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Math Help - Two integration questions

  1. #1
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    Two integration questions

    integral_pi^0(x*sinx)/(1+cosx^2) dx =

    integral_6^4(x)/(x-1)*(x+2)*(x-3)dx =
    Last edited by mr fantastic; October 23rd 2009 at 07:36 PM. Reason: Changed post title.
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  2. #2
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    Quote Originally Posted by fearsking View Post
    integral_pi^0(x*sinx)/(1+cosx^2) dx =

    integral_6^4(x)/(x-1)*(x+2)*(x-3)dx =
    Are you sure you don't mean

    \int_0^{\pi}{\frac{x\sin{x}}{1 + \cos^2{x}}\,dx}

    and

    \int_4^6{\frac{x}{(x -1)(x + 2)(x - 3)}\,dx}?
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  3. #3
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    I = \int_0^\pi \frac{x\sin(x)}{1+\cos^2(x)} ~dx

    Sub  x = \pi - t

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

     2I = \pi \cdot [ -\arctan(\cos(x)) ]_0^{\pi} = \pi^2/2

     I = \frac{\pi^2}{4}
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