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Math Help - Trig Integral

  1. #1
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    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!
    Last edited by smokesalot; March 18th 2013 at 07:29 AM.
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  2. #2
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    Re: Trig Integral

    Bump!
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