Since the bountied question has changed substantially, now asking for the application of an identity in Legendre's work, I am starting a new answer.   
These are <A HREF="https://archive.org/details/traitedesfonctio01amle/page/n33">Legendre's formulas:</A>
\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},\\ %\;\;k^2=\frac{2+\sqrt{3}}{4},\\
&\Rightarrow F(\phi,k)=\tfrac{1}{2}F(\theta,k).
\end{align}
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 $\phi=\arcsin(\sqrt{3}-1)=\arctan\left(\frac{\sqrt{2}}{\sqrt[4]{3}}\right)$. Hence, Legendre's identity gives
$$F\left(\arctan\left(\frac{\sqrt{2}}{\sqrt[4]{3}}\right),\frac{2+\sqrt{3}}{4}\right)=\frac{1}{2}F\left(2 \arctan 3^{-1/4},\frac{2+\sqrt{3}}{4}\right)$$
or equivalently
$$\int_0^{\arctan\left(\frac{\sqrt{2}}{\sqrt[4]{3}}\right)}\frac{d\phi'}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^2 \phi'}}=\int_0^{\arctan 3^{-1/4}}\frac{d\phi'}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^2 2\phi'}},$$
which is the identity in the OP.