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 $\frac{\pi}{2}$ ?
 A: Just another path. One may observe that
$$
\left(\log\left(\tan \left(\frac{x}{2}\right)\right)\right)'=\frac1{\sin x}
$$ then, integrating by parts gives
$$
\begin{align}
\int_0^{\pi/2}\frac{x}{\sin x}\:dx&=\left[x\log\left(\tan \left(\frac{x}{2}\right)\right)\right]_0^{\pi/2}-\int_0^{\pi/2}\log\left(\tan \left(\frac{x}{2}\right)\right)\:dx
\\\\&=0-2\int_0^{\pi/4}\log\left(\tan u\right)\:du
\\\\&=-2\int_0^1\frac{\log\left(v\right)}{1+v^2}\:dv
\\\\&=-2\int_0^1\sum_{n=0}^\infty (-1)^nv^{2n}\log\left(v\right)\:dv
\\\\&=2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2} \quad (\text{Catalan's constant}).
\end{align}
$$ We have an alternating series with a general term decreasing to  $0$, thus
$$
2\left(1-\frac19 \right)<2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}<2\left(1-\frac19+\frac1{16} \right),
$$ giving

$$
\frac{\pi}2<\color{red}{\frac{16}9}<\int_0^{\pi/2}\frac{x}{\sin x}\:dx<\color{red}{\frac{137}{72}}<\frac{\pi^2}4.
$$

A: 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,
$$\int_{0}^{ \frac{\pi}{2}} 1 < \int_{0}^{ \frac{\pi}{2}} \frac{x}{\sin x} \leq \int_{0}^{ \frac{\pi}{2}} \frac{\pi}{2}$$
A: Recall from elementary geometry that the sine function satisfies the inequalities
$$ \frac{2}{\pi}x \le \sin(x)\le x \tag 1$$
for $0\le x\le \pi/2$.  
We see immediately from $(1)$ that 
$$1 \le \frac{x}{\sin(x)}\le \frac{\pi}{2} \tag 2$$
for $0<x\le \pi/2$ from which the inequality of interest is trivial.  
A: $$f'(x)=\frac{\sin x-x\cos x}{\sin^2x}=\frac{\cos x(\tan x-x)}{\sin^2x}$$
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 $\frac{\pi}{2},\frac{\pi}{2}$.
