Your question is basically whether
$$\int_0^{\pi}\frac{\sin 2\theta \,\mathrm{d}\theta}{5-3\cos\theta}=\int_{-1}^1\frac{2x\,\mathrm{d}x}{5-3x}= \tfrac{4}{9}(5\,\ln 2-3)$$
can be written as an integral about a closed curve of some algebraic function over the rationals. This statement is believed to be false; it is in fact an open problem to demonstrate as much (Kontsevich and Zagier, Problem 5).
The usual method (substituting $x=\cos\theta$) will have to do.
$|1-e^{i\theta}|=2\sin{\frac{\theta}{2}}$, so we need to evaluate $I=\int_0^{2\pi}\log |1-e^{i\theta}|d\theta$ as then the answer is $2\pi\log R+I$.
Now $\log|1-z|=\Re{\log(1-z)}$ is harmonic inside the unit disc, so by the mean value theorem for harmonic function, $\int_0^{2\pi}\log |1-re^{i\theta}|d\theta=0$ for any $r<1$.
But $\log|1-e^{i\theta}|$ is obviously integrable on the unit circle (since near $0$, $\log 2\sin{\frac{\theta}{2}}-\log \theta$ is continuos and the latter is clearly integrable, while near $2\pi$ we can use $\sin{\frac{\theta}{2}}=\sin{\frac{2\pi-\theta}{2}}$ and the result at zero) and $\log|1-re^{i\theta}| \to \log|1-e^{i\theta}|$ a.e.
But now if say $0 \le \theta \le \frac{\pi}{100}$ or $0 \le 2\pi-\theta \le \frac{\pi}{100}$, by drawing the perpendicular from $1$ to the $\theta$ ray which has length $|\sin \theta|$ it follows from elementary geometry that $|1-re^{i\theta}| \ge |\sin \theta|$, while for the rest $|1-re^{i\theta}| \ge c >0$ and since $|1-re^{i\theta}| \le 2$, we get that $|\log |1-re^{i\theta}|| \le \max {(\log 2, \log^- c-\log |\sin \theta|)}$ and that is integrable on the unit circle as before, hence we can apply the Lebesgue dominated convergence and conclude that $0=\int_0^{2\pi}\log |1-re^{i\theta}|d\theta \to I$, so $I=0$ and the final answer is $2\pi\log R$
As an aside, there is also a classic real variables proof that $I=0$ using the doubling formula for the $\sin$ and various changes of variable while the above can be expressed in terms of contour integrals if one wishes using $f(z)=\frac {\log (1-z)}{z}$ which is analytic inside the unit disc, or treat the original integral using $f(z)=\frac {\log R(1-z)}{z}$ which has $\log R$ as residue at $0$ etc, but of course it may not quite be what OP had in mind.
(Edit) As asked let's quickly use Cauchy rather than Poisson:
$I_1=\int_0^{2\pi}\log (2R\sin{\frac{\theta}{2}})d\theta=\int_0^{2\pi}\log R|1-e^{i\theta}|d\theta= \int_0^{2\pi} \Re {\log R(1-e^{i\theta})}d\theta=\Re {\int_0^{2\pi} \log R(1-e^{i\theta})d\theta}$
with the last equality holding because $d\theta$ is a real (positive) measure.
But now with the usual $e^{i\theta}=z, d\theta=\frac{1}{iz}dz$ we have;
$I_1=\Re {\int_{|z|=1} \frac{\log R(1-z)}{iz}dz}$
Now by Cauchy ${\int_{|z|=1} \frac{\log R(1-rz)}{iz}dz}= 2\pi \log R$ as the residue at zero of the integrand is $\frac{\log R}{i}$, while it is analytic anywhere else on the closed unit disc when $0 < r <1$. So we need to be able to pass to the limit $ r \to 1$ to conclude that the above unit circle integral (and hence its real part) is $2\pi \log R$ and the same argument as above works since the only problem comes from $\log |1-rz|$ near the boundary as everything else is obviously bounded so the same estimates work to show that we can use the Lebesgue dominated convergence and conclude that $I_1= 2\pi \log R$
(for $|z|=1$ we have $|\frac{\log R(1-rz)}{iz}|\le |\log R|+ |\log (1-rz)| \le |\log R|+ |\Re \log (1-rz)|+ |\Im \log (1-rz)| $ and $|\Re \log (1-rz)|=|\log |1-rz||$ as above, while $|\Im \log (1-rz)|=|\arg (1-rz)| \le \frac{\pi}{2}$ since $\Re (1-rz) >0, |z| =1$)
Best Answer
Let $z=e^{i \theta}$, then $d\theta=dz/(i z)$ and $\sin{\theta} = (z-1/z)/(2 i)$. Then the integral becomes
$$\frac12\frac{-i}{(2 i)^{2 n}} \oint_{|z|=1} \frac{dz}{z} \left( z-z^{-1}\right)^{2 n} =\frac12 \frac{-i}{(2 i)^{2 n}} \oint_{|z|=1} dz \frac{(z^2-1)^{2 n}}{z^{2 n+1}}$$
As you can see, you have a pole of order $2 n+1$ in the integrand. To apply the residue Theorem, you need to evaluate $i 2 \pi$ times the residue at the pole at $z=0$, which is
$$\frac{\pi}{(2 i)^{2 n}} \frac{1}{(2 n)!} \left[\frac{d^{2 n}}{dz^{2 n}}\left ( z^2-1\right)^{2 n}\right]_{z=0}$$
Now, by Rodrigues' formula for Legendre polynomials, the latter expression is
$$\left[\frac{d^{2 n}}{dz^{2 n}}\left ( z^2-1\right)^{2 n}\right]_{z=0} = 2^{2 n} (2n)! P_n(0)$$
ADDENDUM
We can also use the binomial theorem to extract an explicit expression for the residue. Note that
$$\left ( z^2-1\right)^{2 n} = \sum_{k=0}^{2 n} \binom{2 n}{k} z^{2 k} (-1)^k$$
Taking the $2 n$th derivative and setting $z=0$ leaves only the $n$th term in the sum, so we get
$$\left[\frac{d^{2 n}}{dz^{2 n}}\left ( z^2-1\right)^{2 n}\right]_{z=0} = (-1)^n \frac{(2 n)!^2}{(n!)^2}$$
Therefore, the integral is
$$\frac{\pi}{2^{2 n}} \binom{2 n}{n}$$