# Integral with a root in the denominator

1. Jun 3, 2012

### Don'tKnowMuch

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. Jun 3, 2012

### Don'tKnowMuch

the whole denominator is under the square root.

3. Jun 3, 2012

### Infinitum

Did you try the substitution,

$$z=s\cdot tan\theta$$

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

4. Jun 3, 2012

### 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).

5. Jun 3, 2012

### vela

Staff Emeritus
Keep going. Finish the substitution and simplify.

6. Jun 3, 2012

### SammyS

Staff Emeritus
If $z=s\,\tan(\theta)\,,$ then what is $dz\ ?$

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

7. Jun 3, 2012

### 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))

8. Jun 3, 2012

### SammyS

Staff Emeritus
Then use the identity, $1+\tan^2(\theta)=\sec^2(\theta)$ & simplify.

9. Jun 3, 2012

### Don'tKnowMuch

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

10. Jun 3, 2012

### sharks

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