How Do You Solve This Special Relativity Integral?

[yungman]

Homework Statement

I am trying to solve this given $$c=10^8m/s$$.

$$\int^1_{-1} \frac {dz} { [1- \frac {v^2} {c^2} + \frac {v^2} {c^2} z^2 ]^{\frac {3} {2}}}$$

The answer is:

$$\frac 1 {( \frac {v^3}{c^3})} \frac {z} {( \frac {c^2}{v^2}-1 ) \sqrt{( \frac {v^2}{c^2} -1 +z^2})} |^{1}_{-1}$$

The Attempt at a Solution

I tried substitude with $$tan \theta= \frac{z}{\sqrt{(\frac {c^2}{v^2})-1}}$$ but it won't work because if v is small, the answer is zero.

Please give me advice how to solve this.

Thanks

Alan

 

[tiny-tim]

hi yungman!

(you keep miscounting the {} )

try differentiating z/√(a² + z²)

 

[yungman]

hi yungman!

(you keep miscounting the {} )

try differentiating z/√(a² + z²)

Yes I fixed the latex problems I have.

I'll try and come back later.

Thanks