How to show that $\int_{0}^{\pi/2}{\arctan(\tan^2 x)\over \sin^2 x\sqrt{\tan x}}\cdot(3\pm\tan x)\mathrm dx=2\pi\sqrt{2\pm \sqrt{2}}?$

Question

A bit of messy integral but seem to yield a simple closed form

Given that:

$$\int_{0}^{\pi/2}{\arctan(\tan^2 x)\over \sin^2 x\sqrt{\tan x}}\cdot(3\pm\tan x)\mathrm dx=2\pi\sqrt{2\pm \sqrt{2}}\tag1$$

Simplifying this part doesn't yield a simple from.

$${3+\tan x\over \sin^2 x\sqrt{\tan x}}$$

Else we can split the integral $(1)\implies$

$$3\int_{0}^{\pi/2}{\arctan(\tan^2 x)\over \sin^2 x\sqrt{\tan x}}\mathrm dx+\int_{0}^{\pi/2}{\sqrt{\tan x}\arctan(\tan^2 x)\over \sin^2 x}\mathrm dx=I+J\tag2$$

$$2\sin^2x=1-{1-\tan^2 x\over 1+\tan^2 x}$$

$$\sin^2x={\tan^2x\over 1+\tan^2x}$$

$$I=3\int_{0}^{\pi/2}{1+\tan^2x\over \tan^2x}\cdot{\arctan(\tan^2 x)\over \sqrt{\tan x}}\mathrm dx$$

$$J=\int_{0}^{\pi/2}{1+\tan^2x\over \tan^2x}\cdot{\sqrt{\tan x}\arctan(\tan^2 x)}\mathrm dx$$

$$u=\tan^2x\implies du=2\tan x\sec^2xdx=2u^{1/2}+2u^{3/2}$$

$I+J\implies$

$${3\over 2}\int_{0}^{\infty}{1+u\over u^{7/4}+u^{11/4}}\arctan(u)\mathrm du+{1\over 2}\int_{0}^{\infty}{1+u\over u^{5/4}+u^{9/4}}\arctan(u)\tag3$$

simplify to

$${3\over 2}\int_{0}^{\infty}{1\over u^{7/4}}\arctan(u)\mathrm du+{1\over 2}\int_{0}^{\infty}{1\over u^{5/4}}\arctan(u)\tag4$$

$${1\over 2}\int_{0}^{\infty}(3+u^{1/2}){\arctan(u)\over u^{7/4}}\mathrm du\tag5$$

$$u=v^4\implies du=4v^3dv$$

$$2\int_{0}^{\infty}(3+v^2){\arctan(v^4)\over v^4}\cdot{\mathrm dv}\tag6$$

Q: How can we prove $(1)?$

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{\pi/2}{\arctan\pars{\tan^{2}\pars{x}} \over \sin^{2}\pars{x}\root{\tan\pars{x}}}\,\bracks{3 \pm \tan\pars{x}}\,\dd x = 2\pi\root{2 \pm \root{2}}:\ {\large ?}}$.

\begin{align} &\int_{0}^{\pi/2}{\arctan\pars{\tan^{2}\pars{x}} \over \sin^{2}\pars{x}\root{\tan\pars{x}}}\,\bracks{3 \pm \tan\pars{x}}\,\dd x \\[1cm] & = \int_{x\ =\ 0}^{x\ =\ \pi/2} {\tan^{2}\pars{x} + 1 \over \tan^{2}\pars{x}} {\arctan\pars{\tan^{2}\pars{x}} \over \bracks{\tan^{2}\pars{x}}^{1/4}}\,\bracks{3 \pm \bracks{\tan^{2}\pars{x}}^{1/2}} \,\times \\[3mm] & \phantom{=\int_{x\ =\ 0}^{x\ =\ \pi/2}} {\dd\bracks{\tan^{2}\pars{x}} \over 2\bracks{\tan^{2}\pars{x}}^{1/2}\bracks{\tan^{2}\pars{x} + 1}} \\[1cm] \stackrel{\tan^{2}\pars{x}\ \mapsto\ x}{=}\,\,\,& {1 \over 2}\int_{0}^{\infty}{\arctan\pars{x} \over x^{7/4}} \,\pars{3 \pm x^{1/2}}\,\dd x \\[5mm] = &\ {1 \over 2}\int_{x\ =\ 0}^{x\ \to\ \infty}\!\!\!\!\!\arctan\pars{x} \,\dd\bracks{\pars{-4x^{-3/4}} \pm \pars{-4x^{-1/4}}} \\[5mm] \stackrel{\mbox{IBP}}{=}\,\,\,& 2\int_{0}^{\infty}{x^{-3/4} \pm x^{-1/4} \over x^{2} + 1}\,\dd x \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, \int_{0}^{\infty}{x^{-7/8} \pm x^{-5/8} \over x + 1}\,\dd x \\[5mm] \stackrel{t\ =\ 1/\pars{x + 1} \iff x = 1/t - 1}{=}\,\,\,& \int_{1}^{0}t\bracks{\pars{{1 \over t} - 1}^{-7/8} \pm \pars{{1 \over t} - 1}^{-5/8}}\pars{-\,{\dd t \over t^{2}}} \\[5mm] = &\ \int_{0}^{1}t^{-1/8}\,\pars{1 - t}^{-7/8}\,\dd t \pm \int_{0}^{1}t^{-3/8}\,\pars{1 - t}^{-5/8}\,\dd t \\[5mm] = &\ {\Gamma\pars{7/8}\Gamma\pars{1/8} \over \Gamma\pars{1}} \pm {\Gamma\pars{5/8}\Gamma\pars{3/8} \over \Gamma\pars{1}} = {\pi \over \sin\pars{\pi/8}} \pm {\pi \over \sin\pars{3\pi/8}} \\[5mm] = &\ \pi\bracks{{1 \over \sin\pars{\pi/8}} \pm {1 \over \cos\pars{\pi/8}}} \\[5mm] = &\ 2\pi\,\braces{% {\root{\bracks{1 + \cos\pars{\pi/4}}/2} \over \sin\pars{\pi/4}} \pm {\root{\bracks{1 - \cos\pars{\pi/4}}/2} \over \sin\pars{\pi/4}}} \\[5mm] & = 2\pi\,{\root{2 + \root{2}} \pm \root{2 - \root{2}} \over \root{2}} = \bbx{\root{2 \pm \root{2}}} \end{align}

Answer 2

By setting $x=\arctan t$ we are left with $$ \int_{0}^{+\infty}\frac{\arctan(t^2)}{t^{5/2}}(3\pm t)\,dt \stackrel{t\mapsto\sqrt{u}}{=} \frac{1}{2}\int_{0}^{+\infty}\frac{3\pm\sqrt{u}}{u^{7/4}}\arctan(u)\,du \tag{1}$$ and we may apply Feynman's trick to $\int_{0}^{+\infty}\frac{3\pm\sqrt{u}}{u^{7/4}}\arctan(\alpha u)\,du$. We have: $$ \int_{0}^{+\infty}\frac{(3\pm\sqrt{u})u}{u^{7/4}(1+\alpha^2 u^2)}\,du = \frac{\pi}{2\alpha^{3/4}}\left(3\sqrt{\alpha}\csc\frac{\pi}{8}\pm\sec\frac{\pi}{8}\right) \tag{2}$$ for any $\alpha<0$ by the residue theorem, hence the claim readily follows by integrating $(2)$ over $(0,1)$.