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An integral to be solved

  1. Jun 24, 2009 #1
    1. The problem statement, all variables and given/known data
    I have to do the integral of xdx/[sqrt(1-x^2)*sqrt(1-A(x-b)^2)]


    2. Relevant equations



    3. The attempt at a solution
    I tried integrating by parts..using integral of xdx/[sqrt(1-x^2)] as -sqrt(1-x^2).......but since the other radical had a power 3/2 now, I could not proceed..I tried changing variables as (x-b)^2=t ..no use...tried A(x-b)^2=sin^2 theta...that way I got till sec/{[tan-Bsec]^3}.....but again stuck...

    I have also tried directly putting x=sin t ..and tried putting sin t =2tan (t/2)/[1+tan^2(t/2)] then.. sint= [exp(it)+exp(-it)]/2 also was tried..
     
  2. jcsd
  3. Jun 24, 2009 #2

    Mark44

    Staff: Mentor

    Here's a suggestion. No guarantees that it will work. What I have in mind is akin to decomposition by partial fractions.

    If you can break up the the integrand into the sum of two simpler expressions, you'll have something that's easier to integrate.

    If you can find constants B, C, D, and E so that the following equation is identically true, the two resulting expressions can be integrated fairly easily using trig substitutions.
    [tex]\frac{x}{\sqrt{1 - x^2}\sqrt{1 - A(x - b)^2}} = \frac{Bx + C}{\sqrt{1 - x^2}} + \frac{Dx + E}{\sqrt{1 - A(x - b)^2}}[/tex]

    I started in on this and found a relationship between C and E, namely that E = -C*sqrt(1 - Ab^2).
     
  4. Jul 1, 2009 #3
    No..that wont work..If we multiply throughout with the denominator of the LHS....we get x on LHS and the sum of two unequal radicals on the RHS..the two expressions cant be equal...unless the factors multiplying the radicals help remove the roots..which cannot happen for the linear factors used...
     
  5. Jul 1, 2009 #4
    In general, this is an "elliptic integral" so the result is not elementary.

    For example, Maple says
    [tex]
    \int \!{\frac {x}{\sqrt {1-{x}^{2}}\sqrt {1-(x-1)^2}}}{dx}
    =
    i \left( {\rm F} \left( {\frac {\sqrt {2}\sqrt {x}}{\sqrt {x-1
    }}},1/2 \right) -{\Pi} \left( {\frac {\sqrt {2}\sqrt {x}}{
    \sqrt {x-1}}},1/2,1/2 \right) \right)
    [/tex]
    Here, [tex]F[/tex] and [tex]\Pi[/tex] are two of the standard elliptic integrals.
     
  6. Jul 2, 2009 #5
    Thanks...
     
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