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Math Help - Integral problam

  1. #1
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    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?

    Thanks in advance
    Last edited by Diemo; December 13th 2009 at 03:12 AM.
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  2. #2
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    Quote Originally Posted by Diemo View Post
    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?

    Thanks in advance
    If you're going to use trigonometric substitution, rearrange so that

    \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}.

    Now make the substitution

    r = \frac{1}{2}\sin{\theta}.
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  3. #3
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    Simple. Thanks.
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