Let $\displaystyle\int_{-2}^{2} (|\sin x| + [x \sin x])\,dx = 2(3 - \cos 2) + \beta$, where $[\cdot]$ is the greatest integer function. Then $\beta \sin\left(\dfrac{\beta}{2}\right)$ equals:

Let $I = \displaystyle\int_{-2}^{2} \left(|\sin x| + [x \sin x]\right) dx = I_1 + I_2$, where
$I_1 = \displaystyle\int_{-2}^{2} |\sin x|\, dx$ and $I_2 = \displaystyle\int_{-2}^{2} [x \sin x]\, dx$.

Evaluating $I_1$:
Since $|\sin x|$ is even, $I_1 = 2\displaystyle\int_{0}^{2} |\sin x|\, dx$.
For $x \in [0, 2]$, $2 < \pi$, so $\sin x \geq 0$ and $|\sin x| = \sin x$.
$I_1 = 2\displaystyle\int_{0}^{2} \sin x\, dx = 2\left[-\cos x\right]_0^2 = 2(1 - \cos 2) = 2 - 2\cos 2$

Evaluating $I_2$:
Let $g(x) = x\sin x$. Then $g(-x) = (-x)\sin(-x) = x\sin x = g(x)$, so $g(x)$ is even, and hence $[x\sin x]$ is also even.
$I_2 = 2\displaystyle\int_{0}^{2} [x\sin x]\, dx$

Behaviour of $y = x\sin x$ on $[0, 2]$:
At $x = 0$, $y = 0$. At $x = 2$, $y = 2\sin 2 \approx 1.818$.
$y' = \sin x + x\cos x > 0$ on $[0, 2]$, so $y$ is strictly increasing on this interval.

Therefore, $x\sin x$ takes each value in $[0, 1.818]$ exactly once, and crosses $1$ at some unique $\alpha \in (0, 2)$ with $\alpha\sin\alpha = 1$.

For $x \in [0, \alpha)$: $0 \leq x\sin x < 1 \Rightarrow [x\sin x] = 0$
For $x \in [\alpha, 2]$: $1 \leq x\sin x < 2 \Rightarrow [x\sin x] = 1$

$I_2 = 2\left(\displaystyle\int_{0}^{\alpha} 0\, dx + \displaystyle\int_{\alpha}^{2} 1\, dx\right) = 2(2 - \alpha) = 4 - 2\alpha$

Combining:
$I = I_1 + I_2 = (2 - 2\cos 2) + (4 - 2\alpha) = 6 - 2\cos 2 - 2\alpha$

Given $I = 2(3 - \cos 2) + \beta = 6 - 2\cos 2 + \beta$, so:
$6 - 2\cos 2 - 2\alpha = 6 - 2\cos 2 + \beta \Rightarrow \beta = -2\alpha$

Now:
$\beta \sin\left(\dfrac{\beta}{2}\right) = -2\alpha \sin\left(\dfrac{-2\alpha}{2}\right) = -2\alpha \sin(-\alpha) = 2\alpha\sin\alpha$

Since $\alpha\sin\alpha = 1$:
$\beta \sin\left(\dfrac{\beta}{2}\right) = 2 \cdot 1 = 2$

Hence, the correct option is $(2)\ 2$.