# Integral of $\int\int{\sin(x-y)dxdy}$

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?

• hint: Set $r=x-y$,$R=x+y$ – tired Jan 24 '17 at 19:30
• what are the limits on the original double integral? – Umberto P. Jan 24 '17 at 19:32
• $\sin(x-y)=\sin x \cos y - \cos x \sin y$ – W.R.P.S Jan 24 '17 at 19:41
• Umberto, unit circle $x^2+y^2\le1$ – CodeHoarder Jan 24 '17 at 19:45
• Guys, i don't understand how these clues help in this problem. – CodeHoarder Jan 25 '17 at 5:58