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.
 A: 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.
A: Rewriting
Use the trigonometric identity $\csc^2(x) = \cot^2+1$. Rewrite the integral as
$$\int\limits_{\alpha}^{\beta}\csc^2(x)\left(-\frac{\log\cot x}{\cot^2+1}\right)\,\mathrm{d}x$$
After substituting $u=\cot x$ we get the integral
$$-\int\limits_{\cot(\alpha)}^{\cot(\beta)}\frac{\log u}{u^2+1}\,\mathrm{d}u=-\left[\frac{i}{2}\int\limits_{\cot(\alpha)}^{\cot(\beta)}\frac{\log u}{u+i}\,\mathrm{d}u-\frac{i}{2}\int\limits_{\cot(\alpha)}^{\cot(\beta)}\frac{\log u}{u-i}\,\mathrm{d}u\right].$$
First integral
Sub $v=u+i$. You should get
$$\int\limits_{\cot(\alpha)+i}^{\cot(\beta)+i}\frac{\log v-i}{v}\,\mathrm{d}v = \int\limits_{\cot(\alpha)+i}^{\cot(\beta)+i}\frac{\log iv+1}{v}\,\mathrm{d}v+\log(-i)\int\limits_{\cot(\alpha)+i}^{\cot(\beta)+i}\frac{1}{v}\,\mathrm{d}v.$$
Integral 1
Sub $w=-iv$. This will get us to
$$-\int\limits_{1-i\cot(\alpha)}^{1-i\cot(\beta)}-\frac{\log 1-w}{w}\,\mathrm{d}w = -\Big[\operatorname{Li}_2(1-i\cot(\beta))-\operatorname{Li}_2(1-i\cot(\alpha))\Big]$$
Integral 2
Note that $\log(-i)=-i\pi/2$. We get:
$$-i\pi/2\Bigg[\log(\cot(\beta)+i)-\log(\cot(\alpha)+i)\Bigg].$$
Second Integral
Sub $v=u-i$. You should get
$$\int\limits_{\cot(\alpha)-i}^{\cot(\beta)-i}\frac{\log v+i}{v}\,\mathrm{d}v = \int\limits_{\cot(\alpha)-i}^{\cot(\beta)-i}\frac{\log 1-iv}{v}\,\mathrm{d}v+\log(i)\int\limits_{\cot(\alpha)-i}^{\cot(\beta)-i}\frac{1}{v}\,\mathrm{d}v.$$
Integral 1
Sub $w=iv$. This will get us to
$$-\int\limits_{i\cot(\alpha)-1}^{i\cot(\beta)-1}-\frac{\log 1-w}{w}\,\mathrm{d}w = -\Big[\operatorname{Li}_2(i\cot(\beta)-1)-\operatorname{Li}_2(i\cot(\alpha)-1)\Big]$$
Integral 2
Note that $\log(i)=i\pi/2$. We get:
$$i\pi/2\Bigg[\log(\cot(\beta)-i)-\log(\cot(\alpha)-i)\Bigg].$$
Values for $\alpha$ and $\beta$
For the first orignal integral we have
$$-\frac{{\pi}\left(\log\left(\tan^2\left(\frac{7{\pi}}{36}\right)+1\right)-\log\left(\tan^2\left(\frac{5{\pi}}{36}\right)+1\right)\right)+{i}\left(2\operatorname{Li}_2\left({i}\tan\left(\frac{7{\pi}}{36}\right)+1\right)-2\operatorname{Li}_2\left(1-{i}\tan\left(\frac{7{\pi}}{36}\right)\right)-2\operatorname{Li}_2\left({i}\tan\left(\frac{5{\pi}}{36}\right)+1\right)+2\operatorname{Li}_2\left(1-{i}\tan\left(\frac{5{\pi}}{36}\right)\right)\right)}{4}$$
For the second orignal integral we have
$$-\dfrac{{\pi}\left(\log\left(\tan^2\left(\frac{{\pi}}{12}\right)+1\right)-\log\left(\tan^2\left(\frac{{\pi}}{36}\right)+1\right)\right)+{i}\left(2\operatorname{Li}_2\left({i}\tan\left(\frac{{\pi}}{12}\right)+1\right)-2\operatorname{Li}_2\left(1-{i}\tan\left(\frac{{\pi}}{12}\right)\right)-2\operatorname{Li}_2\left({i}\tan\left(\frac{{\pi}}{36}\right)+1\right)+2\operatorname{Li}_2\left(1-{i}\tan\left(\frac{{\pi}}{36}\right)\right)\right)}{4}$$
A few last words
I added these values numerically and I got the right result. The integral is solved but it still needs some long (algebraic) operations to prove the end result. But the Polylogarithm $\operatorname{Li}_2$ looks promising. Maybe I will add the final derivation later on, but first I have to take a break of this monster challenge.
