$$\int_{0}^{\pi\over 2}\arctan(\tan^n(x))\mathrm dx={\pi^2\over 8}\tag1$$
Is this integral $(1)$ a trivial problem? A easy problem to solve but I can't see it.
$$\int_{0}^{\pi\over 2}\arctan(1)\mathrm dx={\pi^2\over 8}\tag2$$
Why $(1)$ acts like integral $(2)$ for any real values of n?$
An attempt: by enforcing $u=\tan^n(x)$ then $du=n(\tan{x})^{n-1}\sec^2{x}dx$
Recall: $1+\tan^2{x}=\sec^2{x}$
$${1\over n}\int_{0}^{\infty}{\arctan{u}\over u}\cdot{\mathrm dx\over u^{1/n}+u^{1/n}}\tag3$$
Enforcing another substitution: $u=e^v$ then $du=e^vdv$
Recall: $e^x+e^{-x}=2\cosh{x}$
$${1\over 2n}\int_{-\infty}^{\infty}{\arctan(e^v)\over \cosh(v/n)}\mathrm dv\tag4$$
How can we prove $(5)?$
$${1\over n}\int_{-\infty}^{\infty}{\arctan(e^v)\over \cosh(v/n)}\mathrm dv={\left(\pi\over 2\right)^2}\tag5$$
$n\ne 0$