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Frustrating Integral

  1. Mar 10, 2008 #1
    Hi all,
    I have checked many integral tables and done alot of working, but I cant seem to derive a solution for:

    \int \frac{1}{(1+a cos(\theta - \phi))^2} d\theta

    where a is constant.
    Any help would be most appreciated :yuck:
  2. jcsd
  3. Mar 10, 2008 #2
    It would be good for us too see an attempt at the problem...there aren't many people that are kind enough to give you the answer straight up.
  4. Mar 10, 2008 #3
    Possibly Helpful Hint:
    1) expand the denominator
    2) look up the derivative of [tex]tan^{-1}{(\theta + \phi)}[/tex]

    See if you can link the two together...
  5. Mar 10, 2008 #4
    Bet you cant
  6. Mar 10, 2008 #5
    hahahaha :rofl: .
  7. Mar 10, 2008 #6
    yeah I've come across this before in an "applied mathematics" course.
    the integral is going to be really big and ugly and involve a lot of hyperbolic arctangents

    the real crappy part is that while you're writing out reams of expressions you'll be thinking the whole time that there must be some cute simplification for all this. I mean it really looks like there should be, but there just isn't.

    use some software to solve this one. that's why we invented it.
  8. Mar 10, 2008 #7
    oh, by the way the derivative of [tex]tan^{-1}{(\theta + \phi)} = 1/(1 +(\theta - \phi)^2) [/tex]

    p.s. sorry, my latex skills are weak
    Last edited: Mar 10, 2008
  9. Mar 11, 2008 #8
    tried it in Mathematica it is a nasty solution if it is correct

    [tex] -\frac{2 ArcTanh[\frac{(a-1)Tan(\frac{\Theta-\phi}{2})}{\sqrt{a^2-1}}]}{(a^2-1)^{3/2}} + \frac{a * Sin(\Theta-\phi)}{(a^2-1)(1+a*Cos(\Theta-\phi))}[/tex]
  10. Mar 11, 2008 #9
    wow! that's pretty if you ask me
  11. Mar 11, 2008 #10
    Thanks for all the input guys, I used MATLAB and Mathematica, both give different solutions....
    I have heard rumors that Mathematica can sometimes give wrong answers, can anyone support this claim?
  12. Mar 11, 2008 #11
    Oh, by the by, maybe I should have said that a<1 .... this may affect the first term of the solution from Mathematica
  13. Mar 11, 2008 #12
    at least a [tex]tan^{-1}{(...)} [/tex] appears in the answer...too bad I can't figure out how or why.
  14. Mar 12, 2008 #13
    Can you do a partial fraction expansion?
  15. Mar 12, 2008 #14
    just differentiate each answer and see which one is the right one
  16. Mar 12, 2008 #15
    I thought about that too. I think I did attempt it and got stuck somewhere...I can't remember anymore.
  17. Mar 12, 2008 #16
    I will opt for differentiating both
  18. Mar 12, 2008 #17
    Would anyone know an easy method to invert this function

    t(\theta)=-\frac{2 ArcTanh[\frac{(a-1)Tan(\frac{\Theta-\phi}{2})}{\sqrt{a^2-1}}]}{(a^2-1)^{3/2}} + \frac{a * Sin(\Theta-\phi)}{(a^2-1)(1+a*Cos(\Theta-\phi))}


  19. Mar 12, 2008 #18
    Construct a Matrix of [tex]t(\theta)[/tex], do the inverse then the determinant. - i think thats how its done...its been too long.
  20. Mar 12, 2008 #19
    what would be the elements of the matrix [tex] t(\theta) [/tex]?
    I have never heard of this method before....
    Last edited: Mar 12, 2008
  21. Mar 12, 2008 #20
    ???? thats for a matrix not for a function
    why do you need to invert that answer? your integrand in the original post is in terms of theta?
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