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I Equivalent form for arctan?

  1. Feb 17, 2017 #1
    Hi, I was just looking at an example for a certain problem and noticed that in the second step they went to arctan(epsilon). I know there's a form that is equal to arctan but am a little unsure.
    I've come across formulas on the web such as
    arctan(x) = ∫(dt)/(a2+t2)
    but nothing else that would get to arctan.

    Can somebody please direct me to the correct formula?

    CnCbra6.jpg
     
  2. jcsd
  3. Feb 17, 2017 #2

    jedishrfu

    Staff: Mentor

  4. Feb 17, 2017 #3

    fresh_42

    Staff: Mentor

    I think the ##\varepsilon## in the term with ##\tan^{-1}## is misleading. What if you substitute ##u := \frac{t}{\varepsilon}## and solve the integral?
     
  5. Feb 17, 2017 #4
    If you don't mind me asking, where did you get t/ε from?
     
  6. Feb 17, 2017 #5

    fresh_42

    Staff: Mentor

    For the integral, ##\varepsilon## is only a disturbing constant as ##\pi## is. I simply tried to get the integral in the form ##\int \frac{1}{1+x^2} dx## which means to pull out ##\varepsilon^2## in the denominator.
     
  7. Feb 17, 2017 #6
    The only way I see taking ε2 out of the denominator is dividing that number by itself for numerator and denominator.
    But you would get ∫(1/ε)⋅( (dt) / ( (t22)+1) )
     
  8. Feb 17, 2017 #7

    fresh_42

    Staff: Mentor

    Yes, and now substitute ##u := \frac{t}{\varepsilon}## and replace ##dt## by ##du##.
     
  9. Feb 17, 2017 #8
    Wouldn't that get you arctan(t) and not arctan(ε)?
     
  10. Feb 17, 2017 #9

    fresh_42

    Staff: Mentor

    ##\arctan u## as I see it. I don't understand how the ##\varepsilon## get's into the argument of ##\arctan##. As such it is the variable where the ##\pm \infty## apply to, not the ##\varepsilon## from the limit. I said I find the notation misleading. But it's only a temporary result anyway, so I didn't bother too much.
     
  11. Feb 17, 2017 #10
    Thank you so much for your help! I also got the same. Possibly their answer is incorrect...
     
  12. Feb 17, 2017 #11

    fresh_42

    Staff: Mentor

    Why? It's ##1## in the end, so the result is correct, only the ##\varepsilon## in between is odd.
     
  13. Feb 17, 2017 #12
    Woops didnt mean to say the answer was wrong. ?:)
    But yeah the ε in the arctan in that step is odd. Think they forgot a 't' in the numerator.
     
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