The area of the region A = x , y : cos x - sin x ≤ y ≤ sin x , 0 ≤ x ≤ π 2

Given,$\left|\cos x-\sin x\right|\le y\le \sin x, 0⩽x⩽\frac{\mathrm{\pi }}{2}$

For intersection point:

$\cos x-\sin x=\sin x$

$\Rightarrow \tan x=\frac{1}{2}$

Let $\psi ={\tan }^{-1}\left(\frac{1}{2}\right)$, so $$\begin{gathered} \tan \psi =\frac{1}{2} \\ \sin \psi =\frac{1}{\sqrt{5}} \\ \cos \psi =\frac{2}{\sqrt{5}} \end{gathered}$$

Required area is $={\int }_{\psi }^{\pi /2}\left(\sin x-\left|\cos x-\sin x\right|\right)dx$

$={\int }_{\psi }^{\pi /4}\left(\sin x-\left(\cos x-\sin x\right)\right)dx+{\int }_{\pi /4}^{\pi /2}\left(\sin x+\left(\cos x-\sin x\right)\right)dx$

$={\int }_y^{\pi /4}\left(2\sin x-\cos x\right)dx+{\int }_{\pi /4}^{\pi /2}\left(\cos x\right)dx$

$={\left[-2\cos x-\sin x\right]}_{\psi }^{\pi /4}+{\left[\sin x\right]}_{\pi /4}^{\pi /2}$

$=-\sqrt{2}-\frac{1}{\sqrt{2}}+2\cos \psi +\sin \psi +\left(1-\frac{1}{\sqrt{2}}\right)$

$=1-\sqrt{2}-\frac{2}{\sqrt{2}}+2\left(\frac{2}{\sqrt{5}}\right)+\left(\frac{1}{\sqrt{5}}\right)$

$=1-2\sqrt{2}+\frac{5}{\sqrt{5}}$

$=\left(\sqrt{5}-2\sqrt{2}+1\right)$ sq. units