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I'm trying to calculate this integral : $$\int_{D}\int{\sin(x-y)dxdy} $$ where $D=${$(x,y)|x^2+y^2\le1$} ,I've changed the parameters to: $$x(u,v)=r\cos(\theta+\frac{\pi}4),\ y(u,v)=r\sin(\theta+\frac{\pi}4)$$ and finally I got stuck at the integral: $$\int_{0}^{2\pi}\int_{0}^{1}|r|\sin(-\sqrt2r\sin(\theta))drd\theta$$ which I believe is very hard to solve.

Can anyone help?

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    hint: Set $r=x-y$,$R=x+y$ – 2017-01-24
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    what are the limits on the original double integral? – 2017-01-24
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    $\sin(x-y)=\sin x \cos y - \cos x \sin y$ – 2017-01-24
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    Umberto, unit circle $x^2+y^2\le1$ – 2017-01-24
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    Guys, i don't understand how these clues help in this problem. – 2017-01-25
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    @CodeHoarder: the contribute given by $(x,y)$ and the contribute given by $(-x,-y)$ cancel out, so the integral is simply zero. – 2017-01-25

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The integral of an odd integrable function over a symmetric (with respect to the origin) domain is always zero.