3. Given the function $f(x)=\sin \omega x+\cos \omega x(\omega>0), x \in \mathbf{R}$, 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 the value of $\omega$ is
Given $f(x)=\sin \left(\omega x+\frac{\pi}{4}\right)$, according to the problem, $\omega^{2}+\frac{\pi}{4}=k \pi+\frac{\pi}{2} \Rightarrow \omega=\sqrt{k \pi+\frac{\pi}{4}}, k \in \mathbf{N}$. Also, $\left(-\omega^{2}+\frac{\pi}{4}, \omega^{2}+\frac{\pi}{4}\right) \subseteq\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \Rightarrow k=0$, so $\omega=\frac{\sqrt{\pi}}{2}$.