<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] Call the given integral as I, then $$I=\int_{\pi/6}^{\pi/2} \frac{x}{\sin x} dx, \mbox{use}~ x=2y~ ~ \mbox{ and get} ~ I= \int_{\pi/12}^{\pi/4} ~\frac{2y}{\sin 2y} 2 dy= 2 \int_{\pi/12}^{\pi/4} \frac{y}{\sin y \cos y} dy.$$Next, let $y=\tan^{-1} z$. then $$I=2 \int_{2-\sqrt{3}}^{1} \frac{\tan^{-1}z}{z} dz < 2(\sqrt{3}-1) <\frac{\pi^2}{6}~~~ (\tan^{-1}z <z, z>0, ~~~\mbox{used here}).$$
Note that $2(\sqrt{3}-1)$ is a better bound to this integral.
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