1. Interpret the region

We need the area of

$$A=\{(x,y): |\cos x-\sin x|\le y\le \sin x, \ 0\le x\le \tfrac{\pi}{2}\}.$$

For a fixed $x$, the vertical length exists only when

$$|\cos x-\sin x|\le \sin x.$$

So first we find the $x$-values where the region is non-empty.

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2. Solve the inequality

We need

$$|\cos x-\sin x|\le \sin x.$$

Since on $[0,\pi/2]$, both $\sin x,\cos x\ge 0$, we can square safely:

$$(\cos x-\sin x)^2\le \sin^2 x.$$

Expand:

$$\cos^2 x-2\sin x\cos x+\sin^2 x\le \sin^2 x.$$

So

$$\cos^2 x-2\sin x\cos x\le 0.$$

Factor:

$$\cos x(\cos x-2\sin x)\le 0.$$

On $[0,\pi/2]$, $\cos x\ge 0$, hence we need

$$\cos x-2\sin x\le 0 \quad\Longrightarrow\quad \tan x\ge \frac12.$$

Thus the valid interval is

$$x\in [\alpha,\pi/2], \quad \text{where } \alpha=\tan^{-1}\left(\frac12\right).$$

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3. Break according to the absolute value

The sign of $\cos x-\sin x$ changes at

$$\cos x=\sin x \iff x=\frac\pi4.$$

So:

• For $x\in[\alpha,\pi/4]$, $\cos x\ge \sin x$, hence $|\cos x-\sin x|=\cos x-\sin x.$
• For $x\in[\pi/4,\pi/2]$, $\sin x\ge \cos x$, hence $|\cos x-\sin x|=\sin x-\cos x.$

Therefore area is

$$\int_{\alpha}^{\pi/4}\bigl[\sin x-(\cos x-\sin x)\bigr]dx + \int_{\pi/4}^{\pi/2}\bigl[\sin x-(\sin x-\cos x)\bigr]dx.$$

Simplify integrands:

$$=\int_{\alpha}^{\pi/4}(2\sin x-\cos x)dx +\int_{\pi/4}^{\pi/2}\cos x\,dx.$$

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4. Evaluate the integrals

First integral:

$$\int (2\sin x-\cos x)dx=-2\cos x-\sin x.$$

So

$$I_1= \left[-2\cos x-\sin x\right]_{\alpha}^{\pi/4}.$$

Now

$$\sin\frac\pi4=\cos\frac\pi4=\frac1{\sqrt2}.$$

Also, if $\tan\alpha=\frac12$, then from a right triangle,

$$\sin\alpha=\frac1{\sqrt5}, \qquad \cos\alpha=\frac2{\sqrt5}.$$

Hence

$$I_1= \left(-2\cdot \frac1{\sqrt2}-\frac1{\sqrt2}\right) - \left(-2\cdot \frac2{\sqrt5}-\frac1{\sqrt5}\right).$$

So

$$I_1=-\frac{3}{\sqrt2}+\frac{5}{\sqrt5} = -\frac{3}{\sqrt2}+\sqrt5.$$

Second integral:

$$I_2=\int_{\pi/4}^{\pi/2}\cos x\,dx =\left[\sin x\right]_{\pi/4}^{\pi/2} =1-\frac1{\sqrt2}.$$

Thus total area is

$$I_1+I_2= \left(\sqrt5-\frac{3}{\sqrt2}\right)+\left(1-\frac1{\sqrt2}\right).$$

So

$$\text{Area}=\sqrt5+1-\frac{4}{\sqrt2}.$$

Since

$$\frac{4}{\sqrt2}=2\sqrt2,$$

we get

$$\boxed{\sqrt5-2\sqrt2+1}.$$

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5. Match with options

$$\sqrt5-2\sqrt2+1$$

corresponds to Option C.

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6. Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.