(This continues the post on integrals that use roots of reciprocal polynomials.)
Given $N=7$. First, how do we show that the algebraic number $\beta$ that solves,
$$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt[4]{1-x}\ \sqrt{1-x\,\beta^2}}=\frac{1}{7}\,\frac{2\pi}{\sqrt{2}\;|\beta|}\tag1$$
is a root of,
$$\beta^2-\frac{2}{1+2\cos\tfrac{\pi}{14}}\beta+\frac{4}{2-\sqrt{2}\,\sec\tfrac{13\pi}{28}}=0$$
Second, the algebraic number $\gamma$ for the related integral,
$$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{1-x\,\gamma^2}}=\frac{1}{7}\,\frac{2\pi}{\sqrt{2\gamma}}\tag2$$
is a root of,
$$\gamma^2-2\cos\tfrac{2\pi}{7}\,\gamma+\frac{2-\sqrt{2}\,\csc\tfrac{9\pi}{28}}{4}=0$$
Equivalently, these roots $\beta$ and $\gamma$ satisfy,
$$I\left(\beta^2;\ \tfrac12,\tfrac14\right)=I\left(\gamma^2;\ \tfrac14,\tfrac14\right)=\frac{1}{7}$$
with regularized beta function $I(z;a,b)$.
Lastly, why is the minpoly of $\beta$ and $\gamma$ which are $12$-deg equations actually reciprocal polynomials?
P.S. The cases $N=2,3,5$ can also be expressed by trigonometric functions, such as for $N=5$, $$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt[4]{1-x}\ \sqrt{1-x\big(1-\tan\tfrac{3\pi}{20}\big)^2}}=\frac{\sqrt{2}\,\pi}{5-5\tan\tfrac{3\pi}{20}}$$ though to do so for $N=11$ given in this post would be difficult.
