3. Given the function
$$
f(x)=\sin \omega x+\cos \omega x(\omega>0)(x \in \mathbf{R}) \text {. }
$$

If the function $f(x)$ is monotonically increasing in the interval $(-\omega, \omega)$, and the graph of the function $y=f(x)$ is symmetric about the line $x=\omega$, then $\omega=$ . $\qquad$
3. $\frac{\sqrt{\pi}}{2}$.

Notice,
$$
f(x)=\sin \omega x+\cos \omega x=\sqrt{2} \sin \left(\omega x+\frac{\pi}{4}\right),
$$

and the graph of the function $y=f(x)$ is symmetric about the line $x=\omega$.
$$
\begin{array}{l}
\text { Then } f(\omega)=\sqrt{2} \sin \left(\omega^{2}+\frac{\pi}{4}\right)= \pm \sqrt{2} \\
\Rightarrow \omega^{2}+\frac{\pi}{4}=\frac{\pi}{2}+k \pi(k \in \mathbf{Z}) \\
\Rightarrow \omega^{2}=\frac{\pi}{4}+k \pi(k \in \mathbf{Z}) .
\end{array}
$$

Also, the function $f(x)$ is monotonically increasing in the interval $(-\omega, \omega)$, thus,
$$
\omega^{2}+\frac{\pi}{4} \leqslant \frac{\pi}{2} \Rightarrow \omega^{2} \leqslant \frac{\pi}{4} .
$$

Taking $k=0$, we get $\omega^{2}=\frac{\pi}{4} \Rightarrow \omega=\frac{\sqrt{\pi}}{2}$.