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Math Help - ( integral )

  1. #1
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    Thumbs up ( integral )

    find :

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  2. #2
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    Quote Originally Posted by a.h.m.a.d View Post
    find :

    the integrand is an even function. so the integral is equal to I=2 \int_0^1 \ln(\sqrt{1-x} + \sqrt{1+x}) \ dx. now by parts gives us:

    I=\ln 2 + \int_0^1 \frac{x^2 \ dx}{\sqrt{1-x^2}(1 + \sqrt{1-x^2})}. finally put x = \sin \theta, to get: I=\ln 2 + \int_0^{\frac{\pi}{2}} (1 - \cos \theta) \ d \theta=\ln 2 + \frac{\pi}{2} - 1.
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  3. #3
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    \int_{0}^{1}{\ln \left( \sqrt{1+x}+\sqrt{1-x} \right)\,dx}=\int_{0}^{1}{\ln \left( \sqrt{1+x}\left( 1+\sqrt{\frac{1-x}{1+x}} \right) \right)\,dx}, besides \frac{1}{2}\int_{0}^{1}{\ln (1+x)\,dx} and \int_{0}^{1}{\ln \left( 1+\sqrt{\frac{1-x}{1+x}} \right)\,dx} are pretty straightforward to compute.
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