Evaluate $\int\limits_{-r}^{0}\int\limits_{0}^{\theta}\sin(2\omega\xi)\cos(\omega\theta)\delta(\theta+r)d\xi d\theta$

I know some basic properties of the Dirac delta 'function', and I'm familiar with the sifting property. In any case, I'm stuck trying to evaluate this double integral$$\int\limits_{-r}^{0}\int\limits_{0}^{\theta}\sin(2\omega\xi)\cos(\omega\theta)\delta(\theta+r)d\xi d\theta,$$where $r,\omega\in\mathbb{R}^{+}$. Here's my initial attempt.$$\int\limits_{-r}^{0}\int\limits_{0}^{\theta}\sin(2\omega\xi)\cos(\omega\theta)\delta(\theta+r)d\xi d\theta=\left(\int\limits_{-r}^{0}\left(\int\limits_{0}^{\theta}\sin(2\omega\xi)d\xi\right)\cos(\omega\theta)\delta(\theta+r)d\theta\right)=\left\{\begin{array}{ccl}-\frac{1}{2\omega}\sin(\omega r)\sin(2\omega r) &,&~~\text{if}~r\in(-r,0)\\0 &,&~r\notin [-r,0]\end{array}\right.$$This does not make any sense to me since $r\in\mathbb{R}^{+}$. My guess is that the value of this integral is zero. All suggestions are appreciated.