Answer
$\dfrac{\pi}{2}$
Work Step by Step
Green's Theorem states that:
$\oint_C M \,dx+ N \,dy=\iint_{D}(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y})dA$
We will use polar coordinates as follows: $x=r \cos \theta; y= r \sin \theta$
We need to work out with the line integral and evaluate the integrand of the double integral as follows:
$\oint_CP\,dx+Q\,dy=\iint_{D} (2y-2x) dA = \int_{-\pi/2}^{\pi/2} \int_{0}^{\cos x} (2y-2x) \ dy \ dx \\= \int_{-\pi/2}^{\pi/2} [y^2-2xy]_{0}^{\cos x} \ dx \\ = \int_{-\pi/2}^{\pi/2} \cos^2 x dx$
Since, $\cos^2 x= \dfrac{1+\cos 2 x}{2}$
Consider $I=2 \int_{0}^{\pi/2} \dfrac{1+\cos 2 x}{2} \ dx =[x+\dfrac{\sin 2x}{2}]_0^{\pi/2} =\dfrac{\pi}{2}$
