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Thread: MIT INTEGRATION Bee problem

  1. #1
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    MIT INTEGRATION Bee problem




    ... someone solve this problem by complex integration, but i don't know it;

    please help me to solve this problem ..!
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  2. #2
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    Re: MIT INTEGRATION Bee problem

    Since you mention complex integration, try the following. The complex sine is defined as

    $\displaystyle \sin w = \frac{1}{2i}(e^{iw}-e^{-iw})$

    Your integrand becomes

    $\displaystyle \sin(nx)\sin^m x = \frac{e^{inx}-e^{-inx}}{2i}\frac{(e^{ix}-e^{-ix})^m}{(2i)^m}$

    Let $\displaystyle z = e^{ix}$. To find the integral we also need to see how $\displaystyle dx$ transforms. $\displaystyle dz = d(e^{ix}) = ie^{ix} dx = iz dx$, so $\displaystyle dx = dz/(iz)$ and so the integrand becomes

    $\displaystyle 2^{-m-1}i^{-m-2}(z^n-z^{-n})(z-z^{-1})^m$

    In your situation you have $\displaystyle n= m +2$. Does this help?
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  3. #3
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    Re: MIT INTEGRATION Bee problem

    Actually, here is a much easier way to perform the integration. Your integrand is of the form

    $\displaystyle \sin((m+1)x)\sin^{m-1}x $

    where $\displaystyle m$ is a positive integer. You have that

    $\displaystyle = (\sin^{m-1}x)\sin(mx+x) = (\sin^{m-1}x)(\cos(mx)\sin x - \cos x\sin(mx))$

    Expanding this, we get

    $\displaystyle \cos(mx)\sin^m x - \sin(mx)\sin^{m-1}x \cos x$

    Now, can you recognize this as the derivative of something?
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