Animated Solution for Mathematics - Definite Integration: The area of the region $A=\{(x,y):|\cos x-\sin x|\le y\le \sin x,0\le x\le \frac{\pi }{2}\}$ is

The region $A$ is defined by the inequality $|\cos x-\sin x|\le y\le \sin x$ within the interval $x\in [0,\frac{\pi }{2}]$. The absolute value function acts as a piecewise operator that changes behavior at the point where $\cos x=\sin x$, which occurs at $x=\frac{\pi }{4}$.

We must split our analysis into two distinct intervals: $[0,\frac{\pi }{4}]$ and $[\frac{\pi }{4},\frac{\pi }{2}]$.

To find the starting point of our region, we equate the upper boundary $y=\sin x$ with the lower boundary $y=\cos x-\sin x$ (valid for $x\in [0,\frac{\pi }{4}]$). Setting these equal yields:

Let $\alpha =\arctan (\frac{1}{2})$. In a right-angled triangle with opposite side $1$ and adjacent side $2$, the hypotenuse is $\sqrt{5}$.

Consequently, we determine the trigonometric values:

The total area $A$ is the sum of two integrals, $I_1$ and $I_2$. For $I_1$, covering the range from $\alpha$ to $\frac{\pi }{4}$, the integrand is the upper curve minus the lower curve:

For $I_2$, covering the range from $\frac{\pi }{4}$ to $\frac{\pi }{2}$, the lower boundary is $\sin x-\cos x$. The integral becomes:

For $I_1$, the anti-derivative of $2\sin x-\cos x$ is $-2\cos x-\sin x$. Evaluating this from $\alpha$ to $\frac{\pi }{4}$:

For $I_2$, the anti-derivative of $\cos x$ is $\sin x$:

Summing these results, $A=(\sqrt{5}-\frac{3}{\sqrt{2}})+(1-\frac{1}{\sqrt{2}})$, we arrive at the final result: