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Integral with a root in the denominator

  • #1

Homework Statement



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

Homework Equations





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.
 

Answers and Replies

  • #2
the whole denominator is under the square root.
 
  • #3
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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...
 
  • #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).
 
  • #5
vela
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Keep going. Finish the substitution and simplify.
 
  • #6
SammyS
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I used the substitution that you suggested to get...

(μ/4∏ε)(1/s) ∫ dz/[√1+√tan^2(θ)]
...

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).
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]
 
  • #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))
 
  • #8
SammyS
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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))
Then use the identity, [itex]1+\tan^2(\theta)=\sec^2(\theta)[/itex] & simplify.
 
  • #9
I"ve got it! Sheesh, it's like i removed splinter. Thank you so much. I love physics forums.
 
  • #10
DryRun
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Just another suggestion for the substitution: [tex]z=s.\sinh \theta[/tex]
 

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