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Integral with double root

  1. Sep 11, 2013 #1
    I'm trying to compute following integral (Wolfram doesn't give answer):
    where [itex]A,B,C,D,E,F,k[/itex] are constants.
    Substitution [itex]t=sin(kr)[/itex] leads to
    The last integral is equal to
    I don't have any idea what to do now. Maybe the substitution [itex]t=sin(kr)[/itex] is not appropriate here?
    Any help is appreciated.
  2. jcsd
  3. Sep 11, 2013 #2


    Staff: Mentor

    It's not obvious to me that you did the substitution correctly. If t = sin(kr), then dt = kcos(kr)dr. I might be wrong, but it appears that you replaced sin(kr) by t and dr by dt.

    Even if you did the substitution correctly, it might not be helpful to do.

    Also, homework and homework-type problems should be posted in the Homework & Coursework sections, not in the technical math sections. I am moving your post to Calculus & Beyond section under Homework & Coursework.
  4. Sep 11, 2013 #3
    Sorry for the wrong forum.
    I think I have done substitution correctly - extracting [itex]k^{2}cos^{2}(kr)[/itex] from the expression in the first integral gives:
    [itex]\int kcos(kr)\sqrt{\frac{E}{k^{2}cos^{2}(kr)}-\frac{B}{sin^{2}(kr)}-\frac{1}{ksin(kr)cos(kr)}\sqrt{D+Fk^{2}\frac{cos^{2}(kr)}{sin^{2}(kr)}}-\frac{A}{sin^{2}(kr)cos^{2}(kr)}}dr[/itex]
    So the expression [itex]kcos(kr)dr[/itex] is present.
  5. Sep 11, 2013 #4


    Staff: Mentor

    I don't see that that is much help. Something different you might try is to rewrite this:

    $$\int\sqrt{E-Bk^{2}\frac{cos^{2}(kr)}{sin^{2}(kr)}-k\frac{cos(kr)}{sin(kr)}\sqrt{D+Fk^{2}\frac{cos^{2}(kr)}{sin^{2}(kr)}}-\frac{Ak^{2}}{sin^{2}(kr)}}dr $$
    as this
    $$\int \sqrt{E-Bk^2cot^2(kr) - k cot(kr) \sqrt{D+Fk^2cot^2(kr)-Ak^2csc^2(kr)}}dr $$

    I don't guarantee that this will be helpful, either. Where I'm going with this is possibly factoring the first three terms in the outer radical, and replacing the cot2(kr) term in the inner radical by csc2(kr) - 1.

    Where did this problem come from? It looks complicated enough that it might not be solvable by analytic means.
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