How to prove that $\int_{0}^{\pi/2}{\arctan(2\cos^2 x)\over \cos^2 x}\mathrm dx={\pi\over \sqrt{\phi}}?$

Question

Given that

$$\int_{0}^{\pi/2}{\arctan(2\cos^2 x)\over \cos^2 x}\mathrm dx={\pi\over \sqrt{\phi}}\tag1$$ where $\phi={\sqrt{5}+1\over 2}$

$t=2\cos^2 x\implies\mathrm dt=-4\sin x\cos x\,\mathrm dx$

$${1\over 4}\int_{0}^{1}{\arctan t\over t\sqrt{t-t^2}}\mathrm dt\tag2$$

$t=\tan v\implies\mathrm dt=\sec^2v\,\mathrm dv$

$${1\over 4}\int_{0}^{\pi/4}{v\sec^2 v\over \tan v\sqrt{\tan v-\tan^2 v}}\mathrm dv\tag3$$

$${1\over 2}\int_{0}^{\pi/4}{v\over \sin(2v)\sqrt{\tan v-\tan^2 v}}\mathrm dv\tag4$$

Or, if we leave it in terms of $\tan t$,

$${1\over 4}\int_{0}^{\pi/4}{v(1+\tan^2 v)\over \tan v\sqrt{\tan v-\tan^2 v}}\mathrm dv\tag5$$

Answer 1

In $(2)$, you should've gotten

$$I=\int_0^2\frac{\arctan(t)}{t\sqrt{2t-t^2}}~\mathrm dt$$

By letting $t=1/u$, we get

$$I=\int_{1/2}^{+\infty}\frac{\arctan(1/u)}{\sqrt{2u-1}}~\mathrm du$$

And by integration by parts,

$$I=\int_{1/2}^{+\infty}\frac{\sqrt{2u-1}}{1+u^2}~\mathrm du$$

which is the real part of

$$I_1=\int_{-\infty}^{+\infty}\frac{\sqrt{2u-1}}{1+u^2}~\mathrm du$$

And by taking a semi-circle contour and the residue theorem,

$$I=\Re(I_1)=\Re(\pi\sqrt{2i-1})=\frac\pi{\sqrt\phi}$$

Answer 2

Let $$I(a)=\int_{0}^{\pi/2}{\arctan(a\cos^2 x)\over \cos^2 x}\mathrm dx.$$ Then $I(0)=0,I(2)=I$ and \begin{eqnarray} I'(a)&=&\int_{0}^{\pi/2}{1\over{1+a^2\cos^4 x}}\mathrm dx\\ &=&\int_{0}^{\pi/2}{\sec^2x\over{a^2+\sec^4 x}}\sec^2x\mathrm dx\\ &=&\int_{0}^{\infty}{1+u^2\over{a^2+(1+u^2)^2}}\mathrm du\\ &=&\int_{0}^{\infty}{1+u^2\over{(u^2+1+ai)(u^2+1-ai)}}\mathrm du\\ &=&\frac12\int_{0}^{\infty}\left({1\over{u^2+1+ai}}+{1\over{u^2+1-ai}}\right)\mathrm du\\ &=&\frac12\bigg[\frac{1}{\sqrt{1+ai}}\arctan\frac{u}{\sqrt{1+ai}}+\frac{1}{\sqrt{1-ai}}\arctan\frac{u}{\sqrt{1-ai}}\bigg]\bigg|_{0}^{\infty}\\ &=&\frac{\pi}{4}\bigg(\frac{1}{\sqrt{1-ai}}+\frac{1}{\sqrt{1+ai}}\bigg) \end{eqnarray} So \begin{eqnarray} I(2)&=&\frac{\pi}{4}\int_{0}^{2}\bigg(\frac{1}{\sqrt{1-ai}}+\frac{1}{\sqrt{1+ai}}\bigg)\\ &=&\frac{\pi}{4}\bigg(\frac{\sqrt{1+ai}}{i}+\frac{\sqrt{1-ai}}{-i}\bigg)\bigg|_0^2\\ &=&\frac{\pi}{4}\frac{\sqrt{1+2i}-\sqrt{1-2i}}{i}\\ &=&\frac{\sqrt{2\sqrt5-2}}{4}\pi=\frac{\pi}{\sqrt\phi} \end{eqnarray}