Also, if someone finds a faster way and all of this is useless I would be twice as happy. Nonetheless I hope it has helped you at least a bit.
A: Too long for comments.
Let
$$f(a)=\int_{5a}^{7a}\log (\cot (t))\,dt+\int_{a}^{3a}\log (\cot (t))\,dt$$ leads to
$$f(a)=a \Big[-\log (\cot (a))+3 \log (\cot (3 a))-5 \log (\cot (5 a))+7 \log (\cot (7 a))\Big]+$$
$$\frac{\pi}{2}  \Big[ \log (\cot (a))-\log (\cot (3 a))+\log (\cot (5 a))-\log (\cot
   (7 a))\Big]+$$
$$\frac{i}{2}  (\text{Li}_2(i \cot (a))-\text{Li}_2(-i \cot (a)))-$$
$$\frac{i}{2}  (\text{Li}_2(i \cot (3a))-\text{Li}_2(-i \cot (3a)))+$$
$$\frac{i}{2}  (\text{Li}_2(i \cot (5a))-\text{Li}_2(-i \cot (5a)))-$$
$$\frac{i}{2}  (\text{Li}_2(i \cot (7a))-\text{Li}_2(-i \cot (7a)))$$
At this point, I am stuck.
Expanding $f(a)$ as a (long) Taylor series around $a=0$ and making $a=\frac \pi {36}$ leads to
$$f\left(\frac{\pi}{36}\right)=
0.407095819634319562246490451081059604788510834$$ which is recognized as $\frac{4 }{9}C$ by the inverse symbolic calculator (notice that all digits are exact).
One point on the side : $\frac{4 }{9}C$ is the maximum value of $f(a)$.
A: Here is an attempt at an elementary solution. The identity is readily seen to be the same as $$\int_{\pi/4}^{3\pi/4}\log\cot\frac v9\,dv+\int_{5\pi/4}^{7\pi/4}\log\cot\frac v9\,dv=4G.$$ We can combine both integrals into one through the linear substitution $2u=v-3\pi/4\pm\pi/2$ to get $$\int_0^{\pi/4}\log\cot\left(\frac{2u}9+\frac\pi{36}\right)+\log\cot\left(\frac{2u}9+\frac{5\pi}{36}\right)\,du=2G.$$ Taking advantage of symmetry over $t=2u/9+\pi/12$, the integral can be written as \begin{align}\frac49G&=\int_{\pi/12}^{5\pi/36}\log\left(\cot\left(t-\frac\pi{18}\right)\cot\left(t+\frac\pi{18}\right)\right)\,dt\\&=\int_{\pi/12}^{5\pi/36}\log\frac{-\cos2t-\cos\pi/9}{\cos2t-\cos\pi/9}\,dt\end{align} or through a Taylor series expansion,$$\frac89G=\int_{\pi/6}^{5\pi/18}\log\frac{\cos\pi/9+\cos x}{\cos\pi/9-\cos x}\,dx=\int_{\pi/6}^{5\pi/18}\sum_{k=1}^\infty\frac2{2k-1}\left(\frac{\cos x}{\cos\pi/9}\right)^{2k-1}\,dx.$$ We can exchange operators to get \begin{align}\frac49G&=\sum_{k=1}^\infty\frac{(\sec\pi/9)^{2k-1}}{2k-1}\int_{\pi/6}^{5\pi/18}(\cos x)^{2k-1}\,dx\end{align} and since \begin{align}\int_{\pi/6}^{5\pi/18}(\cos x)^{2k-1}\,dx&=\int_{\pi/6}^{5\pi/18}\frac1{2^{2k-2}}\sum_{m=0}^{k-1}\binom{2k-1}m\cos(2k-2m-1)x\,dx\\&=\frac4{4^k}\sum_{m=0}^{k-1}\binom{2k-1}m\frac{\sin\frac{5(2k-2m-1)\pi}{18}-\sin\frac{(2k-2m-1)\pi}6}{2k-2m-1}\end{align} the identity is equivalent to $$\frac1{18}G=\sum_{k=1}^\infty\sum_{m=0}^{k-1}\binom{2k-1}m\frac{\sin\frac{(2k-2m-1)\pi}9\cos\frac{4(2k-2m-1)\pi}9}{4^k(2k-1)(2k-2m-1)(\cos\pi/9)^{2k-1}}.$$ The trig expressions should reduce to something simpler (and not involving trig functions) — need help!

