Doing $\int_V^{\alpha} \frac{dE}{\sqrt{(\alpha - E)(E - V)}} = \pi$ without Work Is there a way to see that
$$\int_V^{\alpha} \frac{dE}{\sqrt{(\alpha - E)(E - V)}} = \pi$$
should be true without doing any work?
Is there a quick way to brute-force this without long computations?
Hoping people can explain maybe physics or maths thinking that makes this so obvious.
Thanks!
 A: As per the suggestion of Will Jagy given in the comments above, this all comes down to how we perform the change of variables of $E$.  The easiest way to do this, is that we can perform a sequence of linear transformations
$$
[V,a] \;\; \to \;\; [0, a-V] \;\; \to \;\; [0,2] \;\; \to \;\;[-1,1]
$$
by performing the following operations on $E$:
$$
E \;\; \to \;\; E-V \;\; \to \;\; \frac{2}{a-V} \cdot (E-V) \;\; \to \;\; \frac{2(E-V)}{a-V} - 1 \;\; = \;\; \frac{2E-a-V}{a-V} \;\; =\;\; u.
$$
We can immediately see that $E = \frac{1}{2}[(a-V)u + a +V]$ and $du = \frac{2}{a-V}dE$.  The hardest part is in the substitution in the integrand.  Observe that 
\begin{eqnarray*}
\frac{1}{\sqrt{(a-E)(E-V)}} & = & \frac{1}{\sqrt{\left [a - \frac{1}{2}(a-V)u - \frac{1}{2}a - \frac{1}{2}V \right ]\cdot \left [\frac{1}{2}(a-V)u + \frac{1}{2}a + \frac{1}{2}V - V \right ] }} \\
& = & \frac{1}{\sqrt{\left [\frac{1}{2}(a-V) - \frac{1}{2}(a-V)u \right ] \cdot \left [\frac{1}{2}(a-V) + \frac{1}{2}(a-V)u \right ]}} \\
& = & \frac{2}{a-V} \cdot \frac{1}{\sqrt{1-u^2}}.
\end{eqnarray*}
We therefore see that the change of variables in the original integral is given by 
$$
\int_V^a \frac{dE}{\sqrt{(a-E)(E-V)}} \;\; =\;\; \int_{-1}^1 \frac{a-V}{2} \cdot \frac{2}{a-V} \frac{du}{\sqrt{1-u^2}} \;\; =\;\; \int_{-1}^1 \frac{du}{\sqrt{1-u^2}}.
$$
Since I started writing this, it looks like xpaul chimed in with the same suggestion.
