Prove $ \int_{5\pi/36}^{7\pi/36} \ln (\cot t )dt +\int_{\pi/36}^{3\pi/36} \ln (\cot t )dt = \frac49G $ I have the conjecture for the integral
$$ \int_{\frac{5\pi}{36}}^{\frac{7\pi}{36}} \ln (\cot t )\>dt +\int_{\frac{\pi}{36}}^{\frac{3\pi}{36}} \ln (\cot t )\>dt = \frac49G $$
where $G$ is the Catalan constant, following some heuristic effort. But, I am unable to derive it formally despite having tried for some time. The derivation seems tougher than the tools in my toolbox.
I stick to real methods, though, trying to work it out mainly with elementary approaches. The standard procedures do not help that much. For instance, the substitution $u=\tan t $ is not viable given the inordinate limits; integration-by-parts coverts the integrand to $\frac t{\sin 2t}$, which is not revealing either. The proof for a much simpler integral
$$\int_0^\frac{\pi}{12}\ln(\cot x)=\frac23G
$$
is known, yet perhaps with limited relevance. Maybe, there is no avoidance of resorting to infinite series, complex methods, etc.
Edit:
Taking the cue from the comments below, the integral can be equivalently expressed as
$$\int_{\frac{\pi}{12}}^{\frac{5\pi}{36}} \ln\left( \cot (t+\frac\pi{36} )\cot (t-\frac\pi{36} )\right)=\frac49G$$
which may not reduce the difficulty, albeit appearing compact.

Using the third of these equations we can express your integrals in terms of the Clausen function:
\begin{align}
I &\equiv \int \limits_{\frac{\pi}{36}}^{\frac{3\pi}{36}} \log (\cot(t))\, \mathrm{d} t + \int \limits_{\frac{5\pi}{36}}^{\frac{7\pi}{36}} \log (\cot(t))\, \mathrm{d} t \\
&= \frac{1}{2} \left[- \operatorname{Cl}_2 \left(\vphantom{\operatorname{Cl}_2 \left(\frac{11\pi}{18}\right)}\frac{\pi}{18}\right) + \operatorname{Cl}_2 \left(\frac{3\pi}{18}\right) - \operatorname{Cl}_2 \left(\frac{5\pi}{18}\right) + \operatorname{Cl}_2 \left(\frac{7\pi}{18}\right) \right. \\
&\phantom{= \frac{1}{2} \left[ \vphantom{\operatorname{Cl}_2 \left(\frac{11\pi}{18}\right)} \right.} \left.+ \operatorname{Cl}_2 \left(\frac{11\pi}{18}\right) - \operatorname{Cl}_2 \left(\frac{13\pi}{18}\right) + \operatorname{Cl}_2 \left(\frac{15\pi}{18}\right) - \operatorname{Cl}_2 \left(\frac{17\pi}{18}\right)\right] \\
&= \frac{1}{2} \left[\operatorname{Cl}_2 \left(\frac{\pi}{2}\right) - \sum \limits_{k=0}^{8} (-1)^k \operatorname{Cl}_2 \left(\frac{(2k+1)\pi}{18}\right) \right] \, .
\end{align}
The Fourier series of $\operatorname{Cl}_2$ and its special value $\operatorname{Cl}_2 \left(\frac{\pi}{2}\right) = \mathrm{G}$ allow us to write this result as
$$ I = \frac{1}{2} \left[\mathrm{G} - \sum \limits_{n=1}^\infty \frac{\alpha(n)}{n^2}\right]$$
with $\alpha \colon \mathbb{N} \to \mathbb{R}$ given by
$$\alpha (n) = \sum \limits_{k=0}^8 (-1)^k \sin \left(\frac{(2k+1)n\pi}{18}\right) = \operatorname{Im} \left[\mathrm{e}^{\mathrm{i} \pi \frac{n}{18}} \sum \limits_{k=0}^{8} \mathrm{e}^{\mathrm{i} \pi \left(1+\frac{n}{9}\right) k} \right] \, .$$
Clearly, we have
$$\alpha (9 + 18 m) = \operatorname{Im} \left[\mathrm{i} (-1)^m \sum \limits_{k=0}^{8} 1 \right] = 9 (-1)^m $$
for $m \in \mathbb{N}_0$, while for all other $n \in \mathbb{N}$ we find
$$ \alpha (n) = \operatorname{Im} \left[\mathrm{e}^{\mathrm{i} \pi \frac{n}{18}} \frac{1 - \mathrm{e}^{\mathrm{i} \pi (9+n)}}{1 - \mathrm{e}^{\mathrm{i} \pi \left(1+\frac{n}{9}\right)}} \right] = \operatorname{Im} \left[2 \frac{1 + (-1)^n}{\left \lvert 1 + \mathrm{e}^{\mathrm{i} \pi n/9} \right \rvert^2} \cos \left(\frac{n \pi}{18}\right)\right] = 0 \, .$$
Therefore,
$$ I = \frac{1}{2} \left[\mathrm{G} - \sum \limits_{m=0}^\infty \frac{9 (-1)^m}{(9 + 18 m)^2}\right] = \frac{1}{2} \left[\mathrm{G} - \frac{1}{9}\sum \limits_{m=0}^\infty \frac{(-1)^m}{(2m+1)^2}\right] = \frac{4}{9} \mathrm{G} \, .$$

The same method leads to the more general result
\begin{align}
\sum \limits_{\mu = 0}^{\nu - 1} (-1)^{\nu - 1 - \mu} \int \limits_0^{\frac{2\mu +1}{2\nu+1} \frac{\pi}{4}} \log(\cot(t)) \, \mathrm{d} t &= \frac{1}{2} \left[\mathrm{G} + (-1)^{\nu - 1} \sum \limits_{k=0}^{2\nu} (-1)^k \operatorname{Cl}_2 \left(\frac{2k+1}{2\nu+1} \frac{\pi}{2}\right)\right] \\
&= \frac{\nu + 1_{2\mathbb{N} - 1} (\nu)}{2\nu+1} \mathrm{G}
\end{align}
for $\nu \in \mathbb{N}$ ($1_{2\mathbb{N}-1}$ is the indicator function of the odd positive integers). The first three cases are
\begin{align}
\int \limits_0^{\frac{\pi}{12}} \log(\cot(t)) \, \mathrm{d} t &= \frac{2}{3} \mathrm{G} \, , \\
\int \limits_{\frac{\pi}{20}}^{\frac{3\pi}{20}} \log(\cot(t)) \, \mathrm{d} t &= \frac{2}{5} \mathrm{G} \, , \\
\int \limits_0^{\frac{\pi}{28}} \log(\cot(t)) \, \mathrm{d} t + \int \limits_{\frac{3\pi}{28}}^{\frac{5\pi}{28}} \log(\cot(t)) \, \mathrm{d} t &= \frac{4}{7} \mathrm{G}
\end{align}
and $\nu = 4$ corresponds to the original question.
