Answer
$$
\begin{aligned}
\iint_{D} \cos \sqrt{x^{2}+y^{2}} d =2 \pi(2 \sin 2+\cos 2-1).
\end{aligned}
$$
Work Step by Step
$$
\iint_{D} \cos \sqrt{x^{2}+y^{2}} d A
$$
The region $D$ is the disk with center the origin and radius 2
$$
D=\left\{(x, y) | \quad a \leq x^{2}+y^{2} \leq 4 \right\}
$$
and in polar coordinates it is given by
$$
D=\left\{(r, \theta) | 0 \leq r \leq 2 , \quad 0 \leq \theta \leq 2 \pi \right\}
$$
Therefore,
$$
\begin{aligned}
\iint_{D} \cos \sqrt{x^{2}+y^{2}} d A &=\int_{0}^{2 \pi} \int_{0}^{2} \cos \sqrt{r^{2}} r d r d \theta \\
&=\int_{0}^{2 \pi} d \theta \int_{0}^{2} r \cos r d r .
\end{aligned}
$$
For the second integral, integrate by parts with
$$
\quad\quad \left[\begin{array}{c}{u=r, \quad\quad dv= \cos r d r } \\ {d u= dr, \quad\quad v= \sin r }\end{array}\right].
$$
Then
$$
\begin{aligned}
\iint_{D} \cos \sqrt{x^{2}+y^{2}} d A&=[\theta]_{0}^{2 \pi}[r \sin r+\cos r]_{0}^{2} \\
&=2 \pi(2 \sin 2+\cos 2-1).
\end{aligned}
$$
