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Thread: Integration

  1. #1
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    Question Integration

    Hi All,
    Can any one help me solve the following integration. I tried many times but I always get exactly the same thing I started with. It's like I = I. It would be very helpful if any one can shed some light on it.

    cos(i*acos(x)) sin(j*acos(ax+b)) / sqrt(1-x^2)

    i, j, a, b are constants. I tried integration by parts, with 'cos(i*acos(x)) / sqrt(1-x^2)' in one part and 'sin(j*acos(ax+b))' in another part. But I cannot solve it.

    Thanks,
    ~Reaz
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  2. #2
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    Make substitutions along the lines of:

    $\displaystyle \sin (\arccos x) = \cos(\arcsin x) = \sqrt{1-x^2}$
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    $\displaystyle \sin(j\arccos(ax+b))=j(ax+b)\sqrt{1-(ax+b)^2} $

    And assuming $\displaystyle i\in\mathbb{N} $, $\displaystyle \cos(i\arccos(x))=T_i(x) $, where $\displaystyle T_i(x) $ is The Chebyshev polynomials of the first kind.
    Last edited by chiph588@; Jul 13th 2010 at 12:50 PM.
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  4. #4
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    Hi,
    Thanks for coming up.

    I know $\displaystyle \sin(\arccos(x))=\sqrt(1-x^2)$. But I don't know why $\displaystyle \sin(j\arccos(ax+b))=j(ax+b)\sqrt(1+(ax+b)^2)$. And I actually replaced the Chebyshev's polynomial with it's equivalent so that I can integrate the product. Following is what I tried-



    $\displaystyle
    \int \frac{\cos(i\arccos(x))\sin(j\arccos(ax+b))}{\sqrt {1-x^2}}
    $

    $\displaystyle
    I = \sin(j\arccos(ax+b)) \int \frac{\cos(i\arccos(x))}{\sqrt{1-x^2}} - \int \frac{d}{dx}\sin(j\arccos(ax+b)) \int \frac{\cos(i\arccos(x))}{\sqrt{1-x^2}}
    $

    $\displaystyle
    =-\frac{\sin(j\arccos(ax+b))\sin(i\arccos(x))}{i} - \frac{ja}{i} \int \frac{\cos(j\arccos(ax+b))\sin(i\arccos(x))}{\sqrt {1-(ax+b)^2}}
    $

    $\displaystyle
    I = K - \frac{ja}{i} I2
    $

    Now,
    $\displaystyle
    I2 = \int \frac{\cos(j\arccos(ax+b))\sin(i\arccos(x))}{\sqrt {1-(ax+b)^2}}
    $

    $\displaystyle
    =\sin(i\arccos(x)) \int \frac{\cos(j\arccos(ax+b))}{\sqrt{1-(ax+b)^2}} - \int \frac{d}{dx}\sin( i\arccos(x)) \int \frac{\cos(j\arccos(ax+b) )}{\sqrt{1-(ax+b)^2}}
    $

    $\displaystyle
    = -\frac{\sin(j\arccos(ax+b))\sin(i\arccos(x))}{ja} - \frac{i}{ja} \int \frac{\cos(i\arccos(x))\sin(j\arccos(ax+b))}{\sqrt {1-x^2}}
    $

    $\displaystyle
    I2 =\frac{i}{ja} K - \frac{i}{ja} I
    $

    From the two equations of I and I2 we get I = I.

    Thanks for your time.
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    I don't think this can be solved in general. See here.
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  6. #6
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    Thanks. That also helps.
    If I know it doesn't have a closed form then I can go for numerical methods.
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  7. #7
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by mreazuddin View Post
    Thanks. That also helps.
    If I know it doesn't have a closed form then I can go for numerical methods.
    What is this for anyway?
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  8. #8
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    I was trying to compute double integral of product of Chebyshev's polynomial, with y dependent on x. And got this complex integral.
    Thanks anyway.
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