Answer
$$1-\frac{\sqrt{2}}{2}$$
Work Step by Step
Given $$y=\sin x, \quad y=\csc ^{2} x, \quad x=\frac{\pi}{4}$$
Since
\begin{aligned}
\sin x &=\csc ^{2} x \\
\sin ^{3} &=1 \\
\sin x &=1 \\
x &=\frac{\pi}{2}
\end{aligned}
and $\csc^2 x\geq \sin x$ for $\pi/4\leq x\leq \pi/2$, so
\begin{aligned}
A &=\int_{\pi / 4}^{\pi / 2}\left[\csc ^{2} x-\sin x\right] d x \\
&=[-\cot x+\cos x]\bigg|_{\pi / 4}^{\pi / 2} \\
&=1-\frac{\sqrt{2}}{2}
\end{aligned}
