Calculate double integral using Polar coordinate system Need to calculate $\int_{0}^{R}dx\int_{-\sqrt{{R}^{2}-{x}^{2}}}^{\sqrt{{R}^{2}-{x}^{2}}}cos({x}^{2}+{y}^{2})dy$
My steps:


*

*Domain of integration is the circle with center (0,0) and radius R;

*$x = \rho \cos \varphi ,\: y = \rho \sin \varphi,\: \rho \in \left[0,R \right],\: \varphi \in [0, 2\pi )$;

*$\int_{0}^{R}dx\int_{-\sqrt{{R}^{2}-{x}^{2}}}^{\sqrt{{R}^{2}-{x}^{2}}}\cos\left( {x}^{2} + {y}^{2}\right) dy = \int_{0}^{R}d\rho \int_{0}^{2\pi }cos({\rho }^{2})d\varphi = 2\pi \int_{0}^{R}cos({\rho }^{2})d\rho$
As I know from WolframAlpha last integral can not be calculated using Elementary functions.

In polar coordinates the integral is $\int_{-\pi /2}^{\pi /2}d \varphi \int_{0}^{R}cos({\rho}^{2} ) \rho d \rho =\pi \left[ \frac{\sin \left({\rho}^{2} \right)}{2}\right]_0^R=\frac{\pi }{2}\sin {R}^{2}$
