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Thread: Integration question

  1. #1
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    Integration question

    Prove that $\displaystyle \int\limits_{0}^{1}\frac{\sqrt(1-x^2)}{1+x^2}\, dx = \frac{\pi}{2}(\sqrt2-1)$

    Hints would be appreciated. Thanks.
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  3. #3
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    Quote Originally Posted by anthill View Post
    Prove that $\displaystyle \int\limits_{0}^{1}\frac{\sqrt(1-x^2)}{1+x^2}\, dx = \frac{\pi}{2}(\sqrt2-1)$

    Hints would be appreciated. Thanks.
    substitute 1-x^2=t^2 and express x^2 in denominator using that substitution. You should get a rational function in variable t.
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by anthill View Post
    Prove that $\displaystyle \int\limits_{0}^{1}\frac{\sqrt(1-x^2)}{1+x^2}\, dx = \frac{\pi}{2}(\sqrt2-1)$

    Hints would be appreciated. Thanks.
    $\displaystyle \displaystyle \int \frac{\sqrt{1-x^2} \cdot \sqrt{1-x^2}}{(1+x^2)\cdot \sqrt{1-x^2}} dx $

    $\displaystyle \displaystyle \int \frac{-1 -x^2 }{(1+x^2) \cdot \sqrt{1-x^2}} + \frac{2}{(1+x^2) \cdot \sqrt{1-x^2}} dx $

    $\displaystyle \displaystyle \int \frac{-1 -x^2 }{(1+x^2) \cdot \sqrt{1-x^2}} dx +\int \frac{2}{(1+x^2) \cdot \sqrt{1-x^2}} dx $

    $\displaystyle \displaystyle \int \frac{-1 }{\sqrt{1-x^2}} dx +\int \frac{2}{(1+x^2) \cdot \sqrt{1-x^2}} dx $

    first integration sub
    $\displaystyle \sin u = x $
    in the second sub
    $\displaystyle \displaystyle \tan u = \frac{\sqrt{2} x }{\sqrt{1-x^2}} $
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  5. #5
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    Let $\displaystyle x = \sin(y) $

    $\displaystyle dx = \cos(y)~dy $


    The integral

    $\displaystyle = \int_0^{\frac{\pi}{2}} \frac{\cos^2(y)}{ 1+ \sin^2(y) }~dy $

    $\displaystyle = \int_0^{\frac{\pi}{2}} \frac{\cos^2(y)}{ \cos^2(y) + 2\sin^2(y) }~dy$

    $\displaystyle = \int_0^{\frac{\pi}{2}} \frac{dy}{ 1+ 2\tan^2(y) }$

    Consider $\displaystyle 1 = \sec^2(y) - \tan^2(y) - 1/2 + 1/2 $ thus we have

    $\displaystyle \frac{1}{2} = \sec^2(y) - \frac{1}{2} ( 1+ 2\tan^2(y) ) $


    The integral

    $\displaystyle = 2 \int_0^{\frac{\pi}{2}} \frac{\sec^2(y) }{ 1+ 2\tan^2(y) }~dy - \int_0^{\frac{\pi}{2}} ~dy $


    $\displaystyle = \left[ \sqrt{2} \tan^{-1}[\sqrt{2}\tan(y)] - y \right]_0^{\frac{\pi}{2}}$


    $\displaystyle = \frac{\pi}{2}( \sqrt{2} - 1 ) $
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