Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} \cos x\sin 2y{\rm{d}}A = 1$
Work Step by Step
We have $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} \cos x\sin 2y{\rm{d}}A$ and ${\cal R} = \left[ {0,\frac{\pi }{2}} \right] \times \left[ {0,\frac{\pi }{2}} \right]$.
Write
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} \cos x\sin 2y{\rm{d}}A = \mathop \smallint \limits_{y = 0}^{\pi /2} \left( {\mathop \smallint \limits_{x = 0}^{\pi /2} \cos x{\rm{d}}x} \right)\sin 2y{\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^{\pi /2} \left( {\sin x|_0^{\pi /2}} \right)\sin 2y{\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^{\pi /2} \sin 2y{\rm{d}}y$
$ = - \frac{1}{2}\left( {\cos 2y|_0^{\pi /2}} \right)$
$ = - \frac{1}{2}\left( { - 1 - 1} \right)$
$ = 1$
