How does one show that $\int_{0}^{\pi/4}\sin(2x)\ln{(\ln^2{\cot{x}})}\mathrm dx=\ln{\pi\over4}-\gamma?$ Vardi's type integral
Consider $(1)$

$$\int_{0}^{\pi/4}\sin(2x)\ln{(\ln^2{\cot{x}})}\mathrm dx=\ln{\pi\over4}-\gamma\tag1$$

An attempt : maybe?
$u=\ln{(\ln{\cot{x}})}$ then $du=-{\cos{x}\over \sin^3{x}}dx$
 A: The integral of interest is (through the substitutions $x=\arctan u$, then $u=e^{-t}$) 
$$ I = \int_{0}^{1}\frac{2u\log(\log^2 u)}{(1+u^2)^2}\,du=\int_{0}^{+\infty}\frac{\log(t)}{\cosh^2(t)}\,dt=\frac{d}{ds}\left.\int_{0}^{+\infty}\frac{t^s}{\cosh(t)^2}\,dt\,\right|_{s=0^+}\tag{1}$$
Now we may recall that, through the (inverse) Laplace transform, for any $s>0$ we have:

(Jack's Lemma about the integral representation of the $\zeta$ function in terms of the $\text{sech}$ function)
  $$ \begin{eqnarray*}\zeta(s) &=& \left(1-\frac{2}{2^s}\right)^{-1}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s}\\&=&\left(1-\frac{2}{2^s}\right)^{-1}\frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{t^{s-1}}{e^t+1}\,dt\\(IBP)\quad&=&\left(1-\frac{2}{2^s}\right)^{-1}\frac{2^{s+1}}{4\,\Gamma(s+1)}\int_{0}^{+\infty}\frac{t^s}{\cosh^2(t)}\,dt\end{eqnarray*}$$
  from which it follows that
  $$ g(s)=\int_{0}^{+\infty}\frac{t^s}{\cosh^2(t)}\,dt = \frac{(2^s-2)\,\Gamma(s+1)\,\zeta(s)}{2^{2s-1}}\tag{2} $$

If now we represent $g'(s)$ as $g(s)\cdot \frac{d}{ds}\log g(s)$ we get:
$$ g'(s) = \int_{0}^{+\infty}\frac{t^s\log t}{\cosh^2(t)}\,dt = \left(\frac{2^s\log(2)}{2^s-2}+\psi(s+1)-2\log(2)+\frac{\zeta'(s)}{\zeta(s)}\right)\frac{(2^s-2)\,\Gamma(s+1)\,\zeta(s)}{2^{2s-1}}\tag{3}$$
and the claim boils down to the standard evaluations
$$ \psi(1)=-\gamma,\qquad \zeta(0)=-\frac{1}{2},\qquad\frac{\zeta'(0)}{\zeta(0)}=\log(2\pi).$$
