Use horizontal strips for an inverse-sine boundary

The inverse-sine graph runs from $(-1,-\pi/2)$ to $(1,\pi/2)$ and passes through the origin. Rewrite $y=\sin^{-1}x$ as $x=\sin y$ on the principal inverse-sine range. Horizontal strips run between this curve and the y-axis.

For $-\pi/4\le y\le0$, $\sin y\le0$, so negate that integral. For $0\le y\le\pi/3$, $\sin y\ge0$. Thus $A=-\int_{-\pi/4}^0\sin y\,dy+\int_0^{\pi/3}\sin y\,dy$.

The Teacher gives $A=0.793$ square units to three significant figures. Independently, using $-\cos y$ gives $A=(1-\sqrt2/2)+(1-1/2)=3/2-\sqrt2/2\approx0.792893219$. The source calculator remark allows $-\int_{-\pi/4}^0\sin x\,dx+\int_0^{\pi/3}\sin x\,dx$: $x$ is just a renamed dummy variable with the same limits. Use radian mode.