Answer
$c=0$
Work Step by Step
Given $$y=x \sin x, \quad\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$
Since
$$f'(x)=x \cos x+ \sin x $$
Then by MVT, there exists a constant $c\in (-\pi/2,\pi/2)$ such that
\begin{align*}
f'(c)&= \frac{f(b)-f(a) }{b-a}\\
c \cos c+ \sin c &=\frac{ \pi/2-\pi/2}{\pi}\\
c \cos c+ \sin c &=0
\end{align*}
Then $c=0 \in(-\pi/2,\pi/2)$
