<s>[INST] an identity between two elliptic integrals I would like a direct change of variable proof of the identity
$$\int_0^{\arctan\frac{\sqrt{2}}{\sqrt{\sqrt{3}}}} \frac{1}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^2\phi }}d\phi=\int_0^{\arctan\frac{1}{\sqrt{\sqrt{3}}}}\frac{1}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^22\phi }}d\phi\,.$$ 
I need it as part of a paper on Legendre's proof of the "third singular modulus."
 [/INST] Not yet an answer, but a bit too long for a comment. The Legendre normal form of these elliptic integrals might be a first step, at least by introducing simpler integration limits:
\begin{align}
&I_1=\int_0^{\arctan\frac{\sqrt{2}}{\sqrt{\sqrt{3}}}} \frac{d\phi}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^2\phi }}=\int_0^{\sqrt{3}-1}\frac{dt}{\sqrt{(1-t^2)(1-\frac{2+\sqrt{3}}{4}t^2)}}, \\
&I_2=\int_0^{\arctan\frac{1}{\sqrt{\sqrt{3}}}}\frac{d\phi}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^22\phi }}=\frac{1}{2}\int_0^{3^{1/4}(\sqrt{3}-1)}\frac{dt}{\sqrt{(1-t^2)(1-\frac{2+\sqrt{3}}{4}t^2)}}.
\end{align}
</s>