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Math Help - Beta Integral

  1. #1
    Member kjchauhan's Avatar
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    Beta Integral

    Please help me to solve this..

    \int_{0}^{\frac{\pi}{2}} \frac{{\cos}^{2m-1}\theta{\sin}^{2n-1}\theta}{(a\cos^2 \theta +b\sin^2 \theta)^{m+n}}d\theta = \frac{\beta(m,n)}{2a^mb^n}

    I started from R.H.S.
    i.e.

    \frac{\beta(m,n)}{2a^mb^n} = \frac{1}{2a^mb^n} \int_{0}^{1} x^{m-1}(1-x)^{n-1}dx

    and then put x=sin^2\theta
    but i couldn't reach at the result..

    I also tried from LHS, but then also i couldn't.

    LHS = \int_{0}^{\frac{\pi}{2}} \frac{{\cos}^{2m-1}\theta{\sin}^{2n-1}\theta}{(a\cos^2 \theta +b\sin^2 \theta)^{m+n}}d\theta

    = \int_{0}^{\frac{\pi}{2}} \frac{({\cos}^{2}\theta)^m ({\sin}^{2}\theta)^n (sin\theta cos\theta)}{(a\cos^2 \theta +b\sin^2 \theta)^{m+n}}d\theta

    = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{(1-{\sin}^{2}\theta )^m({\sin}^{2}\theta)^n (2sin\theta cos\theta)}{(a +(b-a)\sin^2 \theta)^{m+n}}d\theta

    then I tried to put x=sin^2\theta, but...

    so pl help..
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  2. #2
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    omg
    what is this :/
    I hope there is some1 will help you.

    Scary integral
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kjchauhan View Post
    Please help me to solve this..

    \int_{0}^{\frac{\pi}{2}} \frac{{\cos}^{2m-1}\theta{\sin}^{2n-1}\theta}{(a\cos^2 \theta +b\sin^2 \theta)^{m+n}}d\theta = \frac{\beta(m,n)}{2a^mb^n}

    I started from R.H.S.
    i.e.

    \frac{\beta(m,n)}{2a^mb^n} = \frac{1}{2a^mb^n} \int_{0}^{1} x^{m-1}(1-x)^{n-1}dx

    and then put x=sin^2\theta
    but i couldn't reach at the result..

    I also tried from LHS, but then also i couldn't.

    LHS = \int_{0}^{\frac{\pi}{2}} \frac{{\cos}^{2m-1}\theta{\sin}^{2n-1}\theta}{(a\cos^2 \theta +b\sin^2 \theta)^{m+n}}d\theta

    = \int_{0}^{\frac{\pi}{2}} \frac{({\cos}^{2}\theta)^m ({\sin}^{2}\theta)^n (sin\theta cos\theta)}{(a\cos^2 \theta +b\sin^2 \theta)^{m+n}}d\theta

    = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{(1-{\sin}^{2}\theta )^m({\sin}^{2}\theta)^n (2sin\theta cos\theta)}{(a +(b-a)\sin^2 \theta)^{m+n}}d\theta

    then I tried to put x=sin^2\theta, but...

    so pl help..
    How much do you know before I type this all up?
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  4. #4
    Moo
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    Hello,

    I'd suggest you use this formula of the beta function :

    \beta(m,n)=\int_0^{\pi/2} \cos^{2m-1}\theta \sin^{2n-1}\theta ~d\theta
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  5. #5
    Super Member
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    Substitute

     (b/a) \tan^2 \theta = \tan^2(x)
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