Chapter 15 - Multiple Integrals - 15.3 Double Integrals in Polar Coordinates - 15.3 Exercises - Page 1055: 12

$$2\pi(2\sin2+\cos2-1)$$

We write the region $D$ in polar coordinate then $\{D:0\leq\theta\leq2\pi, 0\leq r\leq2\}$, therefore the integral becomes $$\int_{0}^{2\pi}\int_{0}^{2}\cos(r^2)r\,dr\,d\theta=\int_{0}^{2\pi}\,d\theta\int_{0}^{2}\cos(r)r\,dr=2\pi\biggl[r\sin r+\cos r\biggr]_{0}^{2}=2\pi(2\sin2+\cos2-1)$$