1. Required area

The area bounded by $y=\cos x$ and $y=\sin x$ from $x=0$ to $x=\dfrac{3\pi}{2}$ is

$A=\int_0^{3\pi/2} |\cos x-\sin x|\,dx.$

So we must find where $\cos x=\sin x$ in this interval.

2. Points of intersection

$\cos x=\sin x \implies \tan x=1 \implies x=\frac{\pi}{4}+n\pi.$

In $\left[0,\frac{3\pi}{2}\right]$, the solutions are

$x=\frac{\pi}{4},\quad x=\frac{5\pi}{4}.$

These points split the interval into three parts.

3. Check which curve is above

• On $\left[0,\frac{\pi}{4}\right]$, take $x=0$: $\cos 0=1,\ \sin 0=0 \implies \cos x>\sin x.$

• On $\left[\frac{\pi}{4},\frac{5\pi}{4}\right]$, take $x=\frac{\pi}{2}$: $\cos \frac{\pi}{2}=0,\ \sin \frac{\pi}{2}=1 \implies \sin x>\cos x.$

• On $\left[\frac{5\pi}{4},\frac{3\pi}{2}\right]$, take $x=\frac{3\pi}{2}$: $\cos \frac{3\pi}{2}=0,\ \sin \frac{3\pi}{2}=-1 \implies \cos x>\sin x.$

Hence,

$A=\int_0^{\pi/4}(\cos x-\sin x)dx+\int_{\pi/4}^{5\pi/4}(\sin x-\cos x)dx+\int_{5\pi/4}^{3\pi/2}(\cos x-\sin x)dx.$

4. Evaluate each integral

First part

$I_1=\int_0^{\pi/4}(\cos x-\sin x)dx=[\sin x+\cos x]_0^{\pi/4}.$

Now,

$I_1=\left(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\right)-(0+1)=\sqrt2-1.$

Second part

$I_2=\int_{\pi/4}^{5\pi/4}(\sin x-\cos x)dx=[-\cos x-\sin x]_{\pi/4}^{5\pi/4}.$

At $x=\frac{5\pi}{4}$,

$-\cos \frac{5\pi}{4}-\sin \frac{5\pi}{4}= -\left(-\frac{\sqrt2}{2}\right)-\left(-\frac{\sqrt2}{2}\right)=\sqrt2.$

At $x=\frac{\pi}{4}$,

$-\cos \frac{\pi}{4}-\sin \frac{\pi}{4}= -\frac{\sqrt2}{2}-\frac{\sqrt2}{2}=-\sqrt2.$

Therefore,

$I_2=\sqrt2-(-\sqrt2)=2\sqrt2.$

Third part

$I_3=\int_{5\pi/4}^{3\pi/2}(\cos x-\sin x)dx=[\sin x+\cos x]_{5\pi/4}^{3\pi/2}.$

So,

$I_3=( -1+0)-\left(-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}\right)=-1+\sqrt2=\sqrt2-1.$

5. Total area

$A=I_1+I_2+I_3=(\sqrt2-1)+2\sqrt2+(\sqrt2-1)=4\sqrt2-2.$

6. Compare with options

The obtained area is

$\boxed{4\sqrt2-2}$

which matches Option D.