Both $|\sin x|$ and $x\sin x$ are even functions, so

${\int }_{-2}^2(|\sin x|+[x\sin x])dx=2{\int }_0^2|\sin x|dx+2{\int }_0^2[x\sin x]dx.$

For $x\in [0,2]\subset [0,\pi ]$, $\sin x\ge 0$, so ${\int }_0^2\sin xdx=1-\cos 2$ and the first piece contributes $2(1-\cos 2)$.

For the greatest integer integral, note $g(x)=x\sin x$ satisfies $g(0)=0$, $g(2)=2\sin 2\approx 1.82$. Let $\alpha \in (0,2)$ be the unique root of $x\sin x=1$. Then $[x\sin x]=0$ on $[0,\alpha ]$ and $[x\sin x]=1$ on $[\alpha ,2]$, so

${\int }_0^2[x\sin x]dx=2-\alpha ,$ contributing $2(2-\alpha )=4-2\alpha$.

Total integral $=2-2\cos 2+4-2\alpha =2(3-\cos 2)-2\alpha$, so $\beta =-2\alpha$ and $\alpha =-\frac{\beta }{2}$.

$\beta \sin \left(\frac{\beta }{2}\right)=(-2\alpha )\sin (-\alpha )=2\alpha \sin \alpha =2\cdot 1=2.$