The area of the region $$\{(x,y): y \leq \pi - |x|, \; y \leq |x \sin x|, \; y \geq 0\}$$ is :

The region is defined by the three simultaneous inequalities
  $$y \ge 0,$$
  $$y \le \pi-|x|,$$
  $$y \le |x\sin x|.$$

For any fixed $$x$$ the upper boundary is the smaller of the two curves
  $$f_1(x)=\pi-|x| \qquad\text{and}\qquad f_2(x)=|x\sin x|.$$

Both $$f_1(x)$$ and $$f_2(x)$$ are even functions of $$x$$, so the required area is twice the area in the right half-plane $$x\ge 0$$.

For $$x\ge 0$$ the equations simplify to
  $$f_1(x)=\pi-x, \qquad f_2(x)=x\sin x \quad(0\le x\le\pi).$$

Set the two curves equal to locate their intersection(s):
  $$\pi-x = x\sin x \;\Longrightarrow\; \pi = x(1+\sin x).$$

This equation is satisfied at
  $$x=0,\qquad x=\frac{\pi}{2},\qquad x=\pi.$$ Ignoring the trivial point $$x=0$$ (where $$f_2=0$$) we note the non-trivial intersection at $$x=\frac{\pi}{2}$$ and the common zero at $$x=\pi$$.

Test values show
  $$f_2(x) \le f_1(x)\quad\text{for}\; 0\le x\le\frac{\pi}{2},$$
  $$f_1(x) \le f_2(x)\quad\text{for}\; \frac{\pi}{2}\le x\le\pi.$$
Hence on $$[0,\pi/2]$$ the upper boundary is $$y=x\sin x$$; on $$[\pi/2,\pi]$$ it is $$y=\pi-x$$.

Area in the first quadrant (call it $$A_{\text{half}}$$):
$$ A_{\text{half}} =\int_{0}^{\pi/2}x\sin x\,dx +\int_{\pi/2}^{\pi}(\pi-x)\,dx. $$

First integral (integration by parts):
$$ \int_{0}^{\pi/2}x\sin x\,dx = -x\cos x\Big|_{0}^{\pi/2}+\int_{0}^{\pi/2}\cos x\,dx = (0-0)+\left[\sin x\right]_{0}^{\pi/2} =1. $$

Second integral:
$$ \int_{\pi/2}^{\pi}(\pi-x)\,dx = \left[\pi x-\frac{x^{2}}{2}\right]_{\pi/2}^{\pi} = \left(\frac{\pi^{2}}{2}\right)-\left(\frac{3\pi^{2}}{8}\right) = \frac{\pi^{2}}{8}. $$

Thus
$$ A_{\text{half}} = 1+\frac{\pi^{2}}{8}. $$

By symmetry the total area is
$$ A = 2A_{\text{half}} = 2\Bigl(1+\frac{\pi^{2}}{8}\Bigr) = 2+\frac{\pi^{2}}{4}. $$

Therefore, the required area equals $$2+\dfrac{\pi^{2}}{4}$$.
Option B which is: $$2 + \dfrac{\pi^2}{4}$$.