TITLE: Integral: $\int_0^{\pi/12} \ln(\tan x)\,dx$
QUESTION [28 upvotes]: 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!

REPLY [7 votes]: Let $I(a)=-\int_0^{\frac\pi{12}}\tanh^{-1}\frac{2\cos2x}{a+a^{-1}}dx$. Then
$$ I’(a) = \int_0^{\frac\pi{12}}\frac{2(a^2-1)\cos2x}{a^4+1-2a^2\cos4x}dx
=\frac{\tan^{-1}\frac a{a^2-1}}{2a}
$$
and
\begin{align}
\int_0^\frac\pi{12}\ln(\tan x)~dx
&= -\int_0^\frac\pi{12}\tanh^{-1}(\cos2x)dx 
=\int_0^1 I’(a)da \\
&=\int_0^1\frac{\tan^{-1}\frac a{a^2-1}}{2a}da
=-\int_0^1\left(\frac{\tan^{-1}a}{2a}\right.
+\underset{a^3\to a}{\left.\frac{\tan^{-1}a^3}{2a}\right)}da\\
&=-\left(\frac12+\frac16\right) \int_0^1\frac{\tan^{-1}a}{a}da=-\frac23G
\end{align}