# 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)

## 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).

Staff Emeritus
Homework Helper
Keep going. Finish the substitution and simplify.

Staff Emeritus
Homework Helper
Gold Member
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))

Staff Emeritus
Then use the identity, $1+\tan^2(\theta)=\sec^2(\theta)$ & simplify.
Just another suggestion for the substitution: $$z=s.\sinh \theta$$