Results 1 to 1 of 1

Thread: Trig integral with x^2

  1. #1
    Junior Member
    Joined
    Feb 2009
    Posts
    38

    Trig integral with x^2

    I want to evaluate the following 2 integrals: $\displaystyle \frac{2}{a} \int_{\frac{-a}{4}}^{\frac{3a}{4}} x \sin^2 \left[\frac{n\pi}{a} \left(x + \frac{a}{4}\right) \right] \, dx = \frac{a}{4}$

    I think this is correct.


    But $\displaystyle x^2$ is more difficult: $\displaystyle \frac{2}{a} \int_{\frac{-a}{4}}^{\frac{3a}{4}} x^2 \sin^2 \left[ \frac{n\pi}{a} \left(x + \frac{a}{4}\right) \right] \, dx$


    Using integration by parts I let:

    $\displaystyle u=x^2 \Rightarrow du = 2x dx$


    $\displaystyle dv = sin^2 \left[ \frac{n\pi}{a} \left(x + \frac{a}{4} \right) \right] \, dx$


    $\displaystyle \Rightarrow v = \frac{x}{2} - \frac{a}{2n\pi} \sin \left[\frac{n\pi}{a} \left(x + \frac{a}{4} \right) \right] \cos \left[ \frac{n\pi}{a} \left(x + \frac{a}{4} \right) \right] = \frac{x}{2} - \frac{a}{4n\pi} \sin \left[ \frac{2n\pi}{a} \left(x + \frac{a}{4}\right) \right]$

    (double angle formula used)


    $\displaystyle < x^2 > = uv - \int v du = \frac{2}{a} \left[x^2 \, \frac{x}{2} - \frac{a}{4n\pi} \sin \left[ \frac{2n\pi}{a} \left( x + \frac{a}{4} \right) \right] \right]_{\frac{-a}{4}}^{\frac{3a}{4}}$ $\displaystyle \left. - \int_{\frac{-a}{4}}^{\frac{3a}{4}} 2x \left( \frac{x}{2}-\frac{a}{4n\pi} \sin \left[ \frac{2n\pi}{a} \left(x+\frac{a}{4} \right) \right] \right)dx\right ]$


    $\displaystyle =\frac{2}{a}\left [ \frac{27a^3}{128}-\frac{-a^3}{128}-\int_{\frac{-a}{4}}^{\frac{3a}{4}}x^3dx+\frac{a}{4n\pi}\int_{\f rac{-a}{4}}^{\frac{3a}{4}}2xsin[\frac{2n\pi}{a}(x+\frac{a}{4})])dx\right ]$


    $\displaystyle =\frac{2}{a}\left [ \frac{7a^3}{32}-\frac{5a^4}{64}+\frac{a}{2n\pi}\left ( \frac{a^2}{4n^2\pi^2}sin[\frac{2n\pi}{a}(x+\frac{a}{4})])-\frac{a}{2n\pi}xcos[\frac{2n\pi}{a}(x+\frac{a}{4})])\right )_{\frac{-a}{4}}^{\frac{3a}{4}}\right ]$


    Sine term vanishes at both limits while cosine term as 1 at both limits:

    $\displaystyle =\frac{2}{a}\left [ \frac{7a^3}{32}-\frac{5a^4}{64}+\frac{a}{2n\pi}\left ( \frac{-3a^2}{8n\pi}+\frac{a^2}{8n\pi} \right )\right ]=\frac{7a^2}{16}-\frac{a^2}{4n^2\pi^2}-\frac{5a^3}{32}=\frac{a^2(7n^2\pi^2-4)}{16n^2\pi^2}-\frac{5a^3}{32}$


    This is obviously wrong since the units don't add up. How can there be $\displaystyle a^2$ and $\displaystyle a^3$ at the same time? They both have the dimension of meters. My answer should be in meters squared. I spend hours checking and still can't find my mistake.

    I plug the equation into a graphic calculator, I think the answer should be: $\displaystyle \frac{a^2(7n^2\pi^2-24)}{64n^2\pi^2}$
    Last edited by mr fantastic; Sep 19th 2009 at 05:34 PM. Reason: Improved readability (added spacing, larger brackets) and fixed some latex
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trig Integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Dec 1st 2010, 06:09 PM
  2. Need help with Trig Integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Feb 23rd 2010, 02:39 AM
  3. trig integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Apr 11th 2009, 02:26 PM
  4. a trig integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Aug 30th 2008, 02:56 PM
  5. trig integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Mar 7th 2006, 04:12 AM

Search Tags


/mathhelpforum @mathhelpforum