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How do you tackle this integral

  1. Sep 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Find F(x)

    f(x) = 1/(1+x2)2)

    2. Relevant equations



    3. The attempt at a solution

    It's actually a subproblem of another huge annoying surface integral. I tried u-sub but that landed me nowhere and partial integration led me nowhere aswell... Any pointers?
     
  2. jcsd
  3. Sep 6, 2013 #2

    Zondrina

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    Perfect candidate for a trig sub. Use ##x = tan(θ)## so that ##dx = sec^2(θ)dθ##.
     
  4. Sep 6, 2013 #3
    well, 1/1+X^2 is the derivative of arctan i believe. i think there has to be a relation. u use integration by parts left u nowhere?
     
  5. Sep 6, 2013 #4
    Zondrina: Aha, thanks. Obviously it just so happens that in the problem, x had an upper bound of 1. Would we have been out of luck if that hadn't been the case?
     
    Last edited: Sep 6, 2013
  6. Sep 6, 2013 #5
    its a nasty integral u sure you are doing everything correctly? so the integral is 1/1+x^2 only?
     
  7. Sep 6, 2013 #6
    No, I was referring to Zondrinas trig sub suggestion!
     
  8. Sep 6, 2013 #7

    Zondrina

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    I don't see any discontinuity in the integrand at all, so I don't think it would matter.
     
  9. Sep 6, 2013 #8
    Right right, we're using tan and not some other trig function. My bad again. Thanks a ton :)
     
  10. Sep 7, 2013 #9

    verty

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    That is a very good question, I don't know the answer to it. I'm going to investigate this.
     
  11. Sep 7, 2013 #10

    verty

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    Sorry, the original question has no trouble because the range of tan is just R. Here is a more applicable problem:

    ##y = \int {dx \over (1 - x^2)^2}##

    ##x =? \; sin\theta##
    ##dx = cos\theta \; d\theta##

    ##y = \int sec^3\theta \; d\theta##
    ## = {1 \over 2} [sec\theta \; tan\theta + \int sec\theta \; d\theta] + C##
    ## = {1 \over 2} [sec\theta \; tan\theta + ln| sec\theta + tan\theta | ] + C##
    ## = {1\over 2}{x \over 1 - x^2} + {1\over 4} ln | {1+x\over 1-x}| + C##


    On the other hand, one can make a translation:

    ##x = u-1##
    ##dx = du##

    ##y = \int {dx \over (1 - x^2)^2} = \int{ dx \over (1+x)^2 (1-x)^2} = \int {du \over u^2(2-u)^2}##

    After some partial fraction magic:

    ##y = \int {1 \over 4(2-u)^2} + {1 \over 4u^2} + {1 \over 4u} + {1 \over 4(2-u)} du##
    ## = {1 \over 4} [ {1 \over 2-u} - {1\over u} + ln|u| - ln|2-u| ] + C##
    ## = {1 \over 4} [ {2(u-1) \over u(2-u)} + ln| {u \over 2-u} | ] + C##
    ## = {1 \over 2} {x \over 1 - x^2} + {1 \over 4} ln| {1+x \over 1-x} | + C##

    I can't explain it but it's the same answer. Probably it has to do with the behaviour of sin(z) and cos(z) for complex z.
     
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