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Math Help - Definite Integration.

  1. #1
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    Definite Integration.

    Evaluate

    I = \int_{0}^{\pi} \frac{x sin x}{1+cos^2 x} dx
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  2. #2
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    Hello,
    Quote Originally Posted by varunnayudu View Post
    Evaluate

    I = \int_{0}^{\pi} \frac{x sin x}{1+cos^2 x} dx
    Let t=\pi-x :

    I=-\int_\pi^0 \frac{(\pi-t) \sin(\pi-t)}{1+\cos^2(\pi-t)} ~ dt=\int_0^\pi \frac{(\pi-t) \sin(\pi-t)}{1+\cos^2(\pi-t)} ~ dt

    But \sin(\pi-t)=\sin(t) and \cos(\pi-t)=-\cos(t)
    Hence :

    I=\int_0^\pi \frac{(\pi-t) \sin(t)}{1+\cos^2(t)}~dt

    I=\pi \int_0^\pi \frac{\sin(t)}{1+\cos^2(t)} ~dt-\underbrace{\int_0^\pi \frac{t \sin(t)}{1+\cos^2(t)} ~ dt}_{=I}

    2I=\pi \int_0^\pi \frac{\sin(t)}{1+\cos^2(t)} ~ dt

    I=\frac{\pi}{2} \int_0^\pi \frac{\sin(t)}{1+\cos^2(t)} ~ dt

    Now substitute \varphi=\cos(t) and it'll be easy.
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  3. #3
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    hw did u do that

    hw did u get that
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  4. #4
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    Quote Originally Posted by zorro View Post
    hw did u get that
    You will need to be more specific. What part of Moo's reply don't you understand?
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  5. #5
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    Please tell me if its correct.

    Quote Originally Posted by Moo View Post
    Hello,

    Let t=\pi-x :

    I=-\int_\pi^0 \frac{(\pi-t) \sin(\pi-t)}{1+\cos^2(\pi-t)} ~ dt=\int_0^\pi \frac{(\pi-t) \sin(\pi-t)}{1+\cos^2(\pi-t)} ~ dt

    But \sin(\pi-t)=\sin(t) and \cos(\pi-t)=-\cos(t)
    Hence :

    I=\int_0^\pi \frac{(\pi-t) \sin(t)}{1+\cos^2(t)}~dt

    I=\pi \int_0^\pi \frac{\sin(t)}{1+\cos^2(t)} ~dt-\underbrace{\int_0^\pi \frac{t \sin(t)}{1+\cos^2(t)} ~ dt}_{=I}

    2I=\pi \int_0^\pi \frac{\sin(t)}{1+\cos^2(t)} ~ dt

    I=\frac{\pi}{2} \int_0^\pi \frac{\sin(t)}{1+\cos^2(t)} ~ dt

    Now substitute \varphi=\cos(t) and it'll be easy.
    --------------------------------------------------------------------


     cos(t) = \varphi
    therefore
     - sin t dt= d\varphi

    <br />
- \frac{\pi}{2} \int_0^\pi \frac{1}{1+\varphi^2}d\varphi<br />


    what to do next ................
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  6. #6
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    Quote Originally Posted by zorro View Post
    --------------------------------------------------------------------


     cos(t) = \varphi
    therefore
     - sin t dt= d\varphi

    <br />
- \frac{\pi}{2} \int_0^\pi \frac{1}{1+\varphi^2}d\varphi<br />


    what to do next ................
    \int \frac{1}{1+\varphi^2} \, d\varphi is a standard form and is equal to \tan^{-1} \varphi (I have omitted the arbitrary cnostant of integration).
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    \int \frac{1}{1+\varphi^2} \, d\varphi is a standard form and is equal to \tan^{-1} \varphi (I have omitted the arbitrary cnostant of integration).

    --------------------------------

    = -  \frac{\pi}{2} (tan^{-1} \pi - tan^{-1}0)

    = - \frac{\pi}{2} (0-0)

    = - \frac{\pi}{2} .0

    = 0

    is this right .........
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  8. #8
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    You forgot to change the limits.
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  9. #9
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    The straight dashed line diff's / anti-diff's with respect to the dashed balloon expression.

    Don't integrate - balloontegrate! http://www.ballooncalculus.org/examples]
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  10. #10
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    hw did t conert to pi

    Quote Originally Posted by tom@ballooncalculus View Post




    The straight dashed line diff's / anti-diff's with respect to the dashed balloon expression.

    Don't integrate - balloontegrate! http://www.ballooncalculus.org/examples]

    -------------------------------------------------------------------

    Here hw did t= \pi turn into 1 and -1.
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  11. #11
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    Quote Originally Posted by zorro View Post
    -------------------------------------------------------------------

    Here hw did t= \pi turn into 1 and -1.
    Recall that there was a change of variable: \varphi = \cos t.

    t = 0 \Rightarrow \varphi = \cos 0 = 1.

    t = \pi \Rightarrow \varphi = \cos \pi = -1.
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  12. #12
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    Quote Originally Posted by zorro View Post
    -------------------------------------------------------------------

    Here hw did t= \pi turn into 1 and -1.
    they are the limits for the new variable
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