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$\begingroup$

I have an example of an equation where I don't know how the right hand side is derived from left hand side. It looks very weird as the differential changes. Notice that $m$ and $c$ are constants.

$\frac{m}{2} \int \frac{\textrm{d}v^2}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{-mc^2}{2} \int \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \mathrm d\left(1-\frac{v^2}{c^2}\right) $

Which rules apply here? Could anyone explain this or provide me with links to any good websites? Than you.

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    This is a part of a derivation i took from Wikipedia article "Relativistic kinetic energy" [here](http://en.wikipedia.org/wiki/Kinetic_energy#Relativistic_kinetic_energy_of_rigid_bodies)2012-11-26

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Added: The answer belows tries to address the OP as it was, assuming it was $\,dv\,$ and not $\,dv^2\,$ as it happened to be later.

Imo, it must be

$\frac{m}{2}\int\frac{v^2}{\sqrt{1-\frac{v^2}{c^2}}}dv=\frac{m}{2}\int\frac{v}{\sqrt{1-\frac{v}{c^2}}}\left(-\frac{2v}{c^2}\right)\left(-\frac{c^2}{2}\right)dv=$

$=-\frac{mc^2}{4}\int\frac{v}{\sqrt{1-\frac{v^2}{c^2}}}d\left(1-\frac{v^2}{c^2}\right)$

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    The differential on the LHS is $\mathrm d v^2$ and not $\mathrm d v$ you used. I am sorry for the confusion, but it was a typo i fixed now.2012-11-26
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$ \mathrm dv^2=c^2\mathrm d\frac{v^2}{c^2}=-c^2\mathrm d\left(-\frac{v^2}{c^2}\right)=-c^2\mathrm d\left(1-\frac{v^2}{c^2}\right)\;. $

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    There is ony one disturbing thing to me... Why does it hold that $\textrm d \left(-\frac{v^2}{c^2}\right) = \textrm{d} \left(1 - \frac{v^2}{c^2} \right )$2012-11-26
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What you want is to get a diferential that contains whats inside the square root in the denominator, so you create a derivative that contains both $dv^2$ and $d\left(1-\frac{v^2}{c^2}\right)$. It is easy to do so, and we see that this one can be used: $\frac{d\left(1-\frac{v^2}{c^2}\right)}{d(v^2)}=-\frac{1}{c^2}$ Isolating $d(v^2)$ we get: $d(v^2)=-c^2 d\left(1-\frac{v^2}{c^2}\right)$

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    Yes, its just a simple derivative.2012-11-27