1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook