Legendre's development of elliptic integrals $E(k), K(k) $ for the third singular modulus $k=k_3=\sin(\pi/12)$ in his Exercices de Calcul Integral is elementary in nature but involves quite a bit of integral transformations and also makes use of addition formulas for elliptic integrals of first and second kind.
To begin with let $$F(\phi, k) =\int_0^{\phi}\frac{dx}{\sqrt {1-k^2\sin^2x}},E(\phi,k) =\int_0^{\phi}\sqrt{1-k^2\sin^2x}\,dx\tag{1}$$ where $k\in(0,1)$ is called modulus, $\phi\in\mathbb{R} $ is called amplitude. The integral $F(\phi, k) $ is the incomplete elliptic integral of first kind and $E(\phi, k) $ is the incomplete elliptic integral of second kind. Note that both $F, E$ are odd functions of $\phi$. Further if $\phi=\pi/2$ the above integrals are called complete and we use the notation $$K(k) =F(\pi/2, k) ,E(k)=E(\pi/2,k)\tag{2}$$ When $k$ is evident from context one just writes $F(\phi), E(\phi), K, E$. The number $k'=\sqrt{1-k^2}$ is called complementary modulus. Legendre used the symbol $c$ in place of $k$ and $b$ or $c'$ for $k'$. Further he wrote $F^1(c)$ for $K(k) $ and $E^1(c)$ for $E(k) $.
The above integrals satisfy addition formulas of an algebraical nature. We list down these below.
Theorem 1 (Addition Formula): If $\phi, \psi, \mu$ are such that $$F(\phi, k) +F(\psi, k) - F(\mu, k) =0$$ then
\begin{align}
\sin \mu&=\frac{\sin\phi\cos\psi\Delta\psi+\sin\psi\cos\phi\Delta\phi} {1-k^2\sin^2\phi\sin^2\psi} \tag{3}\\
\cos\mu&=\frac{\cos\phi\cos\psi-\sin\phi\sin\psi\Delta \phi\Delta\psi} {1-k^2\sin^2\phi\sin^2\psi}\tag{4}
\end{align} and $$E(\phi, k) +E(\psi, k) - E(\mu, k) =k^2\sin \phi\sin \psi\sin \mu\tag{5}$$ where $\Delta\phi=\sqrt{1-k^2\sin^2\phi}$.
The above is the most common form of addition formulas for these elliptic integrals. Another form of these formulas is \begin{align}
\sin \mu&=\frac{\sin\phi\cos\phi\Delta\psi+\sin\psi\cos\psi\Delta\phi} {\cos\phi\cos\psi+\sin\phi\sin\psi\Delta\phi\Delta\psi} \tag{6}\\
\cos\mu&=\frac{1-\sin^2\phi-\sin^2\psi+k^2\sin^2\phi\sin^2\psi} {\cos\phi\cos\psi+\sin\phi\sin\psi\Delta\phi\Delta\psi }\tag{7}
\end{align}
An interesting case occurs when $\mu=\pi/2$ ie when $$F(\phi, k) +F(\psi, k) =F(\pi/2,k)=K(k)$$ and using formula $(4)$ we get $$\tan\phi\tan\psi\Delta\phi\Delta\psi=1$$ Using $(7)$ we have $$\Delta\phi\Delta\psi=k'$$ and hence $$F(\phi, k) +F(\psi, k) =K(k)\implies k'\tan\phi\tan\psi=1\tag{8}$$
Let $n$ be a positive integer and then using these formulas we can see that if $F(\phi_n, k) =nF(\phi, k) $ then $\sin\phi_n$ is an algebraic function of $\sin\phi$ and for small values of $n$ like $n=2,3$ the formulas can be inverted so as to get $\sin\phi$ in terms of $\sin\phi_n$. Legendre gives the identity $$\sin\phi =\frac{\sin(\phi_2/2)} {\sqrt {(1+\Delta \phi_2)/2} } \tag{9}$$ for $n=2$ and then handles the case $n=3$ for some particular examples.
We have $$F(\phi_{n+1},k)=F(\phi_n,k)+F(\phi,k)$$ and $$F(\phi_{n-1},k)=F(\phi_n,k)-F(\phi,k)$$ and using the addition formula $(3)$ (and corresponding subtraction formula obtained by changing sign of $\psi$) we get $$\sin\phi_{n+1}+\sin\phi_{n-1}=\frac{2\Delta \phi\cos\phi\sin\phi_n} {1-k^2\sin^2\phi\sin^2\phi_n}$$ and $$\cos\phi_{n+1}+\cos\phi_{n-1}=\frac{2\cos\phi\cos\phi_n}{1-k^2\sin^2\phi\sin^2\phi_n}$$ Dividing these equations we get $$\tan\left(\frac{1}{2}\phi_{n+1}+\frac{1}{2}\phi_{n-1}\right)=\Delta\phi\tan\phi_n\tag{10}$$ Let us now make the assumption that $\phi_n=\pi/2$ so that $$nF(\phi, k) =F(\phi_n, k) =K(k) $$ and then $$F(\phi_{n-1} ,k)+F(\phi,k)=K(k)$$ and using $(8)$ we get $$k'\tan\phi\tan\phi_{n-1}=1$$ Replacing $n$ with $n-1$ in $(10)$ we get $$\tan\left(\frac{\pi}{4}+\frac{1}{2}\phi_{n-2}\right)=\Delta\phi\tan\phi_{n-1}=\frac{\Delta \phi} {k'} \cot \phi$$ Putting $n=3$ we get $$\frac{\cos\phi} {1-\sin\phi}=\frac{\Delta\phi} {k'} \cot\phi$$ or $$k'\sin\phi=\Delta\phi(1-\sin\phi)$$ Squaring the above equation we get $$k^2x^4-2k^2x^3+2x-1=0\tag{11}$$ where $x=\sin\phi$. We also have $$\tan\phi_2=\frac{\cot\phi}{k'},\cos\phi_2=1-\sin\phi\tag{12}$$ Equation $(11)$ is the algebraic equivalent of $3F(\phi,k)=K(k)$ and Legendre solves it for two values of $k$ namely $k=\cos (\pi/12)$ and $k=\sin(\pi/12)$. The first case is simpler and the equation $(11)$ for $k=\cos(\pi/12)=\sqrt{2+\sqrt{3}}/2$ becomes $$\frac{2+\sqrt{3}}{4}(x^4-2x^2)+2x-1=0$$ and one can check that $\sin\phi= x=\sqrt{3}-1$ is a root. We also have $\tan^2\phi=2/\sqrt{3}$.
For $k=\sin(\pi/12)=\sqrt {2-\sqrt{3}}/2$ the equation $(11)$ is transformed into $$ k^4y^4 +6k^2k'^2y^2+4k'^2(k'^2-k^2)y-3k'^4=0$$ via substitution $x^2=1-y$ and $y=\cos^2\phi$ is its root. Putting $y=(k'/k) z$ we get $$z^4+6z^2+4\cdot\frac{k'^2-k^2}{kk'}z-3=0$$ Putting values of $k, k'$ we get the equation $$z^4+6z^2+8\sqrt{3}z-3=0$$ and $z=-\sqrt{3}$ is a root. Discarding this (as $z$ is supposed to be positive) we get $$z^3-z^2\sqrt{3}+9z-\sqrt{3}=0$$ Putting $z=(1+2t)/\sqrt{3}$ we get $$t^3+6t+2=0$$ Using Cardano's method we get $$t=\sqrt[3]{2}-\sqrt[3]{4}=\frac{p^2}{2}-p$$ where $p=\sqrt[3]{4}$. Then $$z=\frac{p^2-2p+1}{\sqrt{3}}$$ and $$\cos^2\phi=y=\frac{kk'}{k^2}z =(2+\sqrt {3})z$$ so that $$\cos\phi=(p-1)\sqrt{\frac{2+\sqrt{3}}{\sqrt{3}}}=(\sqrt[3]{4}-1)\sqrt{\frac{2+\sqrt{3}}{\sqrt{3}}}$$ We now summarize the above derivation as follows:
Theorem 2 (Trisection Formula): The solution to $3F(\phi,k)=K(k)$ is given by $$k^2x^4-2k^2x^3+2x-1=0$$ where $x=\sin\phi$. If $k=\sin(\pi/12)$ then the equation $3F(\phi,k)=K(k)$ has the solution given by $$\cos\phi=(\sqrt[3]{4}-1)\sqrt{\frac{2+\sqrt{3}}{\sqrt{3}}}$$ and the equation $3F(\psi,k')=K(k')$ has the solution given by $\sin\psi=\sqrt{3}-1$ or equivalently by $\tan^2\psi=2/\sqrt {3}$.
Next Legendre works out a formula for the integral $$Z=\int_0^x\frac{f+gt^2}{\sqrt{a^2+2abt^2\cos\theta+b^2t^4}}\,dt\tag{13}$$ in terms of elliptic integrals $E(\phi), F(\phi)$. He considers another integral $$X=\int_0^x\left(\frac{f+gt^2}{\sqrt{a^2+2abt^2\cos\theta+b^2t^4}}-\frac{g}{b}\right)\,dt\tag{14}$$ with $Z=X+(gx/b) $ and uses the substitution $$\cos^2u=\frac{\sqrt{a^2+2abt^2\cos\theta+b^2t^4}-(a\cos\theta+bt^2)}{2ak^2}$$ or equivalently $$t=\sqrt {\frac{a} {b}} \tan u\sqrt {1-k^2\sin^2u}$$ where $k=\sin(\theta/2)$ to arrive at $$X=\int_0^{\phi}\frac{bf-ag+2agk^2\sin^2u} {b\sqrt {ab}} \frac{du} {\sqrt{1-k^2\sin^2u}} $$ where $$\cos^2\phi=\frac{\sqrt{a^2+2abx^2\cos\theta+b^2x^4}-(a\cos\theta+bx^2)}{2ak^2} \tag{15}$$ or $$x=\sqrt {\frac{a} {b}} \tan \phi\sqrt {1-k^2\sin^2\phi}\tag{16}$$ and hence we have $$X=\frac{bf+ag} {b\sqrt{ab}} F(\phi, k) - \frac{2ag}{b\sqrt{ab}}E(\phi,k)\tag{17}$$ The substitution is complicated and Legendre doesn't provide the detailed calculations for the same. One can however note that if $k=\sin(\theta/2)$ then $\sin\theta=2kk'$ and $\cos \theta=1-2k^2$. From the equation connecting $u, t$ we can get $$2ak^2\cos^2u(\sqrt {a^2+2abt^2\cos\theta+b^2t^4}+bt^2+a\cos \theta) =a^2\sin^2\theta$$ and the right side can be written as $4a^2k^2k'^2$ so that $$2ak'^2\sec^2u=\sqrt {a^2+2abt^2\cos\theta+b^2t^4}+bt^2+a\cos \theta$$ and thus $$bt^2+a\cos\theta=a(k'^2\sec^2u-k^2\cos^2u)$$ or $$bt^2=a(k'^2\sec^2u-k^2\cos^2u-1+2k^2)$$ and the right side can be simplified as $$ a\tan^2u(1-k^2\sin^2u)$$ so that $$t=\sqrt{\frac{a} {b}} \tan u\sqrt{1-k^2\sin^2u}$$ as mentioned earlier.
Differentiating the relation connecting $u, t$ with respect to $t$ we get $$2ak^2\sin u\cos u\frac{du} {dt} =bt\left(1-\frac{bt^2+a\cos\theta}{\sqrt{a^2+2abt^2\cos\theta+b^2t^4}}\right) $$ and the right side is same as $$\frac{bt(2ak^2\cos^2u)}{\sqrt{a^2+2abt^2\cos\theta+b^2t^4}}$$ so that $$\frac{dt} {\sqrt{a^2+2abt^2\cos\theta+b^2t^4}} =\frac{\tan u \, du} {bt} =\frac{du} {\sqrt{ab}\sqrt{1-k^2\sin^2u} } $$ Next we also note that $$\sqrt{a^2+2abt^2\cos\theta+b^2t^4}=a(k'^2\sec^2u+k^2\cos^2u)$$ and then $$\left(\frac{f+gt^2}{\sqrt{a^2+2abt^2\cos\theta+b^2t^4}}-\frac{g}{b}\right) \, dt=\left(bf+gbt^2-g \sqrt{a^2+2abt^2\cos\theta+b^2t^4} \right)\frac{dt}{b \sqrt{a^2+2abt^2\cos\theta+b^2t^4}} $$ The expression in parentheses on the right side is $$bf+ag\tan^2u(1-k^2\sin^2u)-ag(k'^2\sec^2u+k^2\cos^2u)$$ and this can be simplified to $$bf-ag+2agk^2\sin^2u=bf+ag-2ag(1-k^2\sin^2u)$$ It follows that $$\int_0^x\left(\frac{f+gt^2}{\sqrt{a^2+2abt^2\cos\theta+b^2t^4}}-\frac{g}{b}\right)\,dt=\frac{bf+ag}{b\sqrt {ab}}F(\phi, k) - \frac{2ag}{b\sqrt{ab}}E(\phi,k)$$ as mentioned.
Legendre now provides three examples of the application of this formula which essentially deal with the elliptic integrals for third singular modulus $k=k_3=\sin(\pi/12)$ and $k'=k'_3=\cos(\pi/12)$.
We next discuss his three examples below starting with:
Example 1: Let $$M=\int_0^1\frac{z^{-2/3}\,dz}{\sqrt{1-z^2}},N=\int_0^1\frac{z^{1/3}\,dz}{\sqrt{1-z^2}}\tag{18}$$ Using Beta/Gamma functions one can evaluate these integrals easily and show that $MN=3\pi/2$.
Legendre's goal in this example is to relate these integrals to complete elliptic integrals using the formula $(17)$. For the integral $M$ he makes the substitution $z=(1+x^2)^{-3/2}$ to arrive at $$M=\int_0^{\infty}\frac{3\,dx}{\sqrt{3+3x^2+x^4}}$$ and comparing it with $(17)$ we have $$f=3,g=0,a=\sqrt{3},b=1,\cos\theta=\sqrt {3}/2$$ so that $$k=\sin(\theta/2)=\sin(\pi/12)=k_3, \phi=\frac{\pi} {2}$$ and then we have $$M=3^{3/4}K(k_3)\tag{19}$$ and this also gives an evaluation of $K(k_3)$ in terms of Gamma functions.
For the integral $N$ he makes the substitution $z=(1-x^2)^{3/2}$ to arrive at $$N=\int_0^1\frac{3-3x^2}{\sqrt{3-3x^2+x^4}} \,dx$$ Comparing with equation $(13)$ we have $$f=3,g=-3,a=\sqrt{3},b=1,\cos\theta=-\frac{\sqrt{3}}{2}$$ so that $$k=\sin(\theta/2)=\cos(\pi/12)=k'_3$$ and $$\cos^2\phi=\frac{1+(3/2)-1}{2\sqrt{3}k^2}=2\sqrt{3}-3$$ so that $\sin\phi=\sqrt{3}-1$. Using formula $(17)$ we now have $$N=-3+\frac{3-3\sqrt{3}}{\sqrt[4]{3}}F(\phi,k)+\frac{6\sqrt {3}}{\sqrt [4]{3}}E(\phi,k)$$ Using theorem 2 above we have $$3F(\phi,k)=K(k)=K(k'_3)$$ and then using addition theorem $(5)$ twice we get $$3E(\phi,k)-E(k)=k^2\sin\phi(\sin\phi\sin\phi_2+\sin\phi_2\sin\phi_3)$$ where $\phi_3=\pi/2$ and $F(\phi_2,k)=2F(\phi,k)$. Using equation $(12)$ we can now see that $$3E(\phi,k)-E(k)=k^2\sin\phi\sin\phi_2(1+\sin\phi)=\frac{k^2}{k'}\cos^3\phi$$ and the expression on the right side of above equation equals $$\frac{2+\sqrt{3}}{4}\cdot\frac{2}{\sqrt {2-\sqrt {3}}}\cdot 3^{3/4}(2-\sqrt{3})^{3/2}=\frac{3}{2\sqrt[4]{3}}$$ and hence $$E(\phi, k) =\frac{E(k)}{3}+\frac{1}{2\sqrt[4]{3}}$$ and then $$N=\frac{2\sqrt {3}}{\sqrt[4]{3}}E(k'_3)-\frac{\sqrt{3}-1}{\sqrt[4]{3}}K(k'_3)\tag{20}$$ Multiplying equations $(19),(20)$ and noting that $MN=3\pi/2$ we get $$\frac{\pi} {4}=K(k_3)\left(E(k'_3)-\frac{\sqrt {3}-1} {2\sqrt {3}}K(k'_3)\right)\tag{21}$$
Example 2: Let $$P=\int_0^1\frac{z^{-1/3}\,dz}{\sqrt{1-z^2}},Q=\int_0^1\frac{z^{2/3}\,dz}{\sqrt{1-z^2}}\tag{22}$$ Again using Beta/Gamma functions we can show that $PQ=3\pi/4$.
We now relate these integrals to $K, E$. Using the substitution $z=(1-x^2)^{3/2}$ for $P$ we get $$P=\int_0^1\frac{3\,dx}{\sqrt{3-3x^2+x^4}}$$ Comparing with equation $(17)$ we have $$f=3,g=0,a=\sqrt{3},b=1,\cos\theta=-\frac{\sqrt{3}}{2}$$ so that $$k=\sin(\theta/2)=\cos(\pi/12)=k'_3$$ and like for integral $N$ we have $\sin\phi=\sqrt{3}-1$ and hence $$P=\frac{1}{\sqrt[4]{3}}K(k'_3)\tag{23}$$ For integral $Q$ we put $z^2=1/y^3$ to get $$Q=\frac{3}{2}\int_1^{\infty}\frac{dy}{y^2\sqrt{y^3-1}}$$ Using integration by parts we have $$\int\frac{dy}{y^2\sqrt{y^3-1}} =\frac{\sqrt{y^3-1}}{y}-\frac{1}{2}\int\frac{y\, dy} {\sqrt{y^3-1}}$$ putting $y=1+x^2$ we get $$\int\frac{y\, dy} {\sqrt{y^3-1}}=2\int\frac{1+x^2}{\sqrt {3+3x^2+x^4}}\,dx$$ and hence $$\frac{3}{2}\int\frac{dy}{y^2\sqrt{y^3-1}}=\frac{3}{2}\frac{x\sqrt {3+3x^2+x^4}}{1+x^2}-\frac{3x}{2}+\int\left(\frac{3}{2}-\frac{3(1+x^2)}{2\sqrt{3+3x^2+x^4}}\right)\,dx$$ Putting limits $y=1$ to $y=\infty$ and correspondingly $x=0$ to $x=\infty$ we get $$Q=\int_0^{\infty}\left(\frac{3}{2}-\frac{3(1+x^2) }{2\sqrt{3+3x^2+x^4}}\right)\,dx$$ We now have $$f=-3/2,g=-3/2,a=\sqrt{3},b=1,\cos\theta=\frac{\sqrt{3}}{2}$$ and then $$k=\sin(\theta/2)=\sin(\pi/12)=k_3$$ and $\phi=\pi/2$. Using equation $(17)$ we get $$Q=-\frac{3+\sqrt{3}}{2\sqrt[4]{3}}K(k_3)+\frac{3\sqrt{3}}{\sqrt [4]{3}}E(k_3)\tag{24}$$ Using $PQ=3\pi/4$ and multiplying equations $(23),(24)$ we get $$\frac{\pi} {4}=K(k'_3)\left(E(k_3)-\frac{\sqrt {3}+1}{2\sqrt{3}}K(k_3)\right)\tag{25}$$ In both equations $(21),(25)$ the right side involves a mix of $k_3,k'_3$ and this is fixed by Legendre using next example.
