Only z=0 is singular inside: other zeros satisfy z³=2πk and have modulus at least (2π)^(1/3)>1. Near zero, $[1-\cos(z^3)]^{-1}=2z^{-6}+O(1)$. Multiplying by sinz=z−z³/6+z⁵/120−… gives residue 2/120=1/60. Hence I=1/60 and **120I=2**.
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