Answer
$$2-\frac{\pi}{2}$$
Work Step by Step
Given $$x=\sin y, \quad x=\frac{2}{\pi} y$$
Since
\begin{aligned}
\sin y &=\frac{2}{\pi} y \\
y &=0, \pm \frac{\pi}{2}
\end{aligned}
and $ \sin y\geq 2y/\pi $ for $0\leq y\leq \pi/2$
Then from symmetry,
\begin{aligned}
A &=2\int_{0}^{\pi / 2}\left[\sin y-\frac{2}{\pi} y\right] d y \\
&=2\left[-\cos y-\frac{1}{\pi} y^{2}\right]_{0}^{\pi / 2} \\
&=2\left[-\cos \frac{\pi}{2}-\frac{1}{\pi}\left(\frac{\pi}{2}\right)^{2}+\cos 0+0\right] \\
&=2-\frac{\pi}{2}
\end{aligned}
