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Math Help - Intergral problem

  1. #1
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    Intergral problem

    How would I go about taking the integral of the following with respect to x?

    (cos(n0) * cos(m0)) / SquareRoot(1-x^2)

    where 0 is theta.
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  2. #2
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    Quote Originally Posted by jzellt View Post
    How would I go about taking the integral of the following with respect to x?

    (cos(n0) * cos(m0)) / SquareRoot(1-x^2)

    where 0 is theta.

    Well, then =\cos m\theta\,,\,\cos n\theta are constants wrt x, so you can take them out of the integral and concentrate on the immediate integral:
    \int \frac{1}{\sqrt{1-x^2}}\,dx=\arcsin x+C

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Well, then =\cos m\theta\,,\,\cos n\theta are constants wrt x, so you can take them out of the integral and concentrate on the immediate integral:
    \int \frac{1}{\sqrt{1-x^2}}\,dx=\arcsin x+C

    Tonio
    Actually, I don't think it's that simple...

    Remember that x = r\cos{\theta}.


    Would that not therefore mean that

    \cos{(m\theta)} and \cos{(n\theta)} are functions of x, not constants?


    In fact:

    x = r\cos{\theta} \implies \cos{\theta} = \frac{x}{r}

    and

    y = r\sin{\theta} \implies \sin{\theta} = \frac{y}{r}.



    \cos{(n\theta)} = \sum_{k = 0}^n{{{n}\choose{k}}\cos^k{\theta}\sin^{n - k}{\theta}\sin{\left[\frac{1}{2}(n - k)\pi\right]}}

     = \sum_{k = 0}^n{{{n}\choose{k}}\left[\frac{x}{r}\right]^k\left[\frac{y}{r}\right]^{n - k}\sin{\left[\frac{1}{2}(n - k)\pi\right]}}.



    \cos{(m\theta)} = \sum_{k = 0}^m{{{m}\choose{k}}\cos^k{\theta}\sin^{m - k}{\theta}\sin{\left[\frac{1}{2}(m - k)\pi\right]}}

     = \sum_{k = 0}^m{{{m}\choose{k}}\left[\frac{x}{r}\right]^k\left[\frac{y}{r}\right]^{m - k}\sin{\left[\frac{1}{2}(m - k)\pi\right]}}.



    These are definitely not constants...
    Last edited by Prove It; November 1st 2009 at 09:10 PM.
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    Quote Originally Posted by Prove It View Post
    Actually, I don't think it's that simple...

    Remember that x = r\cos{\theta}.


    Would that not therefore mean that

    \cos{(m\theta)} and \cos{(n\theta)} are functions of x, not constants?


    Only if the OP had said so and he didn't. The use of theta doesn't automatically mean we're using polar coordinates and, deducing from the question, there's no a priori relation between theta and x. In fact, it'd be rather odd to have a problem with x and theta: if we already changed coordinates, what's x doing there? Why wasn't it changed into r cos(theta)? And why does the OP explicitly asks for the integral wrt x if he already changed to polar coordinates?
    Anyway, that's something for the OP to clear out and not for us, imo. I assumed that theta and x are independient and under this understanding my answer is, I think, correct. If the OP meant something else he should write it down clearly.

    Tonio

    In fact:

    x = r\cos{\theta} \implies \cos{\theta} = \frac{x}{r}

    and

    y = r\sin{\theta} \implies \sin{\theta} = \frac{y}{r}.



    \cos{(n\theta)} = \sum_{k = 0}^n{{{n}\choose{k}}\cos^k{\theta}\sin^{n - k}{\theta}\sin{\left[\frac{1}{2}(n - k)\pi\right]}}

     = \sum_{k = 0}^n{{{n}\choose{k}}\left[\frac{x}{r}\right]^k\left[\frac{y}{r}\right]^{n - k}\sin{\left[\frac{1}{2}(n - k)\pi\right]}}.



    \cos{(m\theta)} = \sum_{k = 0}^m{{{m}\choose{k}}\cos^k{\theta}\sin^{m - k}{\theta}\sin{\left[\frac{1}{2}(m - k)\pi\right]}}

     = \sum_{k = 0}^m{{{m}\choose{k}}\left[\frac{x}{r}\right]^k\left[\frac{y}{r}\right]^{m - k}\sin{\left[\frac{1}{2}(m - k)\pi\right]}}.



    These are definitely not constants...
    .
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