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Homework Help: Integration problem?

  1. Mar 6, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\int_0^a x Sin^2\left(\frac{nx\pi}{a}\right) dx[/tex]

    3. The attempt at a solution
    [tex]\int_a^b f(x) g'(x)\, dx = \left[ f(x) g(x) \right]_{a}^{b} - \int_a^b f'(x) g(x)\, dx\[/tex]

    [tex]\left| x \left[ \frac{nx\pi}{2a}-\frac{Sin\left(2nx\pi\right)}{4a}\right] \right|^{a}_{0}- \int_0^a \frac{nx\pi}{2a}-\frac{Sin\left(2nx\pi\right)}{4a}[/tex]

    [tex]a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \int_0^a \frac{nx\pi}{2a}-\frac{Sin\left(2nx\pi\right)}{4a}[/tex]

    [tex]a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \frac{1}{2a} \int_0^a nx\pi-\frac{Sin\left(2nx\pi\right)}{2}[/tex]

    [tex]a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \frac{1}{2a}\left[ \int_0^a nx\pi-\int_0^a\frac{Sin\left(2nx\pi\right)}{2}\right][/tex]

    [tex]a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \frac{1}{2a}\left[ n\pi\int_0^a x- \frac{1}{2}\int_0^a Sin\left(2nx\pi\right)\right][/tex]


    [tex]a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \frac{1}{2a}\left[ n\pi \left | x\right |^{a}_{0} - \frac{1}{2} \left| Sin\left(2nx\pi\right)\right|^{a}_{0}\right][/tex]

    [tex]a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \frac{1}{2a}\left[ n\pi a - \frac{1}{2} Sin\left(2na\pi\right)\right][/tex]

    [tex]a \left[ \frac{n\pi}{2}-\frac{Sin\left(2n\pi\right)}{4}\right] - \left[ \frac{n\pi}{2} - \frac{1}{4a} Sin\left(2na\pi\right)\right][/tex]

    [tex]Sin\left(2n\pi\right) = 0[/tex]
    [tex]Sin\left(2na\pi\right) = 0 [/tex]

    [tex]a \left[ \frac{n\pi}{2}\right] - \left[ \frac{n\pi}{2} \right][/tex]

    I'm kinda stuck here....

    I know the answer is [tex]\frac{a}{2}[/tex]

    but I'm not sure how to get to it...maybe an integration mistake?
     
  2. jcsd
  3. Mar 6, 2008 #2
    first of all you are claiming that

    [tex]\frac{d}{dx}\left[ \frac{nx\pi}{2a}-\frac{Sin\left(2nx\pi\right)}{4a}\right]=Sin^2\left(\frac{nx\pi}{a}\right)[/tex] which is not true
     
  4. Mar 6, 2008 #3
    You need to use the identity [tex]\cos(2x) = 1 - 2\sin^2(x)[/tex] to turn [tex]\sin^2(x)[/tex] into a sum of things you know how to integrate (either directly or using integration by parts).
     
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