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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:

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

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

  3. $\int_{0}^{R}dx\int_{-\sqrt{{R}^{2}-{x}^{2}}}^{\sqrt{{R}^{2}-{x}^{2}}}\cos\left( {x}^{2} + {y}^{2}\right) dy = \int_{0}^{R}d\rho \int_{0}^{2\pi }cos({\rho }^{2})d\varphi = 2\pi \int_{0}^{R}cos({\rho }^{2})d\rho$

As I know from WolframAlpha last integral can not be calculated using Elementary functions.

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2 Answers 2

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You took the wrong bounds for the angle, and forgot to insert the Jacobian determinant:

$$ \int_{0}^{R}\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}}\cos(x^2+y^2)\,dy \,dx = \int_{-\pi/2}^{\pi/2}\int_{0}^{R}\rho \cos(\rho^2)\,d\rho\,d\theta = \frac{\pi}{2}\sin(R^2).$$

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

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