<s>[INST] Proof that $\int_{\pi/6}^{\pi/2} \frac{x}{\sin{x}} \le \frac{\pi^2}{6}$ Proof that $\int_{\pi/6}^{\pi/2} \frac{x}{\sin{x}} \le \frac{\pi^2}{6}$

After few calculations I get that if I take $\frac{3}{2}x$ then after integral I get $\frac{3}{4}x^2+ C$ and 
$$ \int_{\pi/6}^{\pi/2} \frac{3}{4}x^2+ C = \pi^2 / 6$$ so I should show that 
$$\frac{x}{\sin{x}} \le \frac{3x}{2} $$
but last inequality is not true...
 [/INST] The maximum value of the integrand in the given range is when $x=\pi/2$ and the integrand is also $\pi/2$ so we just get that
$$\int_{\pi/6}^{\pi/2}\frac{x}{\sin{(x)}}\mathrm{d}x\le\frac\pi2\left(\frac\pi2-\frac\pi6\right)=\frac{\pi^2}6$$
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