Suppose we have defined the "cochord" of an angle $\theta \in (-\pi,\pi)$ as $$\operatorname{coc}(\theta) := 2\cos\left(\frac \theta 2\right),$$ and set $$c_n := \frac 1 \pi \int_{-\pi}^{\pi} \ln^n (\operatorname{coc} x)\ dx. $$ I have employed the Fourier expansion $$\tag 1 \ln(\operatorname{coc}x) = -\sum_{k=1}^\infty \frac {(-1)^k} k \cos(kx) $$ to obtain, via Parseval's identity w.r.t. to the usual inner product on $L^2(-\pi,\pi)$, the interesting result $$ \tag2 \boxed{c_2 = \zeta(2)} = \frac{\pi^2}6. $$ Even more miraculously, it seems that $$\boxed{-\frac 1 3 c_3 = \zeta(3)} = 1.2020569... \tag3$$ Now I cannot seem to find any similarly satisfying relationship between $c_n$ and $\zeta(n)$, for integer $n \geq 4$. (It does also look like $c_1 = 0$, but I haven't bothered to try to prove it.) Does anyone have any idea as to what the general relation might be, and why it should be the case that these logarithmic integrals are related to $\zeta$ at all?
[Wolfram Mathworld has an article on a similar set of integrals, where $\operatorname{coc}(x)$ has been substituted by the usual cosine, but so far I have not been successful at transforming these into something closer to $c_n$.]
Edit. Here is my follow-up question concerning the closed form for $c_n$.
It can be shown that $$\int^\pi_{-\pi}(\text{coc }x)^a dx=\frac{2\pi\Gamma(a+1)}{\Gamma^2\left(\frac a2+1\right)}$$
Differentiation under the integral sign and the identity $$\psi_n(1)=(-1)^{n+1}n!\zeta(n+1)$$ together lead to a clear relationship between $c_n$ and $\zeta$ function.
I will elaborate very soon.
Real analysis derivation:
Recall that $$\mathcal B(x,y)=2\int^{\pi/2}_{0}\sin^{2x-1}t \cos^{2y-1}t dt$$
Therefore, $$\begin{align} \int^\pi_{-\pi}(\text{coc }x)^a dx &=2^a\cdot 2\int^\pi_{0}\cos^a\left(\frac x2\right) dx \\ &=2^{a+1}\cdot 2\int^{\pi/2}_0\cos^a (u) du \\ &=2^{a+1}\mathcal B\left(\frac12,\frac{a+1}2\right) \\ &=2^{a+1}\frac{\sqrt \pi }{\Gamma\left(\frac a2 +1\right)}\cdot \Gamma\left(\frac{a+1}2\right) \\ &=2^{a+1}\frac{\sqrt \pi }{\Gamma\left(\frac a2 +1\right)}\cdot \frac{2^{1-(a+1)}\sqrt\pi\cdot\Gamma(a+1)}{\Gamma\left(\frac{(a+1)+1}{2}\right)} \qquad{(\star)}\\ &=\frac{2\pi\Gamma(a+1)}{\Gamma^2\left(\frac a2+1\right)} \end{align} $$
$(\star):$ Legendre duplication formula $$\Gamma\left(\frac v2\right)=\frac{2^{1-v}\sqrt\pi\cdot\Gamma(v)}{\Gamma\left(\frac{v+1}{2}\right)}$$ is used.
