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Math Help - Integral

  1. #1
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    Integral

    solve
     \int_{0}^{\pi/2}\frac{1}{2+\cos(x)}dx
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  2. #2
    Member sbhatnagar's Avatar
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    Re: Integral

    I=\int_{0}^{\pi \over 2}\frac{1}{2+\cos(x)}dx

    By the substitution  u=\tan \left( \frac{x}{2}\right), we obtain

    \begin{align*} I &=\int_{0}^{1} \Big( \frac{2}{1+u^2} \Big) \Big(\frac{1}{2+\frac{u^2-1}{u^2+1}} \Big)du=\int_{0}^{1}\frac{2}{3u^2+1}du=\cdots =\left[ \frac{2}{\sqrt{3}} \arctan(u \sqrt{3})\right]_{0}^{1}=\boxed{\frac{\pi}{3\sqrt{3}}} \end{align*}
    Last edited by sbhatnagar; March 4th 2012 at 11:30 PM.
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  3. #3
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    Re: Integral

    thanks! how can i solve this with contour integration?
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  4. #4
    Member sbhatnagar's Avatar
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    Re: Integral

    I hate contour integration! Substitute z=e^{ix}.

    I=\int_{0}^{\pi/2}\frac{1}{2+\cos(x)}dx=\frac{1}{6}\oint_{|z|=1}  \frac{2z}{z^2+4z+1}\frac{dz}{iz}

    =\frac{2}{3i}\pi i \, \mathrm{Res} \left( \frac{1}{z^2+4z+1}, \sqrt{3}-2\right) =\frac{2\pi}{3}\frac{1}{2\sqrt{3}}

    =\boxed{\frac{\pi}{3\sqrt{3}}}
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