Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{\cos x} \sin x{\rm{d}}y{\rm{d}}x \simeq 0.35404$
Work Step by Step
We have $f\left( {x,y} \right) = \sin x$; bounded by $x=0$, $x=1$, $y = \cos x$.
The domain ${\cal D}$ is a vertically simple region (please see the figure attached). So, the domain description is given by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,0 \le y \le \cos x} \right\}$
We evaluate the double integral of $f\left( {x,y} \right) = \sin x$ over ${\cal D}$ as an iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{\cos x} \sin x{\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \sin x\left( {y|_0^{\cos x}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \sin x\cos x{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \sin 2x{\rm{d}}x$
$ = - \frac{1}{4}\cos 2x|_0^1$
$ = - \frac{1}{4}\cos 2 + \frac{1}{4}$
$ \simeq 0.35404$
