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We find very quickly that \begin{equation}  \int_0^\pi \mathrm{ln}\;\mathrm{sin}^{-2\mathrm{sin}^2(nx)}(nx)\;dx=\frac{\pi}{2}(\mathrm{ln}(4)-1)  \end{equation}  for any value of $n$! That is the entropy of each state is the same, independant of the level of excitation.