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.

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What about symmetry considerations...? – Andrew D. Hwang Mar 30 '14 at 17:36
Aside: $D$ is the unit disc: the unit circle is the loop around the boundary of the disc. – Hurkyl Mar 30 '14 at 18:53
UW questions....lol – Jing Apr 3 '14 at 15:45

2 Answers

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.

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Thank you! I appreciate your answer. – tkhu Mar 30 '14 at 20:40
You are welcome. – André Nicolas Mar 30 '14 at 20:49

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.

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Thank you for your answer, it's great to have a deeper understanding of the question too. – tkhu Mar 30 '14 at 20:42