<s>[INST] Prove that $\frac{\pi^3}{48} \le \int_0^{\pi/2}\frac{x^2}{2-\sin(x)}\,dx \le \frac{\pi^3}{24}$ Is it possible to prove that 
$$\frac{\pi^3}{48} \le \int_0^{\pi/2}\frac{x^2}{2-\sin(x)}\,dx \le \frac{\pi^3}{24}$$
without evaluating the integral?
 [/INST] Without evaluating this integral:
Note $1\le 2-\sin x\le 2$ on $\;[0,\frac\pi2]$, hence $\;\dfrac{x^2}2\le \dfrac{x^2}{2-\sin x}\le x^2 $, and by the positivity of the integral:
$$\frac{\pi^3}{48}=\int_0^\tfrac\pi2\dfrac{x^2}2\,\mathrm d\mkern1mu x\le \int_0^\tfrac\pi2\dfrac{x^2}{2-\sin x}\,\mathrm d\mkern1mu x\le \int_0^\tfrac\pi2 x^2\,\mathrm d\mkern1mu x=\frac{\pi^3}{24}.$$
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