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**Drexel28** Yes, but there may be an easier way to ascertain that last value. So, by Cauchy's integral theorem we have that $\displaystyle I=\int_{C}\frac{\sin(z)}{(z-\pi)^{10}}dz=2\pi i\left(\sin(z)\right)^{(9)}\mid_{z=\pi}$. But, notice that $\displaystyle \sin(z-\pi)=-\sin(z-\pi)$. But, remember that the Taylor series for $\displaystyle \sin(z-\pi)=\sum_{n=0}^{\infty}\frac{f^{(n)}(\pi)(z-\pi)^n}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n+1}}{n!}$. Comparing this we see that $\displaystyle f^{(n)}(\pi)=\frac{(-1)^4}{(2(4)+1}=\frac{1}{9!}$. Thus, by previous comment we have that $\displaystyle I=-2\pi i\left(\sin(z)\right)^{(9)}\mid_{z=\pi}=-2\pi i\cdot\frac{1}{9!}$ which agrees with your answer.