Evaluating past exam problem: $\int_C \frac{\sin z}{(z+1)^7} \mathrm{d}z$ I want to evaluate the following:
$$\int_C \frac{\sin z}{(z+1)^7} \mathrm{d}z$$
Where $C$ is the circle of radius $5$, centre $0$, positively oriented.
Now this has one root at $z=-1$. Now I should probably use the Cauchy integral formula:
Then I get $$\int_C \frac{\sin z}{(z+1)^7} \mathrm{d}z=\int_C \frac{\sin z /(z+1)^6}{z-(-1)}\mathrm{d}z$$ But I have the same problem in regards to wanting to avoid $z=1$
How else should I go about this.
 A: $$\int_C \frac{\sin z}{(z+1)^7} \mathrm{d}z$$
Where $C$ is the circle of radius $5$, centre $0$, positively oriented.

By David's advice, we will use the generalized Cauchy integral formula:
$$\int_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz =\frac{2\pi i}{n!}f^{(n)}(z_0)\ ,$$
Where we must note that $z_0=1$ is within our contour, that is otherwise analytic within and on the contour and is positively transversed.
$$\int_C \frac{\sin z}{(z-(-1))^{6+1}} \mathrm{d}z=\frac{2\pi i}{6!} \sin^{(6)}(-1)$$
Where $\sin^{(6)} =-\sin$, and hence we obtain $I=-\frac{2\pi i}{6!}\sin(-1)$
A: A second way of proceeding which is similar to @Icuttrees is to use the residue theorem.   Here, the residue of an $n$'th order pole is given by
$$\text{Res} \left(f(z),z=z_0\right)=\frac{1}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\left((z-z_0)^nf(z)\right)$$
Here, we have $f(z)=\frac{\sin z}{(z+1)^7}$, $z_0=-1$, and $n=7$.  Thus, 
$$\text{Res} \left(\frac{\sin z}{(z+1)^7},z=-1\right)=\frac{1}{6!}\frac{d^{6}}{dx^{6}}\left(\sin z \right)=\frac{\sin 1}{6!}$$
whereupon the value of the integral of interest is

$$\oint_C \frac{\sin z}{(z+1)^7}dz=\frac{2\pi i}{6!}\sin 1$$

A: Since the coefficient of $z^6$ in $\sin(z)$ is $0$ and the coefficient of $z^6$ in $\cos(z)$ is $-\frac1{6!}$, we get
$$
\newcommand{\Res}{\operatorname*{Res}}
\begin{align}
\Res_{z=-1}\left(\frac{\sin(z)}{(z+1)^7}\right)
&=\Res_{z=0}\left(\frac{\sin(z-1)}{z^7}\right)\\
&=\Res_{z=0}\left(\frac{\sin(z)\cos(1)-\cos(z)\sin(1)}{z^7}\right)\\
&=\cos(1)\Res_{z=0}\left(\frac{\sin(z)}{z^7}\right)-\sin(1)\Res_{z=0}\left(\frac{\cos(z)}{z^7}\right)\\
&=\cos(1)\cdot0-\sin(1)\cdot\left(-\frac1{6!}\right)\\
&=\frac{\sin(1)}{720}
\end{align}
$$
Thus, the integral is $2\pi i$ times the sum of the residues (times the winding number of the curve around each residue):
$$
\int_C\frac{\sin(z)}{(z+1)^7}\,\mathrm{d}z=\frac{\pi\sin(1)}{360}\,i
$$
