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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 $\phi=\arcsin(\sqrt{3}-1)=\arctan\left(\frac{\sqrt{2}}{\sqrt[4]{3}}\right)$. 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.

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.

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 Legendre's formulas: \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.

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 $\phi=\arcsin(\sqrt{3}-1)=\arctan\left(\frac{\sqrt{2}}{\sqrt[4]{3}}\right)$. 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.

Since the question has changed substantially, no longer asking for a proof by change of variables, but by application of an identity in Legendre's work, I am starting a new answer. These are Legendre's formulas: \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}},\;\;\Delta=\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 indeed $\phi=\arcsin(\sqrt{3}-1)=\arctan\left(\frac{\sqrt{2}}{\sqrt[4]{3}}\right)$. Hence Legendre's formulas give $$F\left(\arctan\left(\frac{\sqrt{2}}{\sqrt[4]{3}}\right),\frac{2+\sqrt{3}}{4}\right)=\frac{1}{2}F\left(2 \arctan 3^{-1/4},\frac{2+\sqrt{3}}{4}\right).$$ This is the identity in the OP.
_{with the additional transformation $\frac{1}{2}F\left(2 \arctan 3^{-1/4},\frac{2+\sqrt{3}}{4}\right)=\int_0^{\arctan 3^{-1/4}}\frac{d\phi'}{\sqrt{1-k^2\sin^2 2\phi'}}.$}

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 Legendre's formulas: \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.