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 $(1)\implies$

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

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

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

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

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

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


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

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 $(1)?$

  • $\begingroup$ What is the question, anyway? $\endgroup$
    – DonAntonio
    Jun 2, 2017 at 15:34
  • $\begingroup$ Is it possible you use the sign =? (1)=(6)? $\endgroup$
    – FDP
    Jun 2, 2017 at 16:50
  • 1
    $\begingroup$ the function $f(x)=\dfrac{3+x^2}{x^4}$ has an anti-derivative. Never heard of integration by parts? $\endgroup$
    – FDP
    Jun 2, 2017 at 16:53
  • 1
    $\begingroup$ @FDP i always wonder what this user(S) learns from our answers. He always gets stuck at the same points and seems not to remember (or look up) the techniques used to answer his other questions $\endgroup$
    – tired
    Jun 2, 2017 at 17:08
  • $\begingroup$ This is the third question of yours that can be solved in the same way: a substitution and Feynman's trick. It might be a good moment to start learning from experience. $\endgroup$ Jun 2, 2017 at 17:11

2 Answers 2


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

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

\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}


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 $(0,1)$.


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