Answer
$$0 \leq \int_{\pi / 4}^{\pi / 2} f(x) d x \leq \frac{1}{\sqrt{2}}$$
Work Step by Step
Since $\frac{d}{dx}\frac{\sin x}{x}=\frac{x \cos x-\sin x}{x^{2}}<0$ on $[\pi/4,\pi/2]$, then $\cos x $ is decreasing and $$f(\pi/2)\leq \frac{\sin x}{x} \leq f(\pi/4) $$
Hence, by the comparison theorem
\begin{aligned}
& \int_{\pi / 4}^{\pi / 2} f(x) d x \leq \int_{\pi / 4}^{\pi / 2} f\left(\frac{\pi}{4}\right) d x \\
0 \leq & \int_{\pi / 4}^{\pi / 2} f(x) d x \leq \int_{\pi / 4}^{\pi / 2} \frac{\sin (\pi / 4)}{\pi / 4} d x \\
0 & \leq \int_{\pi / 4}^{\pi / 2} f(x) d x \leq \int_{\pi / 4}^{\pi / 2} \frac{1 / \sqrt{2}}{\pi / 4} d x \\
0 & \leq \int_{\pi / 4}^{\pi / 2} f(x) d x \leq \frac{4}{\pi \sqrt{2}}\left(\frac{\pi}{2}-\frac{\pi}{4}\right) \\
0 & \leq \int_{\pi / 4}^{\pi / 2} f(x) d x \leq \frac{4}{\pi \sqrt{2}}\left(\frac{\pi}{4}\right) \\
0 & \leq \int_{\pi / 4}^{\pi / 2} f(x) d x \leq \frac{1}{\sqrt{2}}
\end{aligned}
