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

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

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

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


    Would that not therefore mean that

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


    In fact:

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

    and

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



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

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



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

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



    These are definitely not constants...
    Last edited by Prove It; Nov 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 $\displaystyle x = r\cos{\theta}$.


    Would that not therefore mean that

    $\displaystyle \cos{(m\theta)}$ and $\displaystyle \cos{(n\theta)}$ are functions of $\displaystyle x$, 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:

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

    and

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



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

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



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

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



    These are definitely not constants...
    .
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