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Thread: Integral

  1. #1
    MHF Contributor Amer's Avatar
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    Integral

    how I can integrate this

    $\displaystyle \displaystyle \int_{0}^{\pi} \frac{x\sin x }{1 + \cos ^2 x } dx $

    any clues hints
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  2. #2
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    Using $\displaystyle \displaystyle \ \int_{a}^{b}f(x)\;{dx}=\int_{a}^{b}f(a+b-x)\;{dx}\ $, we have:

    $\displaystyle \begin{aligned}\displaystyle I & = \int_{0}^{\pi}\frac{x\sin{x}}{1+\cos^2{x}}\;{dx} = \int_{0}^{\pi}\frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}}\;{dx} \\ & = \int_{0}^{\pi}\frac{\pi\sin(\pi-x)}{1+\cos^2(\pi-x)}}\;{dx}-\int_{0}^{\pi}\frac{x\sin(\pi-x)}{1+\cos^2(\pi-x)}}\;{dx} \\ & = \pi\int_{0}^{\pi}\frac{\sin{x}}{1+\cos^2{x}}}\;{dx }-\int_{0}^{\pi}\frac{x\sin{x}}{1+\cos^2{x}}}\;{dx} \\ & = \pi\int_{0}^{\pi}\frac{\sin{x}}{1+\cos^2{x}}}\;{dx }-I .\end{aligned}$

    Rearranging this, we get:

    $\displaystyle \displaystyle 2I = \pi\int_{0}^{\pi}\frac{\sin(x)}{1+\cos^2(x)}\;{dx} .$

    Therefore, we have:


    $\displaystyle \displaystyle \begin{aligned} I & = \frac{\pi}{2}\int_{0}^{\pi}\frac{\sin{x}}{1+\cos^2 {x}}\;{dx} = -\frac{\pi}{2}\int_{0}^{\pi}\frac{(\cos{x})'}{1+\co s^2{x}}\;{dx} \\ & = -\frac{\pi}{2}\tan^{-1}\left(\cos{x}\right)\bigg|_{0}^{\pi} = -\frac{\pi}{2}\left(-\frac{\pi}{2}\right) = \frac{\pi^2}{4}.\end{aligned} $
    Last edited by TheCoffeeMachine; Feb 7th 2011 at 03:41 PM.
    Thanks from Amer
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  3. #3
    MHF Contributor Amer's Avatar
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    Tanks very much I appreciate that
    but where I can find more information about this

    $\displaystyle \displaystyle \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx $

    or the proof
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Amer View Post
    $\displaystyle \displaystyle \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx $ or the proof

    Using the substitution

    $\displaystyle t=a+b-x$

    you'll obtain

    $\displaystyle \displaystyle \int_{a}^{b} f(a+b-x) dx =\ldots=\displaystyle \int_{a}^{b} f(t) dt=\displaystyle \int_{a}^{b} f(x) dx $


    Fernando Revilla
    Thanks from Amer
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  5. #5
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by FernandoRevilla View Post
    Using the substitution

    $\displaystyle t=a+b-x$
    thanks very much
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  6. #6
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    integrate by parts and get $\displaystyle \displaystyle\frac{{{\pi }^{2}}}{4}+\int_{0}^{\pi }{\arctan (\cos x)\,dx},$ now the last integral is zero by putting $\displaystyle t=x-\dfrac\pi2$ so the latter one equals $\displaystyle \displaystyle-\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\arctan (\sin x)\,dx}=0,$ since the function is odd.

    or even easier, put $\displaystyle t=\pi-x$ and see that the last integral equals to its negative, hence it must be zero.
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  7. #7
    Super Member Random Variable's Avatar
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    In general, $\displaystyle \displaystyle \int^{\pi}_{0} x f(\sin x) \ dx = \frac{\pi}{2} \int^{\pi}_{0} f(\sin x) \ dx $.

    For this problem $\displaystyle \displaystyle f(\sin x) = \frac{\sin x}{2- \sin^{2} x} $
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  8. #8
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Krizalid View Post
    integrate by parts and get $\displaystyle \displaystyle\frac{{{\pi }^{2}}}{4}+\int_{0}^{\pi }{\arctan (\cos x)\,dx},$ now the last integral is zero by putting $\displaystyle t=x-\dfrac\pi2$ so the latter one equals $\displaystyle \displaystyle-\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\arctan (\sin x)\,dx}=0,$ since the function is odd.

    or even easier, put $\displaystyle t=\pi-x$ and see that the last integral equals to its negative, hence it must be zero.
    Thanks thats great
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