\(\int_0^\pi \frac{x \tan x}{\sec x \cdot \operatorname{cosec} x} d x\) is equals to

Let \(I=\int_0^\pi \frac{x \tan x}{\sec x \cdot \operatorname{cosec} x} d x\)
\(\begin{aligned} \text { Using property } I= & =\int_0^\pi f(x) d x \\ & =\int_0^\pi f(0+\pi-x) d x\end{aligned}\)
Then, \(\begin{aligned} I & =\int_0^\pi \frac{(\pi-x) \tan (\pi-x)}{\sec (\pi-x) \operatorname{cosec}(\pi-x)} d x \\ I & =\int_0^\pi \frac{(\pi-x) \tan x}{\sec x \operatorname{cosec} x} d x\end{aligned}\)
On adding Eqs. (i) and (ii), we get
\(\begin{aligned} & 2 I=\int_0^\pi \frac{\tan x(x+\pi-x)}{\sec x \cdot \operatorname{cosec} x} d x \\ & 2 I=\int_0^\pi \frac{\pi \tan x}{\sec x \operatorname{cosec} x} d x \\ & 2 I=\pi \int_0^\pi \sin ^2 x d x \\ & 2 I=\pi \int_0^\pi \frac{1-\cos 2 x}{2} d x \\ & 2 I=\pi\left[\frac{x}{2}-\frac{\sin 2 x}{4}\right]_0^\pi \\ & 2 I=\pi\left[\frac{\pi}{2}-0\right]\end{aligned}\)
\(I=\frac{\pi^2}{4}\)