Consider the following integral
$$c=\int_0^{\pi/2}\log(1-x\cot x)\, \mathrm{d}x\approx-3.35333726288947201778500718670823032.$$
I suspect it can be analytically computed because by expanding the $\log$ function,
$$
c=\sum_{k=1}^{\infty}\frac1k\int_0^{\pi/2}(x \cot x)^k\, \mathrm{d}x
$$
it can be integrated term by term, albeit in involved form (combination of logarithms and $\zeta$-functions)
$$
\int_0^{\pi/2}x \cot x\, \mathrm{d}x=\frac{1}{2} \pi \log2,\\
\int_0^{\pi/2}x^2 \cot^2 x\, \mathrm{d}x=-\frac{\pi ^3}{24}+\pi \log2,\\
\int_0^{\pi/2}x^3 \cot^3 x\,\mathrm{d}x=-\frac{\pi^3}{16} (1+2\log 2)+\frac{3 \pi}{16} (8\log 2+3 \zeta(3)),
$$
and so on.
Just to provide some background to this question, the integral has some significance in theoretical physics. It enters the high-density asymptotics of the quasiparticle renormalization factor of the 3D homogeneous electron gas, see eq. 35 in Phys. Rev. B 70, 035111 (2004) or eqs. 8 and 9 in Phys. Rev. 120, 2041 (1960):
$$ Z_{qp}=1+\frac{c}{\pi^2}\alpha r_s$$
However, since there is no parametric dependence, and since it is easy to compute numerically, no one cared to find the analytic form. However, I find it is a lovely little problem.
Best Answer
$\color{brown}{\textbf{Alternative expressions for the integral.}}$
Firstly, $$I = \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\ln(1-x\cot x)\,\mathrm dx = \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\ln(\sin x - x\cos x)\,\mathrm dx - \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\ln(\sin x)\,\mathrm dx = \dfrac\pi2\ln2 +I_1,$$
wherein $I_1$ allows integration by parts: $$I_1 = \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\ln(\sin x - x\cos x)\,\mathrm dx = x\ln(\sin x-x\cos x)\bigg|_{\ 0}^{\Large^\pi\hspace{-1pt}/_2} - \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^2\sin x}{\sin x - x\cos x}\,\mathrm dx,$$ $$ I_1 =-\int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^2\sin x}{\sin x- x\cos x}\,\mathrm dx = - \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^2}{1-x\cot x}\,\mathrm dx = - J_{21},\tag1$$ where
On the other hand, $$J_{21} = \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^2(1-x\cot x + x\cot x)}{1-x\cot x}\,\mathrm dx = \dfrac{\pi^3}{24} + I_2,$$ where $$I_2 = \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^3\cot x}{1 - x\cot x}\,\mathrm dx = \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^3}{\tan x - x}\,\mathrm dx.\tag3$$
Formulas $(3)$ are not suitable for numeric calculations.
But integration by parts is possible, $$I_2 = \dfrac14\int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{1}{\tan x - x}\,\mathrm dx^4 = \dfrac14\dfrac{x^4}{\tan x-x}\bigg|_{\,0}^{\Large^\pi\hspace{-1pt}/_2} + \dfrac14\int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^4(1+\tan^2x -1)}{(\tan x - x)^2}\,\mathrm dx,$$ $$ I_2 = \dfrac14\int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^4}{(1 - x\cot x)^2}\,\mathrm dx = \dfrac14 J_{42},$$
Formula $(4)$ provides both suitable numeric calculations via Wolfram Alpha by the expression
with the result
and the further building of the series in the elementary functions via the transformations in the form of $$ J_{42} = \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^4((1 - x\cot x)^2 + 2x\cot x(1 - x\cot x) + x^2\cot^2 x) }{(1 - x\cot x)^2}\,\mathrm dx\\ = \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\left(x^4 + 2\,\dfrac{x^5\cot x}{1-x\cot x} + \dfrac{x^6\cot^2x}{(1 - x\cot x)^2}\right)\,\mathrm dx\\ = \int\limits_0^{\Large^\pi\hspace{-1pt}/_2} x^4\,\mathrm dx + \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\left(\dfrac{2x^5\cot x}{1-x\cot x} + \dfrac{x^6\cot^2x}{(1 - x\cot x)^2}\right)\,\mathrm dx,$$ $$J_{42} = \dfrac{\pi^5}{160} + I_3 + I_4,\tag5$$ where $$I_3 = 2\int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^5\cot x}{1-x\cot x} \,\mathrm dx = 2\int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^5}{\tan x - x}\,\mathrm dx = \dfrac13\int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{1}{\tan x - x}\,\mathrm dx^6\\ = \dfrac13\dfrac{x^6}{\tan x-x}\bigg|_{\,0}^{\Large^\pi\hspace{-1pt}/_2} + \dfrac13\int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^6(1+\tan^2x -1)}{(\tan x - x)^2}\,\mathrm dx = \dfrac13 J_{62},$$ $$I_4 = \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^6\cot^2x}{(1 - x\cot x)^2}\,\mathrm dx = \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^6}{(\tan x - x)^2} \,\mathrm dx = \dfrac17\int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{1}{(\tan x - x)^2} \,\mathrm dx^7\\ = \dfrac27\dfrac{x^7}{(\tan x-x)^3}\bigg|_{\,0}^{\Large^\pi\hspace{-1pt}/_2} + \dfrac27\int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^7(1+\tan^2x -1)}{(\tan x - x)^3}\,\mathrm dx = \dfrac27\int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^7\cot x}{(1 - x\cot x)^3}\,\mathrm dx\\ = \dfrac27\int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^6(1 - (1 - x\cot x))}{(1 - x\cot x)^3}\,\mathrm dx =\dfrac27(J_{63}-J_{62}),$$
Therefore,
Numeric calculations via Mathcad Alpha by the formula $(6)$
leads to the same result, and this confirms the correctness of the approach.
