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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."

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    $\begingroup$ your $d\phi$ are not where they should be. $\endgroup$
    – Dima Pasechnik
    Commented Dec 28, 2019 at 16:27
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    $\begingroup$ Instead of writing $\sqrt{\sqrt3}$, you might use either $3^{1/4}$ or $\sqrt[4]{3}$, LaTeX for the latter is \sqrt[4]{3}. $\endgroup$
    – Joe Silverman
    Commented Dec 28, 2019 at 17:45
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    $\begingroup$ Isn't math.se a right forum for such questions? $\endgroup$
    – user64494
    Commented Dec 28, 2019 at 20:15
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    $\begingroup$ @user64494 Once again, not everything for which a CAS provides an answer (but no explanation!) deserves your "isn't MSE a right forum for such questions". Note that the OP desires a proof, presumably seeking understanding $\endgroup$
    – Yemon Choi
    Commented Dec 30, 2019 at 1:31
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    $\begingroup$ @user64494 I agree with Yemon's response. Even worse, such comments serve to "shame" the OP for even asking here, when in fact the question is perfectly on-topic, for the reasons Yemon gave. Please reflect carefully on this. $\endgroup$
    – Todd Trimble
    Commented Dec 30, 2019 at 2:05

2 Answers 2

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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 $\phi=\arcsin(\sqrt{3}-1)=\arctan\left(3^{-1/4}\sqrt{2}\right)$. 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.

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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}

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