<s>[INST] appropriate bounds involving change of variable Been a long time since I've touched integral so a little help is appreciated!
The integral to be evaluated: $\int_{x = -\sqrt{\frac{2E}{c}}}^{x = +\sqrt{\frac{2E}{c}}} \frac{1}{\sqrt{E-\frac{cx^{2}}{2}}}.dx$
Attempt:
Let $u = E - \frac{cx^{2}}{2}$ so that $\frac{du}{dx} = -cx$ so that $dx = \frac{-1}{cx}du$.
Since the bounds must be $u$ expressed in terms of $x$, $u = E - \frac{cx^{2}}{2} = 0$ for both the lower and upper bound, after substituting $x = -\sqrt{\frac{2E}{c}}$ and $x = +\sqrt{\frac{2E}{c}}$
My bounds appears to be wrong.
 [/INST] Notice that this is an even function.
Let $v=\sqrt{\frac{c}{2E}}x$, then $\frac{dv}{dx}=\sqrt{\frac{c}{2E}}$.
Hence, we have
\begin{align}
\int_{-\sqrt{\frac{2E}{c}}}^{\sqrt{\frac{2E}{c}}} \frac1{\sqrt{E-\frac{cx^2}2}}\, dx &=2 \int_0^{\sqrt{\frac{2E}c}} \frac1{\sqrt{E-\frac{cx^2}2}}\, dx \\
&=\frac{2}{\sqrt{E}} \int_0^\sqrt{\frac{2E}{c}} \frac1{\sqrt{1-\frac{cx^2}{2E}}} \, dx\\
&= \frac{2}{\sqrt{E}}\cdot \sqrt{\frac{2E}{c}} \int_0^1 \frac1{\sqrt{1-v^2 \, }} \, dv \\
&=2\cdot \sqrt{\frac{2}{c}} (\sin^{-1}(1) - \sin^{-1}(0))\\
&=\pi \sqrt{\frac{2}{c}}
\end{align}
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