= Solution
For an arc represented by $0\leq x\leq a$, take $0\leq a\leq2\pi R$. The <Born rule> uses
$$
|\Psi(t,x)|^2=\frac1{\pi R}\cos^2\!\left(\frac{2(x+s\tau)}R\right),\qquad \tau=t-T.
$$
Integrate using $\cos^2u=(1+\cos2u)/2$:
$$
\boxed{\begin{aligned}
\Pr(0\leq x\leq a)&=\frac1{\pi R}\int_0^a\cos^2\!\left(\frac{2(x+s\tau)}R\right)dx\\
&=\frac{a}{2\pi R}+\frac1{8\pi}
\left[\sin\!\left(\frac{4(a+s\tau)}R\right)-\sin\!\left(\frac{4s\tau}R\right)\right].
\end{aligned}}
$$
The result is zero at $a=0$ and one at $a=2\pi R$, checking the <wavefunction normalization>. At $s=0$ it reduces to $a/(2\pi R)+\sin(4a/R)/(8\pi)$, the stationary position <probability>. Longer intervals on the periodically identified coordinate require specifying the corresponding subset of the circle; integrating multiple turns would count the same positions repeatedly.
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