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Math Help - problematic integral

  1. #1
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    problematic integral

    \int {\frac{{x arcsin x}}<br />
{{\sqrt{1 - x^2}}}dx}
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  2. #2
    Ted
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    Quote Originally Posted by Shock View Post
    \int {\frac{{x arcsin x}}<br />
{{\sqrt{1 - x^2}}}dx}
    Substitute x=sin(\theta), to get:
    \int \theta \, sin(\theta) \, d\theta , then use integration by parts.
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  3. #3
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    Hello Shock

    Let I=\int {\frac{{x \arcsin x}}<br />
{{\sqrt{1 - x^2}}}dx}

    Put u = \arcsin x


    Then
    x=\sin u
    and
     du = \frac{dx}{\sqrt{1-x^2}}
    So:
    I = \int \sin u\; u\;du
    =u(-\cos u) - \int 1(-\cos u) \; du, by Parts

    =-u\cos u + \sin u + c


    =-\arcsin x\sqrt{1-x^2} + x + c

    Grandad
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