Remark: if $\int_{\pi/6}^{5\pi/18}(\cos x)^{2k-1}\,dx=a+b\sin\pi/18+c\cos\pi/9+d\cos2\pi/9$ for some rational $a,b,c,d$ then $$\frac49G=\sum_{k=1}^\infty\frac{a+(b-d)\sin\pi/18+(c+d)\cos\pi/9}{(2k-1)(\cos\pi/9)^{2k-1}}$$
A: Here's an alternative method that may address OP's request of avoiding some calculations techniques.
$\mathbf{Preliminaries}:$
$$\begin{align} 1)\ &\cot\left(\frac{\pi}{3}-x\right)\cot\left(x\right)\cot\left(\frac{\pi}{3}+x\right)=\cot\left(3x\right)\\
2)\ & \int_{0}^{{\pi}/{4}}\log\left(\cot(x)\right)dx=G\\
3)\ & \int_{0}^{{\pi}/{12}}\log\left(\cot(x)\right)dx=\frac{2}{3}G
\end{align}$$
With that in mind, let's start the evaluation of the integral:
$$
\int_{5\pi/36}^{7\pi/36}\log\left(\cot(x)\right)dx+\int_{\pi/36}^{3\pi/36}\log\left(\cot(x)\right)dx \\
=\int_{5\pi/36}^{7\pi/36}\log\left(\cot(x)\right)dx+\underbrace{\int_{\pi/36}^{3\pi/36}\log\left(\cot(3x)\right)dx}_{3x\rightarrow x}-\underbrace{\int_{\pi/36}^{3\pi/36}\log\left(\cot\left(\frac{\pi}{3}-x\right)\right)dx}_{\frac{\pi}{3}-x\rightarrow x}-\underbrace{\int_{\pi/36}^{3\pi/36}\log\left(\cot\left(\frac{\pi}{3}+x\right)\right)dx}_{\frac{\pi}{3}+x\rightarrow x}\\
=\frac{1}{3}\int_{\pi/12}^{\pi/4}\log\left(\cot\left(x\right)\right)dx+\int_{5\pi/36}^{7\pi/36}\log\left(\cot(x)\right)dx-\underbrace{\int_{9\pi/36}^{11\pi/36}\log\left(\cot(x)\right)dx}_{x\rightarrow \frac{\pi}{2}-x}-\underbrace{\int_{13\pi/36}^{15\pi/36}\log\left(\cot(x)\right)dx}_{x\rightarrow \frac{\pi}{2}-x}\\
=\frac{1}{3}\int_{\pi/12}^{\pi/4}\log\left(\cot\left(x\right)\right)dx+\int_{5\pi/36}^{7\pi/36}\log\left(\cot(x)\right)dx-\int_{7\pi/36}^{9\pi/36}\log\left(\tan(x)\right)dx-\int_{3\pi/36}^{5\pi/36}\log\left(\tan(x)\right)dx
$$
Notice that:
$$\int_{5\pi/36}^{7\pi/36}\log\left(\cot(x)\right)dx-\int_{7\pi/36}^{9\pi/36}\log\left(\tan(x)\right)dx-\int_{3\pi/36}^{5\pi/36}\log\left(\tan(x)\right)dx\\
=\int_{3\pi/36}^{5\pi/36}\log\left(\cot(x)\right)dx+\int_{5\pi/36}^{7\pi/36}\log\left(\cot(x)\right)dx+\int_{7\pi/36}^{9\pi/36}\log\left(\cot(x)\right)dx\\
=\int_{\pi/12}^{\pi/4}\log\left(\cot\left(x\right)\right)dx$$
Therefore, the original integral is equal to:
$$\frac{4}{3}\int_{\pi/12}^{\pi/4}\log\left(\cot\left(x\right)\right)dx\\
=\frac{4}{3}\left(\int_{0}^{\pi/4}\log\left(\cot\left(x\right)\right)dx-\int_{0}^{\pi/12}\log\left(\cot\left(x\right)\right)dx\right)\\
=\frac{4}{3}\left(G-\frac{2}{3}G\right)=\boxed{\frac{4}{9}G}$$
