Need help with Cauchy's integral $\frac{1}{2\pi i} \int_{c_1(1)} \frac{\sin(z)}{z^{n+1}} \mathrm{d}z$ value the integral

Question

I need to calculate this integral with Cauchy's integral $$\frac{1}{2\pi i} \int_{c_1(1)} \frac{\sin(z)}{z^{n+1}} \mathrm{d}z$$ by 1(1) a circle of radius 1.

Answer 1

You titled this "Cauchy's integral" so I presume you know the version of Cauchy's integral formula called "Cauchy's differential formula", $\frac{d^nf(a)}{dx^n}= \frac{n!}{2\pi i}\oint_\gamma \frac{f^n(z)}{(z- a)^{n+1}}dz$

We can write $\frac{1}{2\pi i}\int \frac{sin(z)}{z^{n+1}}dz= \frac{1}{n!}\left(\frac{n!}{2\pi i}\int \frac{sin(z)}{z^{n+1}}dz\right)$ where the integrals of "Cauchy's integral formula" has a= 0 and f(z)= sin(z). So the integral is equal $\frac{1}{n!}\frac{d^n sin(x)}{d^n x}$.