$\color{brown}{\textbf{Recurrence relations.}}$
For the arbitrary $m,n$ $$ J_{mn} = \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\,(x\cot x + (1-x\cot x))^n \dfrac{x^m}{(1 - x\cot x)^n}\,\mathrm dx = \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\sum\limits_{k=0}^n\binom nk\dfrac{x^{m+k}\cot^k x}{(1 - x\cot x)^k}\,\mathrm dx = \dfrac{\pi^{m+1}}{(m+1)2^{m+1}} + \sum\limits_{k=1}^n\dfrac{\dbinom nk}{m+k+1} \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{\mathrm dx^{m+k+1}}{(\tan x - x)^k} = \dfrac{\pi^{m+1}}{(m+1)2^{m+1}}\\ + \sum\limits_{k=1}^n\dfrac{\dbinom nk}{m+k+1} \left(\dfrac{x^{m+k+1}}{(\tan x - x)^{k}}\bigg|_{\,0}^{\Large^\pi\hspace{-1pt}/_2} + k\int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^{m+k+1}(1+\tan^2x-1)}{(\tan x-x)^{k+1}}\,\mathrm dx\right)\\ = \dfrac{\pi^{m+1}}{(m+1)2^{m+1}} + \sum\limits_{k=1}^n\dfrac{k}{m+k+1} \dbinom nk \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^{m+2}(x\cot x)^{k-1}}{(1 -x\cot x)^{k+1}}\,\mathrm dx\\ = \dfrac{\pi^{m+1}}{(m+1)2^{m+1}} + \sum\limits_{k=1}^n\dfrac{k}{m+k+1} \dbinom nk \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^{m+2}(1-(1-x\cot x))^{k-1}}{(1 -x\cot x)^{k+1}}\,\mathrm dx\\ = \dfrac{\pi^{m+1}}{(m+1)2^{m+1}} + \sum\limits_{k=1}^n\dfrac{k}{m+k+1} \dbinom nk \sum\limits_{j=0}^{k-1}(-1)^{k-1-j}\dbinom{k-1}j J_{m+2,\,j+2},$$
where
If $(m,n)=(2,1),\ $ then $$F_{0} = \sum\limits_{k=1}^1 (-1)^{k-1} \dfrac{k}{2+k+1}\dbinom1k \dbinom{k-1}0 =\dfrac14,$$ $$J_{21} = \dfrac{\pi^{3}}{3\cdot2^3} + \sum\limits_{j=0}^0 F_{j} J_{4,\,j+2} = \dfrac{\pi^{3}}{24} + J_{42}.$$
If $(m,n)=(4,2),\ $ then $$F_{0} = \sum\limits_{k=1}^2 (-1)^{k-1} \dfrac{k}{4+k+1}\dbinom2k \dbinom{k-1}0 =\dfrac13 - \dfrac27 = \dfrac{1}{21},$$ $$F_{1} = \sum\limits_{k=2}^2 (-1)^{k} \dfrac{k}{4+k+1}\dbinom2k \dbinom{k-1}1 =\dfrac27,$$ $$J_{42} = \dfrac{\pi^{5}}{5\cdot2^5} + \sum\limits_{j=0}^1 F_{j} J_{2,\,j+2} = \dfrac{\pi^5}{160} + \dfrac1{21}J_{62} + \dfrac27J_{63}.$$
Similarly, $$J_{62} = \dfrac{\pi^7}{896}+\dfrac1{36}J_{82}+\dfrac29J_{83}\tag9$$ (see also Wolfram Alpha test).
Besides, $$J_{63} = \dfrac{\pi^7}{896}+\dfrac1{120}J_{82} + \dfrac1{15}J_{83} + \dfrac3{20}J_{84}.\tag{10}$$
$\color{brown}{\textbf{Simple series.}}$
Obtained results are not the best way to get required series of the arbitrary length.
$$\boxed{ \begin{matrix} I & = & -3.35333726288947201778500718670823032009876022464933939598 \\ \frac\pi2\ln2 & = &1.088793045151801065250344449118806973669291850184643147162 \\ J_{21} & = & 4.442130308041273083035351635930890531086461245854584994170 \\ \frac{\pi^3}{24} & = & 1.291928195012492507311513127795891466759387023578546153922 \\ J_{42} & = & 12.60080845211512230289535403253999625730829688910415536099 \\ \frac{\pi^5}{160} & = & 1.912623029908009082892133187771472540501879416425468690959 \\ J_{62} & = & 9.357325953756236734147158157553707227832359838953032605558 \\ J_{63} & = & 35.84909465209885681432007993043088180418373451454989791084 \\ \frac{\pi^7}{896} & = & 3.370862977429455432493534032446475258836420173320761453966 \\ J_{82} & = & 13.21743446830609099759197972403428192140938899336281280188 \\ J_{83} & = & 25.28690408493225448274231109747825862030555487117486858192 \\ J_{84} & = & 102.2743092725712233044348622015074565154951081384648503713 \\ \end{matrix}}$$
On the other hand, using of the simple Laurent series for the function $$g(y) = \dfrac{35}{1-y\sqrt{15}\cot y\sqrt{15}} = \dfrac7{y^2}-\sum\limits_{i=0}^\infty c_iy^{2i}\tag{11}$$
gives evidently convergent series $$J_{21} = \dfrac1{35}\int\limits_0^{\Large^\pi\hspace{-1pt}/_2} \left(7 - \sum\limits_{i=0}^\infty c_i\left(\dfrac{x^2}{15}\right)^{i+1}\right)\,\mathrm dx,$$
wherein the first $8$ terms provide the accuracy of $8$ decimal digits.