Prove $\lim_{n\to\infty}\int_0^{a} \left(\sqrt{2n/\pi-x^2}-\sqrt{2n/\pi-a^2}\right)dx=1/6$ where $a$ is the largest real root of $4x^6+x^2=2n/\pi$.

Question

I've never seen anything like this before: an unsolvable cubic, within a definite integral, within a limit (which applies to the cubic and the integral), resulting in a simple closed form.

Prove $\lim\limits_{n\to\infty}\int_0^{a} \left(\sqrt{2n/\pi-x^2}-\sqrt{2n/\pi-a^2}\right)dx=1/6$ where $a$ is the largest real root of $4x^6+x^2=2n/\pi$.

It arose in my attempt to answer a question about packing rectangles in a semicircle. Numerical investigation suggests that it is true. That's about all I can make of it.

Answer 1

From the comments by @metamorphy:

In the integral, let $2n/\pi=4a^6+a^2$.

Then integrate to get

$$\frac{a^2}{2}(4a^4+1)\arcsin\frac1{\sqrt{4a^4+1}}-a^4$$

Then take the limit as $a\to\infty$ (for example by using Maclaurin series).