Show that $ \int_0^\pi \cot(x)\arctan(\tanh\pi(\tan(x))) \mathrm{d}x=\pi \log(3)$ 
Show that$$\int_0^\pi  \cot(x)\arctan(\tanh\pi(\tan(x))) \mathrm{d}x=\pi \log(3)$$

A friend sent me this problem. I managed to solve it but can we do it in other ways ?
 A: I found general generalization for this problem
$$\mathrm{Let} \; \mathrm{I} \; = \; \int_0^\pi \cot(x)\arctan(\tanh\theta (\tan(x))) \mathrm{d}x$$ 
$$= \quad 2 \int_0^\infty \frac{\arctan(\tanh(\theta x))}{x(x^2+1)} \mathrm{d}x $$
$$ \mathrm{we\;know\;that}\;\int_0^\infty \frac{\cos(\theta x )}{\cosh(x)} \mathrm{d}x = \frac{\pi}{2} \frac{1}{\cosh(\frac{\pi \theta}{2})} $$
So
$$ \; \int_0^\infty \frac{\sin(\theta x )}{x \cosh(x)} \mathrm{d}x = 2 \arctan(\tanh(\theta \frac{\pi}{4})) $$
$$I = \quad 4 \int_0^\infty \int_0^\infty \frac{\sin(a yx)}{y \cosh(y)x(x^2+1)} \mathrm{d}x\mathrm{d}y \;=\; 2 \pi \int_0^\infty \frac{(1-e^{-ay})}{y \cosh(y)} \mathrm{d}y $$
where $a = \frac{4\theta}{\pi}$
$$ = 2\pi \int_0^\infty \int_0^\infty \frac{e^{-xy}(1-e^{-ay})}{ \cosh(y)} \mathrm{d}y\mathrm{d}x $$
$$ = \pi \int_0^\infty \psi\bigg(\frac{3}{4}+\frac{x}{4}\bigg)-\psi\bigg(\frac{1}{4}+\frac{x}{4}\bigg)-\psi \bigg(\frac{3+a}{4}+\frac{x}{4}\bigg)+\psi\bigg(\frac{1+a}{4}+\frac{x}{4}\bigg)\mathrm{d}x $$
$$= \quad \pi \log\Bigg(\frac{\Gamma(\frac{\theta }{\pi}+\frac{3}{4}) \Gamma(\frac{1}{4})}{\Gamma(\frac{\theta }{\pi}+\frac{1}{4}) \Gamma(\frac{3}{4})} \Bigg) $$
So when $\theta = \pi $
$$ \int_0^\pi \cot(x)\arctan(\tanh\pi (\tan(x))) \mathrm{d}x \; = \; \pi \log(3)$$
