= Solution
The <Cauchy integral formula> for <derivatives>, with $z_0=0$, gives
$$
\oint_C\frac{f(z)}{z^{n+1}}\,dz=\frac{2\pi i}{n!}f^{(n)}(0).
$$
The <derivative> cycles of <sine> and <cosine> give $\sin^{(n)}(0)=\sin(n\pi/2)$ and $\cos^{(n)}(0)=\cos(n\pi/2)$. Therefore
$$
\boxed{\oint_C\frac{\sin z}{z^{n+1}}\,dz=\frac{2\pi i}{n!}\sin\frac{n\pi}{2},\qquad
\oint_C\frac{\cos z}{z^{n+1}}\,dz=\frac{2\pi i}{n!}\cos\frac{n\pi}{2}.}
$$
In particular the <sine> <integral> vanishes for even $n$, and the <cosine> <integral> vanishes for odd $n$. For the remaining cases their signs alternate. The radius is irrelevant provided $R>0$, since the numerator is entire and the origin is enclosed once counterclockwise.
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