1. ## Integral problam(solved)

Heya all

I need to integrate the following function for an exercise in general relativity. The question is: Find the proper distance between two spheres R[sub]1[\sub]and R[sub]2[\sub] given the metric $ds^2=\frac{dr^2}{1-4r^2} +r^2d \theta^2+r^2sin^2 \theta d\phi^2$. So my function is
$\int {\frac{1}{\sqrt{1-4r^2}}dr}$.
First I thought that I could substitute $u=1-4r^2$. However, this doesn't work out. Then I thought that in general what you would actually use here would be let $r = sin \theta , dr = cos \theta$ then your $1 - 4r^2$becomes $1-4cos^2 \theta$, hoping to get $sin\theta$under the line as well. In fact your left with $sin\theta +3cos\theta.$ How would you compute it?

2. Originally Posted by Diemo
Heya all

I need to integrate the following function for an exercise in general relativity. The question is: Find the proper distance between two spheres R[sub]1[\sub]and R[sub]2[\sub] given the metric $ds^2=\frac{dr^2}{1-4r^2} +r^2d \theta^2+r^2sin^2 \theta d\phi^2$. So my function is
$\int {\frac{1}{\sqrt{1-4r^2}}dr}$.
First I thought that I could substitute $u=1-4r^2$. However, this doesn't work out. Then I thought that in general what you would actually use here would be let $r = sin \theta , dr = cos \theta$ then your $1 - 4r^2$becomes $1-4cos^2 \theta$, hoping to get $sin\theta$under the line as well. In fact your left with $sin\theta +3cos\theta.$ How would you compute it?

$\int{\frac{1}{\sqrt{1 - 4r^2}}\,dr} = \int{\frac{1}{\sqrt{4\left(\frac{1}{4} - r^2\right)}}\,dr}$
$= \frac{1}{2}\int{\frac{1}{\sqrt{\frac{1}{4} - r^2}}\,dr}$.
$r = \frac{1}{2}\sin{\theta}$.