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An integral

  1. Nov 21, 2006 #1

    quasar987

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    Does anyone know how to do this integral? Mapple just thinks and think and never gives an answer. I couldn't find a primitive in a table either.

    [tex]\int_{0}^{\pi}\int_{0}^{+\infty}\frac{r^4\sin^3\theta}{(c+br\cos\theta+(\sin^2\theta )r^2)^3}drd\theta[/tex]

    In one attempt, I used a table to reduce the r integral to

    [tex]\int_{0}^{\pi}\int_0^{+\infty}\frac{\sin^3\theta}{(c+br\cos\theta+(\sin^2\theta) r^2)^3}dr[/tex]

    But what's the integral of that? :confused:
     
    Last edited: Nov 21, 2006
  2. jcsd
  3. Nov 21, 2006 #2

    arildno

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    Complete the square within the denominator with rcos(theta) as one of your terms within that square.
    Then, make a change of variables from theta to cos(theta) (i.e, integrate from 1 to -1); probably, you need to use partial fractions decomposition.

    You'll end up (hopefully!) with a rational function in r, which you then are to integrate with respect to r.
     
    Last edited: Nov 21, 2006
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