Integral: $\int_0^{\pi/12} \ln(\tan x)\,dx$ I am trying to evaluate:
$$\int_0^{\pi/12} \ln(\tan x)\,dx$$
I think the integral is quite simple but I am having a hard time evaluating it. I started with the result:
$$\int_0^{\pi/4} \ln(\tan x)\,dx= -G$$
where $G$ is the Catalan's constant. With the change of variables $x\rightarrow 3x$ and using the fact that $\tan(3x)=\tan x\tan\left(\frac{\pi}{3}+x\right)\tan\left(\frac{\pi}{3}-x\right)$, the integral is:
$$\int_0^{\pi/12}\ln(\tan x)\,dx+\int_0^{\pi/12}\ln \tan\left(\frac{\pi}{3}+x\right)\,dx+\int_0^{\pi/12}\ln \tan\left(\frac{\pi}{3}-x\right)\,dx=-\frac{G}{3}$$
$$\Rightarrow \int_0^{\pi/12}\ln(\tan x)\,dx+\int_{-\pi/12}^{\pi/12}\ln \tan\left(\frac{\pi}{3}+x\right)\,dx=-\frac{G}{3}$$
But I do not see how to proceed. 
Help is appreciated. Thanks!

An alternative "elementary" method.
Consider,
\begin{align*}
K&=\int_0^1 \frac{\arctan\left(\frac{x}{1-x^2}\right)}{x}\,dx\\
\end{align*}
Perform the change of variable $x=\tan\left(\frac{t}{2}\right) $,
\begin{align*}
K&=\int_0^{\frac{\pi}{2}} \frac{\arctan\left(\frac{1}{2}\tan t\right)}{\sin t}\,dt
\end{align*}
Défine the function $H$ on $\left[\frac{1}{2};1\right]$ to be,
\begin{align*}H(a)&=\int_0^{\frac{\pi}{2}} \frac{\arctan\left(a\tan t\right)}{\sin t}\,dt\end{align*}
Observe that $K=H\left(\dfrac{1}{2}\right)$ and,
\begin{align*}H(1)&=\int_0^{\frac{\pi}{2}} \frac{t}{\sin t}\,dt\\
&=\Big[t\ln\left(\tan\left(\frac{t}{2} \right)\right)\Big]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\ln\left(\tan\left(\frac{t}{2} \right)\right)\,dt\\
&=-\int_0^{\frac{\pi}{2}}\ln\left(\tan\left(\frac{t}{2} \right)\right)\,dt\\
\end{align*}
Perform the change of variable $x=\dfrac{t}{2}$,
\begin{align*}H(1)&=-2\int_0^{\frac{\pi}{4}}\ln\left(\tan\left(t \right)\right)\,dt\\
&=2\text{G}
\end{align*}
\begin{align*}H^\prime (a)&=\int_0^{\frac{\pi}{2}} \frac{\cos x}{1-(1-a^2)\sin^2 x}\,dt\\
&=\left[\frac{1}{2\sqrt{1-a^2}}\ln\left(\frac{1+\sin(x)\sqrt{1-a^2}}{1-\sin(x)\sqrt{1-a^2}}\right)\right]_0^{\frac{\pi}{2}}\\
&=\frac{1}{2\sqrt{1-a^2}}\ln\left(\frac{1+\sqrt{1-a^2}}{1-\sqrt{1-a^2}}\right)
\end{align*}
Therefore,
\begin{align*}H(1)-H\left(\frac{1}{2}\right)&=\int_{\frac{1}{2}}^1 \frac{1}{2\sqrt{1-a^2}}\ln\left(\frac{1+\sqrt{1-a^2}}{1-\sqrt{1-a^2}}\right)\,da\end{align*}
Perform the change of variable $y=\arctan\left(\sqrt{\dfrac{1+\sqrt{1-a^2}}{1-\sqrt{1-a^2}}}\right)$
\begin{align*}H(1)-H\left(\frac{1}{2}\right)&=-2\int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \ln\left(\tan y\right)\,dy\\
&=-2\int_0^{\frac{\pi}{4}} \ln\left(\tan y\right)\,dy+2\int_0^{\frac{\pi}{12}} \ln\left(\tan y\right)\,dy
\end{align*}
But, it is well known that,
\begin{align*}\int_0^{\frac{\pi}{4}} \ln\left(\tan y\right)\,dy=-\text{G}\\\end{align*}
Thus,
\begin{align*}\int_0^{\frac{\pi}{12}} \ln\left(\tan y\right)\,dy=-\frac{1}{2}K\\\end{align*}
On the other hand,
\begin{align}\int_0^1 \frac{\arctan\left( \frac{x}{1-x^2}\right)}{x}\,dx-\int_0^1 \frac{\arctan x}{x}\,dx=\int_0^1 \frac{\arctan \left(x^3\right)}{x}\,dx\end{align}
In the latter integral perform the change of variable $\displaystyle y=x^3$,
\begin{align}\int_0^1 \frac{\arctan\left( \frac{x}{1-x^2}\right)}{x}\,dx-\int_0^1 \frac{\arctan x}{x}\,dx=\frac{1}{3}\int_0^1 \frac{\arctan x}{x}\,dx\end{align}
Therefore,
\begin{align}\int_0^1 \frac{\arctan\left( \frac{x}{1-x^2}\right)}{x}\,dx&=\frac{1}{3}\int_0^1 \frac{\arctan x}{x}\,dx+\int_0^1 \frac{\arctan x}{x}\,dx\\
&=\frac{4}{3}\int_0^1 \frac{\arctan x}{x}\,dx\\
&=\frac{4}{3}\text{G}
\end{align}
Thus,
\begin{align*}\int_0^{\frac{\pi}{12}} \ln\left(\tan y\right)\,dy&=-\frac{1}{2}\times \frac{4}{3}\text{G} \\
&=\boxed{-\frac{2}{3}\text{G}}
\end{align*}
