TITLE: How to show that $ \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \ln\left(\tan x+\tan\frac{\pi}{6}\right)\tan x\space dx=\frac{\zeta(2)}{6} $
QUESTION [5 upvotes]: I was trying to prove the well known result:
$$
\sum_{k=1}^\infty \frac{1}{\binom{2k}kk^2}=\frac{\zeta(2)}{3}
$$
and it came down to prove the following integral
$$
\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \ln\left(\tan x+\tan\frac{\pi}{6}\right)\tan x\space dx=\frac{\zeta(2)}{6}
$$
Can this integral be proven without using the above mentionend sum? I tried applying the angle sum and difference identities but I didn't get something helpful, so any help is highly appreciated!
Edit:
Can it be shown directly, in other words without knowing $\zeta(2)=\frac{\pi^2}{6}$, that
$$
\sum_{k=1}^\infty \frac{1}{\binom{2k}kk^2}=\frac{\zeta(2)}{3}
$$

REPLY [2 votes]: Substitute $t= \cot\frac{\pi}{6}\>\tan x$
\begin{align}
&\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \ln\left(\tan x+\tan\frac{\pi}{6}\right)\tan x\space dx\\
=& \int_{-1}^1 \frac{t\ln (1+t)}{3+t^2}dt
= \int_{-1}^0 \frac{t\ln (1+t)}{3+t^2}\overset{t\to -t}{dt }+\int_{0}^1 \frac{t\ln (1+t)}{3+t^2}dt\\
=& \int_{0}^1 \frac{t\ln \frac{1+t}{1-t}}{3+t^2}dt
\overset{t\to \frac{1-t}{1+t}}=
-\frac12\int_0^1 \frac{(1-t)\ln t}{(1+t)(1+t+t^2)}dt \\
=& -\frac12\int_0^1 \ln t\>d\left(\ln \frac{(1-t^2)^2}{(1-t)(1-t^3)}\right)\\
=&\>\frac12\left(2\int_0^1 \frac{\ln(1-t^2)}t\overset{t^2\to t}{dt}
-\int_0^1 \frac{\ln(1-t)}tdt -\int_0^1 \frac{\ln(1-t^3)}t \overset{t^3\to t}{dt}\right)\\
=&\>-\frac16\int_0^1 \frac{\ln(1-t)}tdt= -\frac16(-\zeta(2))
=\frac{\zeta(2)}{6} 
\end{align}
where $\int_0^1 \frac{\ln(1-t)}tdt=-\sum_{k\ge 1} \frac1 k \int_0^1 t^{k-1}dt=\sum_{k\ge 1} \frac1{k^2}=-\zeta(2)
$.