Show that $\int_0^{\pi/3}\arccos(2\sin^2 x-\cos x)\mathrm dx=\frac{\pi^2}{5}$

Question

There is numerical evidence that $$ I = \int_0^{\pi/3}\arccos\left(2 \sin^{2}\left(x\right) -\cos\left(x\right)\right)\mathrm dx = \frac{\pi^{2}}{5} $$ How can this be proved ?.

• I was trying to answer another question, and I got it down to this integral.
• I tried integration by parts, substitution and Maclaurin series, but it seems that this integral requires more advanced techniques.
• $\tt Wolfram$ does not evaluate the indefinite integral.
• Here is the graph of $$ y = \arccos\left(2\sin^{2}\left(x\right) -\cos\left(x\right)\right)\quad\mbox{for}\quad 0\le x\le \frac{\pi}{3} $$ It intersects the axes at $\left(\pi/3,0\right)$ and $\left(0,\pi\right)$.

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• Not sure if this helps, but apparently we also have $$ \int_{0}^{\pi/3}\arccos^{\color{red}{2}} \left(2\sin^{2}\left(x\right) - \cos\left(x\right)\right) \mathrm{dx} = \frac{19\pi^3}{135}$$ Edit: I ask about this here.

Answer 1

$$\int_0^\frac{\pi}{3}\arccos(2\sin^2 x-\cos x)dx\overset{\large \tan\frac{x}{2}\to x}=2\int_0^\frac{1}{\sqrt 3}\frac{\arccos\left(\frac{-1+8x^2+x^4}{(1+x^2)^2}\right)}{1+x^2}dx$$

$$=4\int_0^\frac{1}{\sqrt 3} \frac{\arctan\left(\frac{1}{x}\sqrt{\frac{1-3x^2}{5+x^2}}\right)}{1+x^2}dx\overset{\large \frac{1}{x}\to\sqrt x}=2\int_3^\infty \frac{\arctan \left(\sqrt x\sqrt{\frac{x-3}{5x+1}}\right)}{\sqrt{x}(1+x)}dx$$

$$\overset{\large \frac{x-3}{5x+1}\to x}=8\int_0^\frac{1}{5} \frac{1}{(1-x)(1-5x)} \frac{\arctan\left(\sqrt x \sqrt{\frac{3+x}{1-5x}}\right)}{\sqrt{\frac{3+x}{1-5x}}} dx$$

$$=8\int_0^\frac{1}{5} \frac{1}{(1-x)(1-5x)} \left(\int_0^{\sqrt x} \frac{1}{1+\frac{3+x}{1-5x}y^2}dy\right)dx$$

$$=8\int_0^\frac{1}{\sqrt 5}\int_{y^2}^\frac{1}{5} \frac{1}{1-x} \frac{1}{(1+3y^2)-(5-y^2)x}dxdy=2\int_0^\frac{1}{\sqrt 5} \frac{\ln\left(\frac{1-y^2}{4y^2}\right)}{1-y^2}dy$$

$$\overset{y\to\frac{1-y}{1+y}}=\int_{\large\frac{1}{\phi^2}}^1 \frac{\ln\left(\large \frac{y}{(1-y)^2}\right)}{y}dy=\int_{\large\frac{1}{\phi^2}}^1 \frac{\ln\left(y\right)-2\ln\left(1-y\right)}{y}dy$$$$=2 \operatorname{Li}_2(1)-2\operatorname{Li}_2\left(\frac{1}{\phi^2}\right)-2\ln^2\phi=\boxed{\frac{\pi^2}{5}}$$

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

Answer 2

Too long for a comment. This is only a partial answer, but I believe it may help in some manner.

As pointed out by @Po1ynomial, the integral is equivalent to

$$ I = \int_0^{\frac{1}{\sqrt{3}}} \arctan \left(x \frac{\sqrt{x^2+5}}{\sqrt{1-3x^2}}\right) \frac{dx}{x^2+1}$$

Then we let

$$ I(a) = \int_0^{\frac{1}{\sqrt{3}}} \arctan \left(ax \frac{\sqrt{x^2+5}}{\sqrt{1-3x^2}}\right) \frac{dx}{x^2+1}$$

Differentiating with respect to $a$, we get:

$$I'(a) = \int_0^{\frac{1}{\sqrt{3}}} \frac{x\sqrt{x^2+5}\sqrt{1-3x^2}}{(1 + (5a^2-3)x^2 + a^2x^4)(x^2+1)} dx$$

Let

$$ \sqrt{1-3x^2} = u$$

Then

$$ x = \frac{1}{\sqrt{3}}\cdot \sqrt{1-u^2}$$

$$dx = \frac{-1}{\sqrt{3}}\cdot \frac{u}{\sqrt{1-u^2}} du$$

The integral becomes

$$I'(a) = \int_0^1 \frac{\frac{1}{\sqrt{3}}\cdot \sqrt{1-u^2} \cdot \frac{\sqrt{16-u^2}}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}}\cdot \frac{u}{\sqrt{1-u^2}}}{\left(1 + (5a^2-3)\cdot \frac{1-u^2}{3} + a^2\cdot \frac{(1-u^2)^2}{9}\right) \cdot (\frac{4-u^2}{3})}$$

$$I'(a) = 3\sqrt3 \cdot \int_0^1 \frac{u^2 \sqrt{(16-u^2)}}{(16a^2-(17a^2-9)u^2 + a^2u^4)(4-u^2)} du$$

Now substitute $u = 4\mathrm{sin}t$. The integral then becomes

$$I'(a) = 3\sqrt3 \cdot \int_0^{\alpha} \frac{ 16\sin^2t \cdot 16\cos^2t}{(16a^2-(17a^2-9)\cdot 16\sin^2t + a^2 \cdot 256\sin^4t)(4-16\sin^2t)} dt$$

Where $\sin{\alpha} = \frac{1}{4}$

Dividing by $\sec^6t$ throughout and substituting $\operatorname{tan}t = v$, we get:

$$I'(a) = \frac{3\sqrt3}{4} \cdot \int_0^{\tan{\alpha}} \frac{v^2}{[16a^2(v^2+1)^2 - (17a^2-9)\cdot v^2(v^2+1) + 16a^2 \cdot v^4](1-3v^2)}dv$$

$$ = \frac{3\sqrt3}{4}\cdot \int_0^{\frac{1}{\sqrt{15}}} \frac{v^2}{\left[(15a^2+9)v^4 + (15a^2+9)v^2 + 16a^2)\right]\cdot (1-3v^2)}dv$$

I am stuck here. Maybe someone else can find this useful.