$$\int_0^1\frac{3x^3-2x}{(1+x)\sqrt{1-x}}K\left(\frac{2x}{1+x}\right)\,dx\stackrel ?=\frac\pi{5\sqrt2}$$

The integral above comes from the evaluation of the integral $A=\int_0^{\pi/2}\frac{f(\theta)}\pi d\theta$, where

$$f(\theta)=\int_0^\pi\frac{(3\sin^2\theta-2)\sin\theta\,d\phi}{\sqrt{2+2\sin\theta\cos\phi}}=\frac{\sqrt2(3\sin^2\theta-2)}{\csc\theta\sqrt{1+\sin\theta}}\int_0^{\pi/2}\left(1-\frac{2\sin\theta}{1+\sin\theta}\sin^2\gamma\right)^{-1/2}d\gamma,$$

where $\gamma=\phi/2$, and the right side integral is $K\big(\!\frac{2\sin\theta}{1+\sin\theta}\!\big)$ by definition. After substitutions $x=\sin\theta$ and $y=\frac{2x}{1+x}$ we get

$$\begin{align}A&=\frac{\sqrt2}\pi\int_0^1\frac{3x^3-2x}{(1+x)\sqrt{1-x}}K\left(\frac{2x}{1+x}\right)\,dx\\ &=\frac1\pi\int_0^1\frac{y^3+8y^2-8y}{(2-y)^3\sqrt{y^2-3y+2}}K(y)\,dy\stackrel ?=\frac15,\end{align}$$

where the integral has been numerically evaluated to suggest the analytic result on the right to several thousand digits. Is there any way to prove this equality, and are there any generalizations of this conjecture to include other parameters?

**P.S.** Not that it's overly relevant to the question, but for the interested reader, the original integral comes from the following physics problem:

[Morin

Intro to Classical Mechanics, Ex. 10.12] The earth bulges slightly at the equator, due to the centrifugal force in the earth’s rotating frame. The goal of this exercise is to find the shape of the earth, first incorrectly, and then correctly.(a) The common incorrect method is to assume that the gravitation force from the slightly nonspherical earth points toward the center, and to then calculate the equipotential surface (incorporating both the gravitational and centrifugal forces). Show that this method leads to a surface whose height (relative to a spherical earth of the same volume) is given by $h(\theta)=R\big(\!\frac{R\omega^2}{6g}\!\big)(3\sin^2\theta-2)$, where $\theta$ is the polar angle (the angle down from the north pole), and $R$ is the radius of the earth.

(b) The above method is incorrect, because the slight distortion of the earth causes the gravitational force to not point toward the center of the earth (except at the equator and the poles). This tilt in the force direction then changes the slope of the equipotential surface, and it turns out (although this is by no means obvious) that this effect is of the same order as the slope of the surface found in part (a). Your task: Assuming that the density of the earth is constant, and that the correct height takes the form of some constant factor $f$ times the result found in part (a), show that $f=5/2$. Do this by demanding that the potential at a pole equals the potential at the equator.