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Thread: problematic integral

  1. #1
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    problematic integral

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

    Let $\displaystyle I=\int {\frac{{x \arcsin x}}
    {{\sqrt{1 - x^2}}}dx}$

    Put $\displaystyle u = \arcsin x$


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

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


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

    Grandad
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