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...
 A: 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
$$\int_{\pi/6}^{\pi/2}\frac{x}{\sin{(x)}}\mathrm{d}x\le\frac\pi2\left(\frac\pi2-\frac\pi6\right)=\frac{\pi^2}6$$
A: 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.
A: $$(f=)\ \frac{x}{\sin\ x} = \frac{x}{x-x^3/6+\cdots }
=\frac{1}{1-x^2/6+\cdots }=1+Cx^2+\cdots ,\ C>0$$ 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. $$ \int\ f(x)\ dx \leq \frac{ f( \pi/6 ) + f(\pi/2)
}{2}\cdot [ -\pi/6 + \pi/2 ] = \frac{5\pi^2}{36} $$
