Show that $\int_{\arccos(1/4)}^{\pi/2}\arccos(\cos x (2\sin^2x+\sqrt{1+4\sin^4x})) \mathrm dx=\frac{\pi^2}{40}$

Question

There is numerical evidence that $$I=\int_{\arccos(1/4)}^{\pi/2}\arccos\left(\cos x\left(2\sin^2x+\sqrt{1+4\sin^4x}\right)\right)\mathrm dx=\frac{\pi^2}{40}$$

How can this be proved?

I was trying to answer another question, and I got it down to this integral.

Wolfram does not evaluate the indefinite integral.

I tried techniques from a roughly similar integral. Letting $u=\tan \frac{x}{2}$, I got

$$I=\int_{\sqrt{3/5}}^1 \dfrac{2\arccos{\left(\frac{(1-u^2)\left(8u^2+\sqrt{u^8+4u^6+70u^4+4u^2+1}\right)}{(1+u^2)^3}\right)}}{1+u^2}\mathrm du$$

but I don't know what to do with this.

Answer 1

Some algebra to arrive at a manageable integral

Notice that the $\arccos $ argument is the "$+$" solution of a quadratic equation:

$$\cos x\left(2\sin^2x\pm\sqrt{1+4\sin^4x}\right)=\frac{2\sin x \sin(2x)\pm\sqrt{4\sin^2 x\sin^2(2x)+4\cos^2 x}}{2}$$

More exactly, with $b=-2\sin x\sin(2x)$ and $c=-\cos^2x$ in $y_{\pm}=\frac{-b\pm\sqrt{b^2-4c}}{2}$.

We also have $\arccos y_-\pm\arccos y_+=\arccos\left(y_-y_+\mp\sqrt{1-(y_-+y_+)^2 +2y_-y_+ +y_-^2y_+^2}\right)$, and since Vieta's relations allows us to utilize $y_-+y_+=2\sin x\sin(2x)$ and $\ y_-y_+=-\cos^2 x$, it makes sense to consider the following integrals:

$$I_{\pm}=\int_{\arccos(1/4)}^{\pi/2}\arccos\left(\cos x\left(2\sin^2x\pm\sqrt{1+4\sin^4x}\right)\right) dx$$

The original integral can then be rewritten as $I_+=\frac12\left((I_- + I_+)-(I_- - I_+)\right)$, where:

$$I_- \pm I_+=\int_{\arccos(1/4)}^{\pi/2}\arccos\left(-\cos^2 x\mp\sin^2 x\sqrt{1-16\cos^2 x}\right)dx$$

$$\overset{\sqrt{1-16\cos^2 x}\to x}=\int_0^1 \frac{x\arccos \left(-\frac{1-x^2}{16}\mp x\frac{15+x^2}{16}\right)}{\sqrt{1-x^2}{\sqrt{15+x^2}}}dx$$

Moreover, substituting $x\to -x$ in $I_- - I_+$, gives us:

$$I_+=\frac12\int_{-1}^1 \frac{x\arccos \left(-\frac{1-x^2}{16}- x\frac{15+x^2}{16}\right)}{\sqrt{1-x^2}{\sqrt{15+x^2}}}dx=\int_{-1}^1 \frac{x\operatorname{arctan}\sqrt{\frac{1-x^2}{15+x^2} \left(1+\frac{16}{(1-x)^2}\right)}}{\sqrt{1-x^2}{\sqrt{15+x^2}}}dx$$

In the last step it was utilized that $\arccos x =2\arctan \sqrt{\frac{1-x}{1+x}}$.

---

Evaluation of the integral

$$\int_{-1}^1 \frac{x\operatorname{arctan}\sqrt{\frac{1-x^2}{15+x^2} \left(1+\frac{16}{(1-x)^2}\right)}}{\sqrt{1-x^2}{\sqrt{15+x^2}}}dx=\int_{-1}^1 \int_0^{\sqrt{1+\frac{16}{(1-x)^2}}} \frac{x}{(15+x^2)+(1-x^2) y^2}dy dx$$

$$\overset{(*)}=16\int_{-1}^1 \int_0^{x} \frac{x}{(15+x^2)+(1-x^2) \left(1+\frac{16}{(1-y)^2}\right)}\frac{1}{(1-y)^2\sqrt{16+(1-y)^2}} dy dx$$

$$=16\int_{-1}^1 \frac{1}{(1-y)^2\sqrt{16+(1-y)^2}}\int_y^1 \frac{x}{(15+x^2)+\left(1+\frac{16}{(1-y)^2}\right)(1-x^2)} dx dy$$

$$=-\frac12\int_{-1}^1 \frac{\ln\left(\frac{1-y}{2}\right)}{\sqrt{16+(1-y)^2}}dy\overset{\frac{1-y}{2}\to y}=-\frac12 \int_0^1 \frac{\ln y}{\sqrt{4+y^2}}dy$$

$$\overset{y \to \frac{1-y^2}{y}}=\frac12 \int_{\large \frac{1}{\phi}}^1 \frac{\ln\left(\frac{y}{1-y^2}\right)}{y}dy =\frac14\operatorname{Li}_2(1) - \frac14\operatorname{Li}_2\left(\frac{1}{\phi^2}\right) - \frac{1}{4}\ln^2\phi \overset{(**)}=\, \boxed{\frac{\pi^2}{40}}$$

---

In $(**)$, the following dilogarithm values were employed:

$$\operatorname{Li}_2(1)=\frac{\pi^2}{6},\quad \operatorname{Li}_2\left(\frac{1}{\phi^2}\right) = \frac{\pi^2}{15}-\ln^2\phi,\quad \phi=\frac{1+\sqrt 5}{2}$$

Also, the $(*)$ step is not necessary, however I didn't know how to change the order of integration directly, so I had to make the substitution $y^2\to 1+\frac{16}{(1-y)^2}$ after noticing that the $x$-integral is odd which allows to write $\int_{-1}^1 \int_0^{f(x)}dydx = -\int_{-1}^1 \int_{f(x)}^\infty dydx$. Finally, $-\int_{-1}^1 \int_x^1 dydx$ was also rewritten as $\int_{-1}^1 \int_0^x dydx$ in order to change the order of integration easier.