Integrating sin(x) on a unit circle I am trying to integrate $\int\int_{D} sin(x)$ where $D$ is a unit circle centered at $(0,0)$.
My approach is to turn the area into the polar coordinate so I have $D$ as $0\leq r\leq1$ , $0 \leq \theta \leq 2\pi$.
Which turns the integral into:
$$\int^{2\pi}_{0}\int^{1}_{0} \sin(r\cos(\theta)) |r| drd\theta$$
and it is not integrable. 
I also tried the Cartesian approach by evaluating:
$$\int^{1}_{0}\int^{\sqrt{1-x^2}}_{0} \sin(x) dydx$$
which is also not integrable
Which direction or method should I use to integrate this problem?
Edit*
Thank you so much for the answers, I agree that because of symmetry, the answer is zero. The hint says don't do too work work also lol.
 A: By symmetry, the integral is $0$.
Remark: In the Cartesian version, the integral of the post integrates over the quarter disk in the first quadrant. 
A: Additional: From a complex analysis approach we have $\int_{|z|=1} \sin(z) \ dz = 0 $, since $f(z)=\sin(z)$ is analytic on the unit circle.
A: knowing some complex analysis helps dramatically simplify the problem. Just define $f(z)=sin(z)$ then we have that $sin(z)$ is holomorphic/analytic inside the disk $|z|=1$, which is just the unit circle. By Cauchy's Integral Formula (CIF), the integral evaluates to $0$. 
This approach is much stronger than the traditional double integral approach for the following reasons.
i) Integrating functions sometimes can be a tedious bookkeeping exercise, and sometimes may even require referencing some table of integrals to get the solution. There are a lot of techniques and tricks, but they apply to different situations, whereas the CIF applies to any general holomorphic function. 
ii) the composition of holomorphic functions are holomorphic, and by extension, compositions of continuous, complex-analytic functions are continuous. So we can take some nasty function $f(x)=e^xcos(sin(2e^x))$ and integrate over the circle in "one" step (need to check Cauchy-Riemann criteria first). Just rewrite as $f(z)=e^zcos(sin(2e^z))$ and realize that since this function is holomorphic on the complex unit disk, the integral vanishes. 
