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...

$$(f=)\ \frac{x}{\sin\ x} = \frac{x}{x-x^3/6+\cdots }
=\frac{1}{1-x^2/6+\cdots }=1+Cx^2+\cdots ,\ C>0$$ so that
$\frac{x}{\sin\ x}$ is a positive increasing convex function on
$[{\pi/6}, {\pi/2}]$
By drawing a graph of $f$, the integral $\int\ f(x)\ dx$ is smaller than an area of
trapezoid i.e. $$ \int\ f(x)\ dx \leq \frac{ f( \pi/6 ) + f(\pi/2)
}{2}\cdot [ -\pi/6 + \pi/2 ] = \frac{5\pi^2}{36} $$
