{n}$
and I know that:
$\frac{e^{ix}+e^{-ix}}{2} =cosx$
$\frac{e^{ix}-e^{-ix}}{2} =sinx$
therefore the integral becomes $\int_{0}^{\frac{\pi }{2}}\frac{e^{i(2n+1)x}-e^{-i(2n+1)x}}{e^{ix}-e^{-ix}}dx$
But how can I evaluate this integral?
Any help is very much appreciated!

Bump!