Answer
$A=\frac{\pi}{2}$
Work Step by Step
$\sin 2 \theta=r, 0 \leqslant \theta \leqslant \frac{\pi}{2}$
so $A=\int_{0}^{\frac{\pi}{2}} \int_{0}^{\sin 2 \theta} r d r d \theta$
Integrate with respect to $r$
$4 \int_{0}^{\frac{\pi}{2}}\left[\frac{r^{2}}{2}\right]_{0}^{\sin \theta} d \theta=A$
$\int_{0}^{\frac{\pi}{2}}\left[r^{2}\right]_{0}^{\sin 2 \pi} d \theta=A$
Evaluate
$A=2 \int_{0}^{\frac{\pi}{2}} \sin ^{2} 2 \theta d \theta$
Evaluate
$2 \int_{0}^{\frac{\pi}{2}} \sin ^{2} 2 \theta d \theta=A$
and then
$2 \int_{0}^{\frac{\pi}{2}}\left[\frac{1}{2}-\frac{\cos 4 \theta}{2}\right]_{0}^{2} d \theta=A$
$\int_{0}^{\frac{\pi}{2}}(-\cos 4 \theta+1) d \theta=A$
Integrate and evaluate
$A=\left[-\frac{\sin 4 \theta}{4}+\theta \right]_{0}^{\frac{\pi}{2}}$
$A=\left[-\frac{\sin 2 \pi}{4}+\frac{\pi}{2}\right]-[0]$
$A=\frac{\pi}{2}$
