Integral with a root in the denominator

[Don'tKnowMuch]

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.

 

[Don'tKnowMuch]

the whole denominator is under the square root.

 

[Infinitum]

Did you try the substitution,

z=s\cdot tan\theta

The integral will be reduced to the integral of a single trigonometric ratio...

 

[Don'tKnowMuch]

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).

 

[vela]

Keep going. Finish the substitution and simplify.

 

[SammyS]

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 z=s\,\tan(\theta)\,, then what is dz\ ?

The denominator should be \sqrt{1+\tan^2(\theta)}\,, which is not the same as \sqrt{1}+\sqrt{\tan^2(\theta)}\,.

 

[Don'tKnowMuch]

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))

 

[SammyS]

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, 1+\tan^2(\theta)=\sec^2(\theta) & simplify.

 

[Don'tKnowMuch]

I"ve got it! Sheesh, it's like i removed splinter. Thank you so much. I love physics forums.

 

[DryRun]

Just another suggestion for the substitution: z=s.\sinh \theta