Contour integral of $\text{sin}(z)dz$ I'm stuck trying to figure out how to solve the following integral:
$\int_{C(0,1)^+}\sin(z)dz$
I've tried parameterizing z(t) but then I get
$\int_0^{2\pi}\sin(e^{it})ie^{it}$ which I don't know how to integrate.
So then I'm looking to use Cauchy's Integral formula but I'm not sure if I can use it because it doesnt fit the structure.
I'm not sure where to go from here so any help would be greatly appreciated!
 A: First way:
Parametrize the unit circle as $\;\gamma(t):=e^{it}\;,\;\;0\le t<2\pi\;\implies \gamma'(t)=ie^{it}dt $
so that your integral becomes
$$i\int_0^{2\pi}e^{it}\sin e^{it}dt=\left.-i\cos e^{it}\right|_0^{2\pi}=0$$
Second way:
The function $\;\cos z\;$ is a primitive function of $\;\sin z\;$ in the whole complex plane (this is what sometimes's called "a potential function" of a two-variable function in real analysis in the plane, and thus we get a conservative field...), so that directly
$$\oint_{S^1}\sin z\;dz=\left.-\cos z\right|_{z_1=0,z_2=0}=-(\cos 0-\cos 0)=0$$
or, of course, directly noting that when we've a primitive function (potential) and integrate on a closed, simple rectifiable path, the integral is zero.
Third way:
A third way is mentioned in the other answer: using the power series of sine, which has infinite convergence radius and is thus possible to integrate it elementwise in the whole complex plane, and then observing that with the usual parametrization of the unit circle (which you use)
$$\oint_{S^1} z^n dz=\int_0^{2\pi} ie^{it}e^{nit}\,dt=i\int_0^{2\pi} e^{(n+1)it}\,dt=\left.\frac{-i}{n+1}e^{(n+1)it}\right|_0^{2\pi} =0\;\;,\;\;\;n=0,1,2,...$$
The last way:
I only mention it for completeness as you obviously haven't yet studied it, but it is Cauchy's Theorem: the integral over a closed, simple path of an analytic function (analytic on the path and inside the region it encloses) is zero.
A: The integral of any entire function over any closed curve is just zero. That follows from:
$$ \forall n\in\mathbb{Z},\qquad \int_{0}^{2\pi} e^{nit}\,dt = 2\pi\cdot\delta(n). $$
In your case:
$$ \sin z = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)!}z^{2n+1} $$
can be integrated termwise. Can you see what happens, then?
