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Math Help - Integration Query

  1. #1
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    Jul 2009
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    Integration Query

    When I intergrated:

    1/sqrt(x)sqrt(1-x) = I, Ive got sinh sqrtx. This i used a subsitution x=u^2

    However, expanding the bottom;
    I = 1/sqrt(x-x^2), then chaging the denominator to a perfect square:
    I = 1/sqrt (1/4 - (1/2-x)^2), then using the same sin inverse identity, i got a complete different answer.

    Can someone please explain to me this?
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  2. #2
    Super Member
    Joined
    Jan 2009
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    Quote Originally Posted by Lukybear View Post
    When I intergrated:

    1/sqrt(x)sqrt(1-x) = I, Ive got sinh sqrtx. This i used a subsitution x=u^2

    However, expanding the bottom;
    I = 1/sqrt(x-x^2), then chaging the denominator to a perfect square:
    I = 1/sqrt (1/4 - (1/2-x)^2), then using the same sin inverse identity, i got a complete different answer.

    Can someone please explain to me this?
    So you have two different answers , one is  2 \sin^{-1}(\sqrt{x}) + C and the other is  \sin^{-1}(2x-1) + K .


    Consider  \cos(2t) = 1 - 2\sin^2(t)

    Also  \cos(2t) = \sin( \frac{\pi}{2} - 2t ) =  1 - 2\sin^2(t)

    Therefore ,   \frac{\pi}{2} - 2t = \sin^{-1}(1 - 2\sin^2(t))

    Sub  \sin(t) = x so we have  \frac{\pi}{2} - 2 \sin^{-1}(x) = \sin^{-1}(1 - 2x^2 ) = - \sin^{-1}(2x^2-1)

     2 \sin^{-1}(x) =  \frac{\pi}{2} + \sin^{-1}(2x^2-1)  or

     2 \sin^{-1}(\sqrt{x}) =  \frac{\pi}{2} + \sin^{-1}(2x-1)


    Then compare the two answers . Here  K = \frac{\pi}{2} + C .
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