Answer 3

\begin{align}J&=\int_{0}^{\pi/2}{\arctan(2\cos^2 x)\over \cos^2 x}\mathrm dx\\ &\overset{u=\tan x}=\int_0^\infty\arctan\left(\frac{2}{1+u^2}\right)du\\ &\overset{\text{IBP}}=\underbrace{\left[u\arctan\left(\frac{2}{1+u^2}\right)\right]_0^\infty}_{=0}+4\int_0^\infty \frac{u^2}{u^4+2u^2+5}du\\ &\overset{u=5^{\frac{1}{4}}t}=4\times 5^{\frac{3}{4}}\int_0^\infty\frac{t^2}{5t^4+2t^2\sqrt{5}+5}dt\\ &\overset{w=\frac{1}{t}}=4\times 5^{\frac{3}{4}}\int_0^\infty\frac{1}{5w^4+2w^2\sqrt{5}+5}dw\\ 2J&=4\times 5^{\frac{3}{4}}\int_0^\infty\frac{1+w^2}{5w^4+2w^2\sqrt{5}+5}d=4\times 5^{-\frac{1}{4}}\int_0^\infty\frac{1+\frac{1}{w^2}}{w^2+\frac{2}{\sqrt{5}}+\frac{1}{w^2}}dw\\ &=4\times 5^{-\frac{1}{4}}\int_0^\infty\frac{1+\frac{1}{w^2}}{\left(w-\frac{1}{w}\right)^2+\frac{2}{\sqrt{5}}+2}dw\\ &\overset{z=w-\frac{1}{w}}=4\times 5^{-\frac{1}{4}}\int_{-\infty}^{+\infty}\frac{1}{z^2+\frac{2}{\sqrt{5}}+2}dz=\frac{2\sqrt{2}}{\sqrt{1+\sqrt{5}}}\left[\arctan\left(\frac{5^{\frac{1}{4}}z}{\sqrt{2}\sqrt{1+\sqrt{5}}}\right)\right]_{-\infty}^{+\infty}\\ &=\frac{2\pi\sqrt{2}}{\sqrt{1+\sqrt{5}}}=\frac{2\pi}{\sqrt{\frac{1+\sqrt{5}}{2}}}\\ J&=\boxed{\frac{\pi}{\sqrt{\varphi}}} \end{align}

Answer 4

Via Feynman's technique and Glasser's master theorem:

$$\begin{align}\mathcal I(a)&=\int_0^\frac\pi2\tan^{-1}(a\cos^2x)\sec^2x\,\mathrm dx, a\ge0\\&=\int_0^\infty\tan^{-1}\left(\frac{a}{x^2+1}\right)\mathrm dx\\\implies\mathcal I'(a)&=\int_0^\infty\frac{x^2}{x^4+2x^2+a^2+1}\mathrm dx+\int_0^\infty\frac{\frac1{x^2}}{\frac{a^2+1}{x^4}+\frac2{x^2}+1}\frac{\mathrm dx}{x^2}\\&=\int_0^\infty\frac{\mathrm dx}{\left(x-\frac{\sqrt{a^2+1}}x\right)^2+2\left(\sqrt{a^2+1}+1\right)}+\int_0^\infty\frac{\mathrm dx}{\left(x\sqrt{a^2+1}-\frac1x\right)^2+2\left(\sqrt{a^2+1}+1\right)}&x\to\frac1x\text{ in the second integral}\\&=\left(1+\frac1{\sqrt{a^2+1}}\right)\int^\infty_0\frac{\mathrm dx}{x^2+2\left(\sqrt{a^2+1}+1\right)}&\text{Glasser's master theorem}\\&=\frac\pi{2\sqrt2}\frac{\sqrt{\sqrt{a^2+1}+1}}{\sqrt{a^2+1}}\\&=\frac\pi{2\sqrt2}\frac1{\sqrt{\sqrt{a^2+1}-1}}\frac{a}{\sqrt{a^2+1}}\\\implies\mathcal I(2)&=\mathcal I(0)+\frac\pi{2\sqrt2}\int_1^\sqrt5\frac{\mathrm db}{\sqrt{b-1}}&b=\sqrt{a^2+1}\\&=\frac\pi{\sqrt\phi}\end{align}$$

Answer 5

\begin{align} &\int_{0}^{\pi/2}{\arctan(2\cos^2 x)\over \cos^2 x}\mathrm dx\\ =& \int_{0}^{\pi/2}\int_0^{\ln\phi^3}\frac{\cosh y}{1+\sinh^2 y \cos^4 x}dy \overset{t=\tan x}{dx}\\ =& \int_0^{\ln\phi^3}\int_0^\infty \frac{(1+t^2) \cosh y}{t^4+2t^2+\cosh^2y}dt\ dy\\ = & \ \frac\pi2 \int_0^{\ln\phi^3}\cosh\frac y2\ dy =\pi \sinh\frac y2\bigg|_0^ {\ln\phi^3}={\pi\over \sqrt{\phi}} \end{align}