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Double Integral

  1. Oct 4, 2004 #1
    Hi, I've tried to solve this problem over and over and always end up with an enormous second integral that seems to never reduce to simpler terms.


    Where the bounds of the inner integral are [tex][x-1,xcos(2(\pi)x)][/tex] and the outer integral are [tex][1,0][/tex]

    Thank you for any help in advance. Any would be great.
  2. jcsd
  3. Oct 4, 2004 #2


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    I get

    [tex]-\frac {116 \pi^2 + 153}{192 \pi^2}[/tex]

    The inner integral came out to be

    [tex]x(x^2+1) \cos {2\pi x} - \frac {x^3 \sin^2 2\pi x}{2} -x^3 + 2x^2-\frac {3x}{2} + 1[/tex]
  4. Oct 4, 2004 #3
    Thank you Tide.

    For the inner integral I get:


    I can simplify the second half but not the first. Is there some trick like a substitution to going further from this point, because I've tried integrating this and it was a monster... 2 pages and I couldnt reach a solution.
  5. Oct 4, 2004 #4
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