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Integral with a root in the denominator

  1. Jun 3, 2012 #1
    1. The problem statement, all variables and given/known data

    I have been hung up on this integral: (μ /4∏ε) ∫ dz/(√s^2 + z^2)

    2. Relevant equations



    3. The attempt at a solution

    i have tried a couple of different u-substitutions, and none are getting me anywhere. I do not that partial fractions, or by parts would help either. I know what the integral equals (integral table), but i should like to know how to calculate it.
     
  2. jcsd
  3. Jun 3, 2012 #2
    the whole denominator is under the square root.
     
  4. Jun 3, 2012 #3
    Did you try the substitution,

    [tex]z=s\cdot tan\theta[/tex]

    The integral will be reduced to the integral of a single trigonometric ratio...
     
  5. Jun 3, 2012 #4
    I used the substitution that you suggested to get...

    (μ/4∏ε)(1/s) ∫ dz/[√1+√tan^2(θ)]

    I know the answer i should get is...

    (μ/4∏ε) Ln[ z+√s^2+√z^2] (evaluated at the bounds...i think i can do that part)

    I do not see the connection between your suggestion and the expression above (i.e. how does one get a Ln[stuff] term from trig).
     
  6. Jun 3, 2012 #5

    vela

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    Keep going. Finish the substitution and simplify.
     
  7. Jun 3, 2012 #6

    SammyS

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    If [itex]z=s\,\tan(\theta)\,,[/itex] then what is [itex]dz\ ?[/itex]

    The denominator should be [itex]\sqrt{1+\tan^2(\theta)}\,,[/itex] which is not the same as [itex]\sqrt{1}+\sqrt{\tan^2(\theta)}\,.[/itex]
     
  8. Jun 3, 2012 #7
    First off, thanks for the attention. I truly appreciate it!!! Sammy, yes you're right about the notation, i'm with you on that. dz should be s*(sec^2(theta))
     
  9. Jun 3, 2012 #8

    SammyS

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    Then use the identity, [itex]1+\tan^2(\theta)=\sec^2(\theta)[/itex] & simplify.
     
  10. Jun 3, 2012 #9
    I"ve got it! Sheesh, it's like i removed splinter. Thank you so much. I love physics forums.
     
  11. Jun 3, 2012 #10

    sharks

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    Just another suggestion for the substitution: [tex]z=s.\sinh \theta[/tex]
     
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