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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?

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    $\begingroup$ Hint: in the domain $\left[0,\frac{\pi}{2}\right]$ you have $1 \le 2-\sin(x)\le 2$. $\endgroup$ – Galc127 May 2 '16 at 16:46
  • $\begingroup$ What is $\int_0^{\frac{\pi}{2}}\frac{1}{2}x^2\ dx$? So what bounds to you need on $\frac{1}{2-\sin x}$ to get the result? $\endgroup$ – almagest May 2 '16 at 16:46
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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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Since $2-\sin x\in[1,2]$ for $x\in [0,\frac\pi2]$, we have $$ \frac12\int_0^{\pi/2}x^2\,dx \le \int_0^{\pi/2}\frac{x^2}{2-\sin x}\,dx \le \int_0^{\pi/2}x^2\,dx. $$ Computing the integral of $x^2$ gives you the desired inequalities.

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Hint: For $0 \leq x \leq \frac{\pi}{2}$, $0 \leq \sin(x) \leq 1$.

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Notice that $\frac{1}{2} ~\leq ~\frac{1}{2-\sin(x)}~\leq ~1$ for $x \in [0, \pi/2]$.

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The left inequality you'll get by removing $\sin(x)$ from the denominator. The right inequality you'll get by putting $1$ in the denominator.

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