Example 3: Let $$R=\int_0^1\frac{dz}{(1-z^2)^{2/3}}\tag{26}$$ Putting $1-z^2=(z/y)^{3/2}$ the integral is transformed into $$R=\int_0^{\infty}\frac{dy}{\sqrt{4y^3+1}}$$ Putting $p=\sqrt[3]{4}$ and $py=x^2-1$ we get $$R=\frac{2}{3p}\int_{1}^{\infty}\frac{3\,dx}{\sqrt{3-3x^2+x^4}}$$ We have earlier seen that $$P=\int_{0}^{1}\frac{3\,dx}{\sqrt{3-3x^2+x^4}} =\frac{1}{\sqrt[4]{3}}K(k'_3)$$ and using the same technique we have $$ \int_{0}^{\infty}\frac{3\,dx}{\sqrt{3-3x^2+x^4}}=3P$$ and hence $$ \int_{1}^{\infty}\frac{3\,dx}{\sqrt{3-3x^2+x^4}}=2P=\frac{2}{q}K(k'_3)$$ where $q=\sqrt[4]{3}$ and thus $$R=\frac{4}{3pq}K(k'_3)\tag{27}$$ Using another set of substitutions Legendre next relates $R$ to $K(k_3)$. Putting $1-z^2=(1-(z/y)) ^3$ ie $$y=\frac{z} {1-\sqrt[3]{1-z^2}}$$ we get $$R=\sqrt{3}\int_1^{\infty}\frac{dy}{\sqrt{4y^3-1}}$$ Next using substitution $py=1+x^2$ we get $$R=\frac{2\sqrt{3}}{3p}\int_{\sqrt{p-1}}^{\infty}\frac{3\,dx}{\sqrt{3+3x^2+x^4}}$$ Now we can note that $$M=\int_{0}^{\infty}\frac{3\,dx}{\sqrt{3+3x^2+x^4}}$$ and hence $$ \int_{\sqrt{p-1}}^{\infty}\frac{3\,dx}{\sqrt{3+3x^2+x^4}}=M-\int_0^{\sqrt{p-1}}\frac{3\,dx}{\sqrt{3+3x^2+x^4}}$$ The integral on the right can be evaluated just like $M$ and we just need to check using equation $(15)$ that $$\cos^2\phi=\frac{-x^2-(3/2)+\sqrt {3+3x^2+x^4}}{2k_3^2\sqrt{3}}$$ where $x^2=p-1$ and then $$\cos^2\phi=\frac{-2p-1+\sqrt{4p^2+4p+4}}{2\sqrt {3}-3}$$ which simplifies to $$\frac{-2p-1+\sqrt{4p^2+4p+4}}{2\sqrt {3}-3}= \frac{-2p-1+\sqrt{p^4+4p^2+4}}{2\sqrt {3}-3}=\frac{(p-1)^2}{2\sqrt{3}-3}$$ And thus $$\cos\phi=(p-1)\sqrt{\frac{2+\sqrt{3}}{\sqrt {3}}}$$ and by theorem 2 we have $F(\phi, k_3)=K(k_3)/3$. Then $$\int_0^{\sqrt{p-1}}\frac{3\,dx}{\sqrt{3+3x^2+x^4}} =\frac{M} {3}$$ and $$\int_{\sqrt{p-1}}^{\infty}\frac{3\,dx}{\sqrt{3+3x^2+x^4}} =\frac{2M}{3}=\frac{2}{q}K(k_3)$$ and finally $$R=\frac{4\sqrt{3}}{3pq}K(k_3)\tag{28}$$ Using equation $(27),(28)$ we get $$\frac{K(k'_3)}{K(k_3)}=\sqrt{3}\tag{29}$$ and we can recast equations $(21),(25) $ as \begin{align} \frac{\pi\sqrt{3}}{4}&=K(k'_3)\left(E(k'_3)-\frac{\sqrt {3}-1} {2\sqrt {3}}K(k'_3)\right)\tag{30a}\\
\frac{\pi}{4\sqrt{3}}&=K(k_3)\left( E(k_3)-\frac{\sqrt {3}+1} {2\sqrt {3}}K(k_3)\right)\tag{30b}
\end{align}
The above equations should be compared with equation $(6)$ of this answer of mine which is a general relation between $K, E$ for singular modulus.
