1. Let the function $f(x)=x \sin x(x \in \mathbf{R})$ attain an extremum at $x=x_{0}$, then $\left(1+x_{0}^{2}\right)\left(1+\cos 2 x_{0}\right)=$
$\qquad$ .
1. 2 .

It is known that $f^{\prime}\left(x_{0}\right)=\sin x+x \cos x$. Since the function $f(x)$ attains an extremum at $x=x_{0}$, we have $f^{\prime}\left(x_{0}\right)=0$, which means $x_{0}=-\tan x_{0}$. Therefore,
$$
\left(1+x_{0}^{2}\right)\left(1+\cos 2 x_{0}\right)=\left(1+\tan ^{2} x_{0}\right) \cdot 2 \cos ^{2} x_{0}=\sec ^{2} x_{0} \cdot 2 \cos ^{2} x_{0}=2 \text {. }
$$