Why $\dfrac{\sqrt{3}}{8}+\dfrac{1}{10}\leq\displaystyle\int_{\frac{\pi}{2}}^{\pi}\dfrac{\sin x}{x}\ \mathrm{d}x$ ?
2026-03-28 11:29:46.1774697386
An inequality for $\int_{\frac{\pi}{2}}^{\pi}\frac{\sin x}{x}\ \mathrm{d}x$
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$$\frac{\sqrt3}{8}+\frac{1}{10}<\frac{1}{\pi}=\int_{\frac{\pi}{2}}^{\pi} \frac{\sin x}{\pi}dx<\int_{\frac{\pi}{2}}^{\pi} \frac{\sin x}{x}dx$$