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 $\frac{\pi}{2}$ ?

The function is:
$$f(x)=x\csc x$$
Using the Taylor series of $\csc x$ with radius $\pi$,meaning it converges for $x \in (-\pi,0) \cup (0,\pi) \supseteq (0,\frac{\pi}{2}]$:
$$f(x)=x(\frac{1}{x}+\frac{1}{6}x+....)$$
$$=1+\frac{1}{6}x^2+...$$
In the specified interval,
$$1 < \frac{x}{\sin x}$$
Also the Taylor series expansion shows us that our function will be at maximum for the highest value of $x$ in our interval, that is $x=\frac{\pi}{2} $. So,
$$\frac{\pi}{2} \geq  \frac{x}{\sin x}$$
Thus,
$$\int_{0}^{ \frac{\pi}{2}} 1 < \int_{0}^{ \frac{\pi}{2}} \frac{x}{\sin x} \leq \int_{0}^{ \frac{\pi}{2}} \frac{\pi}{2}$$
