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Math Help - Hard integral

  1. #1
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    Hard integral

    Can anyone help with the following integral:

    int[sqrt(2/L)*sin((n*Pi*x)/L)]dL

    The hint given is to use the trig identity:

    sin(alpha)*sin(beta) = (1/2)*cos(alpha - beta) - (1/2)*cos(alpha + beta)

    Thanks.
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  2. #2
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    Hello, Ideasman!

    If no one is responding, it's that the problem makes no sense.
    . . I did an autopsy and came up with a COD . . .


    int[sqrt(2/L)*sin((n*Pi*x)/L)]dL . ??

    The hint given is to use the trig identity:
    . . \sin(\alpha)\cdot\sin(\beta) = \frac{1}{2}\cos(\alpha - \beta) - \frac{1}{2}\cos(\alpha + \beta)

    Could this be the problem?

    . . \int\sin\left(\frac{2}{L}\right)\cdot\sin\left(\fr  ac{n\pi}{L}x\right)\,dx
    . . . . . \uparrow . . . . . . . . . . . . \uparrow


    Apply the identity:
    . . \sin\left(\frac{2}{L}\right)\cdot\sin\left(\frac{n  \pi}{L}x\right) \:=\:\frac{1}{2}\cos\left(\frac{2}{L} - \frac{n\pi}{L}x\right) - \frac{1}{2}\cos\left(\frac{2}{L} + \frac{n\pi}{L}x\right)

    The problem becomes:
    . . \frac{1}{2}\int\cos\left(\frac{2}{L} - \frac{n\pi}{L}x\right)\,dx - \frac{1}{2}\int\cos\left(\frac{2}{L} + \frac{n\pi}{L}x\right)\,dx


    Then use substitution: . u \:=\:\frac{2}{L} - \frac{n\pi}{L}x and v \:=\:\frac{2}{L} + \frac{n\pi}{L}x

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, Ideasman!

    If no one is responding, it's that the problem makes no sense.
    . . I did an autopsy and came up with a COD . . .



    Could this be the problem?

    . . \int\sin\left(\frac{2}{L}\right)\cdot\sin\left(\fr  ac{n\pi}{L}x\right)\,dx
    . . . . . \uparrow . . . . . . . . . . . . \uparrow


    Apply the identity:
    . . \sin\left(\frac{2}{L}\right)\cdot\sin\left(\frac{n  \pi}{L}x\right) \:=\:\frac{1}{2}\cos\left(\frac{2}{L} - \frac{n\pi}{L}x\right) - \frac{1}{2}\cos\left(\frac{2}{L} + \frac{n\pi}{L}x\right)

    The problem becomes:
    . . \frac{1}{2}\int\cos\left(\frac{2}{L} - \frac{n\pi}{L}x\right)\,dx - \frac{1}{2}\int\cos\left(\frac{2}{L} + \frac{n\pi}{L}x\right)\,dx


    Then use substitution: . u \:=\:\frac{2}{L} - \frac{n\pi}{L}x and v \:=\:\frac{2}{L} + \frac{n\pi}{L}x

    It's a good thought. I came up with something similar, but I think we would have to leave the integral as "dL" otherwise the first sin function is simply a constant and we wouldn't need the hint. In fact, the integral becomes rather trivial. The problem I had with this is if we are integrating over L then I can't find a way to do the integral, and I suspect it has no closed form. (If I remember correctly anyway.)

    But you are definitely right about one thing: this problem makes no sense as it is stated.

    -Dan
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  4. #4
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    Hello, Dan!

    I didn't notice that the first sine is constant . . . *blush*


    You're right . . . with dL, we can't integrate it:

    . . \frac{1}{2}\int\cos\left[(2-n\pi x)L^{-1}\right]\,dL - \frac{1}{2}\int\cos\left[(2+n\pi x)L^{-1}\right]\,dL

    That's what convinced me that it must be dx.


    I hope Ideasman clarifies it soon . . .

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