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."
 A: Since the bountied question has changed substantially, now asking for the application of an identity in Legendre's Traite des fonctions elliptiques, I am starting a new answer. Legendre defines
\begin{align}
&F(\phi,k)=\int_0^{\phi}\frac{d\phi'}{\sqrt{1-k^2\sin^2\phi'}},\\
&\sin\phi=\frac{\sin(\theta/2)}{\sqrt{\tfrac{1}{2}+\tfrac{1}{2}\Delta(\theta)}},\;\;\Delta(\theta)=\sqrt{1-k^2\sin^2\theta},
\end{align}
and then derives the identity
$$F(\phi,k)=\tfrac{1}{2}F(\theta,k).$$
Now we apply this to $k^2=\frac{2+\sqrt{3}}{4}$, $\theta=2 \arctan 3^{-1/4}$ and find 
$$\sin\phi=\frac{2 \sin (\theta/2)}{\sqrt{\sqrt{4-\left(\sqrt{3}+2\right) \sin ^2\theta}+2}}=\sqrt{3}-1,$$
and thus $\phi=\arcsin(\sqrt{3}-1)=\arctan\left(3^{-1/4}\sqrt{2}\right)$. Hence, Legendre's identity gives
$$F\left(\arctan\left(3^{-1/4}\sqrt{2}\right),\frac{2+\sqrt{3}}{4}\right)=\frac{1}{2}F\left(2 \arctan 3^{-1/4},\frac{2+\sqrt{3}}{4}\right)$$
or equivalently
$$\int_0^{\arctan\left(3^{-1/4}\sqrt{2}\right)}\frac{d\phi'}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^2 \phi'}}=\int_0^{\arctan 3^{-1/4}}\frac{d\phi'}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^2 2\phi'}},$$
which is the identity in the OP.
A: 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}
