Results 1 to 2 of 2

Math Help - Trig integral

  1. #1
    Junior Member
    Joined
    Feb 2009
    Posts
    38

    Trig integral

    I need to do an integral for my quantum mechanics homework but it has been a while since I've taken calc so my math is a bit rusty.
    \frac{4}{L^2}\int^L_0\sin^2(\frac{m\pi x_2}{L})\int^L_0x_1^2\sin^2(\frac{n\pi x_1}{L})dx_1dx_2 =\frac{4}{L^2}\int^L_0\sin^2(\frac{m\pi x_2}{L})dx_2\int^L_0x_1^2\sin^2(\frac{n\pi x_1}{L})dx_1
    The first integral I can do, it's the second one that I'm having trouble with. I tried by parts and know that I can do it with brute force, but the it involves using by parts multiple times. In physics, we are usually given less involve math problems, so I was just wondering whether there is a more elegant way to do this.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570

    Trig Integral

    Hello synclastica_86
    Quote Originally Posted by synclastica_86 View Post
    I need to do an integral for my quantum mechanics homework but it has been a while since I've taken calc so my math is a bit rusty.
    \frac{4}{L^2}\int^L_0\sin^2(\frac{m\pi x_2}{L})\int^L_0x_1^2\sin^2(\frac{n\pi x_1}{L})dx_1dx_2 =\frac{4}{L^2}\int^L_0\sin^2(\frac{m\pi x_2}{L})dx_2\int^L_0x_1^2\sin^2(\frac{n\pi x_1}{L})dx_1
    The first integral I can do, it's the second one that I'm having trouble with. I tried by parts and know that I can do it with brute force, but the it involves using by parts multiple times. In physics, we are usually given less involve math problems, so I was just wondering whether there is a more elegant way to do this.
    When you replace \sin^2kx by \tfrac{1}{2}(1-\cos 2kx), where k =\frac{n\pi}{L}, and then separate off the cosine term the only bit that needs any work is an integral of the form \int x^2\cos 2kx \,dx. As far as I know, the only way is to use Parts twice, to reduce the x^2 term to a constant, and then finally integrate the remaining trig term. You end up with

    \int x^2\cos 2kx \,dx = \frac{1}{4k^3}\left(2k^2x^2\sin 2kx + 2kx\cos 2kx -\sin 2kx\right)

    I hope that is helpful.

    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Trig Integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 1st 2011, 10:00 AM
  2. trig integral 1/(1-sin(x)cos(x))
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 26th 2010, 07:29 AM
  3. trig integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 24th 2009, 03:35 PM
  4. trig integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 11th 2009, 06:48 PM
  5. Another trig integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 5th 2009, 08:50 PM

Search Tags


/mathhelpforum @mathhelpforum