Estimation the integration of $\frac{x}{\sin (x)} $ How to prove that integral of $\frac{x}{\sin (x)} $ between $0$ to $\frac{\pi}{2}$ lies within the interval $\frac{\pi^2}{4}$ and $\frac{\pi}{2}$ ?

$$f'(x)=\frac{\sin x-x\cos x}{\sin^2x}=\frac{\cos x(\tan x-x)}{\sin^2x}$$
The function is increasing in the given domain since $\tan x>x$ for $x\in\left(0,\frac{\pi}{2}\right)$.
When $x\to 0$, $\frac{x}{\sin x}=1$ and at $x=\frac{\pi}{2}$, $\frac{x}{\sin x}=\frac{\pi}{2}$.
The minimum area is the area of rectangle with sides $1,\frac{\pi}{2}$.
The maximum area is the area of a rectangle with sides $\frac{\pi}{2},\frac{\pi}{2}$.
