I've been given to following inequality to prove:
(The hint given was not to evaluate the integral)
\begin{equation*} \frac{1}{4} \leq \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{sin(x)}{x}dx\leq \frac{\sqrt{3}}{2}. \end{equation*}
(I see that $\frac{\sqrt{3}}{2}$ is $sin(\frac{\pi}{3})$ and I have been trying to prove it using the comparative properties of the Integral)
As the function $\dfrac{\sin x}x$ decreases on every interval on which $\sin x$ and $x$ have the same sign, we can have better bounds: $$ \frac{\sqrt 3}4=\frac3\pi \sin\frac\pi3\Bigl(\frac\pi3-\frac\pi6\Bigr)\le\int_{\tfrac{\pi}{6}}^{\tfrac{\pi}{3}}\frac{\sin x}{x}\,\mathrm d\mkern 1mu x\le \frac6\pi \sin\frac\pi6\Bigl(\frac\pi3-\frac\pi6\Bigr)=\frac12 $$