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

Split the integral into two parts:

$$I=\int_{-2}^{2}\left(|\sin x|+[x\sin x]\right)dx =\underbrace{\int_{-2}^{2}|\sin x|\,dx}_{I_1} \;+\;\underbrace{\int_{-2}^{2}[x\sin x]\,dx}_{I_2}$$

1. Evaluation of $$I_1$$

Since $$|\sin x|$$ is an even function,

$$I_1=2\int_{0}^{2}|\sin x|\,dx$$

On $$0\le x\le 2$$ we have $$\sin x\gt 0$$ (because $$2&lt\pi$$), hence $$|\sin x|=\sin x$$.

$$I_1=2\int_{0}^{2}\sin x\,dx =2\left[-\cos x\right]_{0}^{2} =2(1-\cos 2) =2-2\cos 2$$ $$-(1)$$

2. Evaluation of $$I_2$$

The function $$g(x)=x\sin x$$ is even (product of two odd functions), so

$$I_2=2\int_{0}^{2}[x\sin x]\,dx$$

For $$0\le x\le 2$$ we have $$0\le x\sin x\le 2\sin 2\approx1.8186$$, so $$[x\sin x]$$ can take only the values $$0$$ or $$1$$.

Let $$\alpha$$ be the point in $$(1,2)$$ where

$$\alpha\sin\alpha=1$$ $$-(2)$$

Then

$$[x\sin x]=\begin{cases} 0,&0\le x&lt\alpha\\[2pt] 1,&\alpha\le x\le 2 \end{cases}$$

Hence

$$I_2=2\!\left(\int_{0}^{\alpha}0\,dx+\int_{\alpha}^{2}1\,dx\right) =2(2-\alpha) =4-2\alpha$$ $$-(3)$$

3. Complete value of $$I$$

Adding $$(1)$$ and $$(3)$$:

$$I=(2-2\cos 2)+(4-2\alpha)=6-2\cos 2-2\alpha$$ $$-(4)$$

The question states that

$$I=2(3-\cos 2)+\beta=6-2\cos 2+\beta$$ $$-(5)$$

Comparing $$(4)$$ and $$(5)$$ we get

$$\beta=-2\alpha$$ $$-(6)$$

4. Compute $$\beta\sin\!\left(\dfrac{\beta}{2}\right)$$

Using $$(6)$$:

$$\beta\sin\!\left(\frac{\beta}{2}\right) =(-2\alpha)\sin(-\alpha) =2\alpha\sin\alpha$$

From $$(2)$$, $$\alpha\sin\alpha=1$$, therefore

$$\beta\sin\!\left(\frac{\beta}{2}\right)=2$$

Hence the required value is $$2$$.

Option B which is: $$2$$