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Math Help - Integration

  1. #1
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    Integration

    Hi, can someone help me find the following integration

    exp(x)sin^2(n*pi*x/L)

    the limits of integraion is [0,L]

    Thanks for you help.
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  2. #2
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    In: \int _0^Le^x \text{Sin}\left[\frac{n*\pi * x}{L}\right]^2dx Out: \frac{-4 n^2 \pi ^2+e^L \left(L^2+4 n^2 \pi ^2-L (L \text{Cos}[2 n \pi ]+2 n \pi  \text{Sin}[2 n \pi ])\right)}{2 \left(L^2+4 n^2 \pi ^2\right)}.
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  3. #3
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    Thanks for giving me the answer but i need to know how to get it. I don't understand how to do the working out.
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  4. #4
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    Quote Originally Posted by Louise View Post
    Hi, can someone help me find the following integration

    exp(x)sin^2(n*pi*x/L)

    the limits of integraion is [0,L]

    Thanks for you help.
    Firstly using double angle trig formulas you have
    \frac{1}{2}\int_{0}^{L}e^{x}\left(1-cos\left(\frac{2n\pi x}{L}\right)\right)dx

    This can now be split into 2 integrals
    \frac{1}{2}\int_{0}^{L}e^{x}dx-\frac{1}{2}\int_{0}^{L}e^{x}cos\left(\frac{2n\pi x}{L}\right)dx

    The first integral gives \frac{1}{2}(e^{L}-1) the second integral needs the use of integration by parts twice as follows
    I=\frac{1}{2}\int_{0}^{L}e^{x}cos\left(\frac{2n\pi x}{L}\right)dx
    = \frac{1}{2}\left[e^{x}cos\left(\frac{2n\pi x}{L}\right)\right]_{0}^{L}+\frac{n\pi}{L}\int_{0}^{L}e^{x}sin\left(\  frac{2n\pi x}{L}\right)dx
    =\frac{1}{2}(e^{L}-1)+\frac{n\pi}{L}\left[e^{x}sin\left(\frac{2n\pi x}{L}\right)\right]_{0}^{L}-\frac{4n^{2}\pi^{2}}{L^{2}}I
    \implies I=\frac{1}{2}\frac{L^{2}}{L^{2}+4n^{2}\pi^{2}}(e^{  L}-1)

    Putting this all together we get a final solution
    \frac{1}{2}\frac{4n^{2}\pi^{2}}{L^{2}+4n^{2}\pi^{2  }}\left(e^{L}-1\right)

    Hope that's helpful.
    PS. I have assumed that n is an integer and thus that sin(2n\pi)=0, \ cos(2n\pi)=1
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