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Math Help - Problem with integration

  1. #1
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    Problem with integration

    Hi. How do you integrate this function?
    \int_0^1 \left(\alpha^2 -\beta^2 x^2\right)^{-1/2} \left(\gamma^2 -k^2 \beta^2 x^2 \right)^{-1} dx
    where:
    \alpha, \gamma > 0 , 0 \leq k \leq 1 and 0\leq \beta \leq 2k
    are some constants.
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  2. #2
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    Quote Originally Posted by CyBeaR View Post
    Hi. How do you integrate this function?
    \int_0^1 \left(\alpha^2 -\beta^2 x^2\right)^{-1/2} \left(\gamma^2 -k^2 \beta^2 x^2 \right)^{-1} dx
    where:
    \alpha, \gamma > 0 , 0 \leq k \leq 1 and 0\leq \beta \leq 2k
    are some constants.

    Let's solve the INdefinite integral , to be convenient , i would rewrite the integral

     \int \frac{dx}{ (1-ax^2) \sqrt{ 1 - bx^2 } }

    Sub  x = \frac{1}{t}

     dx = -\frac{dt}{t^2}

     = \int \frac{t^3}{ (t^2-a) \sqrt{ t^2 - b } }\left( \frac{-dt}{t^2} \right)

     = - \int \frac{ t}{ (t^2 - a) \sqrt{t^2 -b} }<br />
~dt

    Sub.  t^2 -b = u^2

     = -\int \frac{u~du}{ (u^2 + b - a )u}


     = -\int \frac{du}{ u^2 + b-a}

    What do you think ?  b-a > 0 or  < 0

     b-a > 0 . Let's continue

     = -\int \frac{du}{ u^2 + \sqrt{b-a}^2 }

     = \frac{-1}{\sqrt{b-a}} \tan^{-1}\left( \frac{u}{\sqrt{b-a}} \right) + C

     u = \sqrt{t^2 - b} = \frac{ \sqrt{ 1 - bx^2} }{x}
    so

    the integral is

       \frac{-1}{\sqrt{b-a}} \tan^{-1}\left( \frac{\sqrt{ 1 - bx^2} }{x \sqrt{b-a}} \right) + C

    or     \frac{1}{\sqrt{b-a}} \tan^{-1}\left( \frac{x \sqrt{b-a}}{\sqrt{ 1 - bx^2} } \right) + C
    Last edited by simplependulum; May 17th 2010 at 01:32 AM.
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  3. #3
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    b-a>0

    Thanks!
    Last edited by CyBeaR; May 18th 2010 at 06:25 AM.
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  4. #4
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    How do You obtain the 2nd solution?
    I've tried to calculate it without any luck. I've found in "Tables of indefinite integrals." [Petit Bois G.,p.50] that the only solution for (b-a)>0 is the later one.
    I'd like to understand what's behind this.
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  5. #5
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    Quote Originally Posted by CyBeaR View Post
    How do You obtain the 2nd solution?
    I've tried to calculate it without any luck. I've found in "Tables of indefinite integrals." [Petit Bois G.,p.50] that the only solution for (b-a)>0 is the later one.
    I'd like to understand what's behind this.
    Do you mean you are confused why i put 'or' to connect them ?


       \frac{-1}{\sqrt{b-a}} \tan^{-1}\left( \frac{\sqrt{ 1 - bx^2} }{x \sqrt{b-a}} \right) + C
    or     \frac{1}{\sqrt{b-a}} \tan^{-1}\left( \frac{x \sqrt{b-a}}{\sqrt{ 1 - bx^2} } \right) + C

    Actually , the second one can be deduced from the first one .

    because we have  \tan( \frac{\pi}{2} - x ) = \frac{1}{\tan{x}}

    so  \frac{\pi}{2} - x = \tan^{-1}(  \frac{1}{\tan{x}} )

    Let  t = \tan{x} and we have  \frac{\pi}{2} - \tan^{-1}(t) = \tan^{-1}(\frac{1}{t})

       \frac{-1}{\sqrt{b-a}} \tan^{-1}\left( \frac{\sqrt{ 1 - bx^2} }{x \sqrt{b-a}} \right) + C

     =  \frac{-1}{\sqrt{b-a}} \left( \frac{\pi}{2} - \tan^{-1}\left( \frac{x \sqrt{b-a}}{\sqrt{ 1 - bx^2} } \right)  \right) + C

     = \frac{1}{\sqrt{b-a}} \tan^{-1}\left( \frac{x \sqrt{b-a}}{\sqrt{ 1 - bx^2} } \right) + C -\frac{\pi}{2} \frac{1}{\sqrt{b-a}}

    I use  C again instead of   C -\frac{\pi}{2} \frac{1}{\sqrt{b-a}}
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  6. #6
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    I didn't have relation: \tan( \frac{\pi}{2} - x ) = \frac{1}{\tan{x}} within arm's reach and I've tried other ones.
    Now, all is clear to me. Thank You very much.
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