Complex analysis derivation:
$$\begin{align} \int^{\pi}_{-\pi}(\text{coc }x)^a dx &=\int^{\pi}_{-\pi}(e^{ix/2}+e^{-ix/2})^a dx \\ &\stackrel{z=e^{ix}}{=}\oint_{|z|=1}\left(\sqrt z+\frac1{\sqrt z}\right)^a\frac{dz}{iz} \\ &=\oint_{|z|=1}\frac{(z+1)^a}{iz^{a/2+1}}dz \\ &=-\lim_{\epsilon\to0^+}\left(\int^{0+i\epsilon}_{-1+i\epsilon}+\int^{-1-i\epsilon}_{0-i\epsilon}\right)\frac{(z+1)^a}{iz^{a/2+1}}dz \\ &=i\left(\int^0_{-1} \frac{(z+1)^a}{e^{i\pi(a/2+1)}|z|^{a/2+1}}dz+\int^{-1}_0 \frac{(z+1)^a}{e^{-i\pi(a/2+1)}|z|^{a/2+1}}dz \right) \qquad(1)\\ &=i\left(-e^{-i\pi a/2}\int^1_{0} \frac{(1-z)^a}{z^{a/2+1}}dz+e^{i\pi a/2}\int^{1}_0 \frac{(1-z)^a}{z^{a/2+1}}dz \right) \\ &=-2\sin\frac{\pi a}{2}\mathcal B\left(a+1,-\frac a2\right) \\ &=\frac{2\pi\Gamma(a+1)}{\Gamma^2\left(\frac a2+1\right)} \qquad{(2)}\\ \end{align} $$
$(1)$: Consider a key hole contour avoiding the principal logarithmic branch cut on the negative real axis, and apply Cauchy's integral theorem (no singularities are enclosed).
$(2)$: By applying the Gamma reflection formula.
Hence, $$c_n = \left(\frac{\partial}{\partial a}\right)^n \underbrace{\frac{2\Gamma(a+1)}{\Gamma^2\left(\frac a2+1\right)}}_{f(a)}\bigg\vert_{a=0} \\$$
For example, differentiating twice gives $$\begin{align} c_2 &=\frac{2\Gamma(a+1)}{\Gamma^2\left(\frac a2+1\right)}\left[\psi_0(a+1)-\psi_0\left(1+\frac a2\right)\right]^2+\frac{2\Gamma(a+1)}{\Gamma^2\left(\frac a2+1\right)}\left(\psi_1(a+1)-\frac12\psi_1\left(1+\frac a2\right)\right)\bigg\vert_{a=0} \\ &=2\cdot\frac12\psi_1(1) \\ &=(-1)^{1+1}1!\zeta(1+1) \\ &=\zeta (2) \end{align} $$
Other $c_n$ can be found similarly. I tried calculating $c_5$ by hand, and it turns out the algebra is pretty tedious.
It it useful to define $p_k=\psi_k(a+1)-\frac1{2^k}\psi_k\left(\frac a2+1\right)$, since $$f’=fp_0\qquad\qquad p_n’=p_{n+1}$$
Doing some algebra, I got $$\frac{f^{(5)}}{f}=p_0^5+10p_0^3p_1+15p_0p_1^2+10p_0^2p_2+10p_1p_2+5p_0p_3+p_4$$
As $p_0(0)=0$, $$c_5=2(10p_1p_2+p_4)=20\cdot\frac{\pi^2}{12}\frac{-3\zeta(3)}2+2\cdot\frac{-45\zeta(5)}2$$ $$\implies c_5=-\frac52\pi^2\zeta(3)-45\zeta(5)$$
It can be seen that the numerical factors grow quite quickly. One may be interested in the asymptotics of $c_n$.
Notice that $\frac{c_n}{n!}$ is the $n$th coefficient of the Maclaurin series of $f$. Due to the nearest pole at $a=-1$, $$\frac{|c_{n+1}/(n+1)!|}{|c_n/n!|}\sim 1\implies |c_{n+1}|\sim (n+1)|c_n|$$ implying factorial growth.
$f(a)$ satisfies $$f'(a)=f(a)\underbrace{\left((\psi_0(a+1)-\psi_0\left(\frac a2+1\right)\right)}_{\gamma_1(a)}$$
In general $f^{(n)}(a)=f(a)\gamma_n(a)$ where $$\gamma_{n+1}=\gamma_1\gamma_n+\gamma_{n}'$$
Since $\gamma_1(0)=0$, $$c_n=2\gamma_n'(0)$$