Answer
$$\frac{{\sin \left( 1 \right)}}{2}\pi $$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^1 {\int_0^{\sqrt {1 - {x^2}} } {\cos \left( {{x^2} + {y^2}} \right)} dydx} \cr
& R{\text{ has the bounds}} \cr
& 0 \leqslant y \leqslant \sqrt {1 - {x^2}} ,{\text{ }} - {\text{1}} \leqslant x \leqslant 1 \cr
& {\text{Is a semicircle of radius }}1,{\text{ centered at the origin}} \cr
& {\text{The region }}R{\text{ in polar coordinates is}} \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant r \leqslant 1,{\text{ }}0 \leqslant \theta \leqslant \pi } \right\} \cr
& {\text{Where: }}x = r\cos \theta ,{\text{ }}y = r\sin \theta {\text{ and }}dydx = rdrd\theta \cr
& {\text{So}},{\text{ we have}} \cr
& = \int_0^\pi {\int_0^1 {\cos \left( {{r^2}{{\cos }^2}\theta + {r^2}{{\sin }^2}\theta } \right)rdrd\theta } } \cr
& = \int_0^\pi {\int_0^1 {\cos \left( {{r^2}} \right)rdrd\theta } } \cr
& {\text{Integrate with respect to }}r \cr
& = \frac{1}{2}\int_0^\pi {\left[ {\sin {r^2}} \right]_0^1} d\theta \cr
& = \frac{1}{2}\int_0^\pi {\left[ {\sin \left( 1 \right) - \sin \left( 0 \right)} \right]} d\theta \cr
& = \frac{{\sin \left( 1 \right)}}{2}\int_0^\pi {d\theta } \cr
& = \frac{{\sin \left( 1 \right)}}{2}\left[ \theta \right]_0^\pi \cr
& = \frac{{\sin \left( 1 \right)}}{2}\pi \cr} $$
