$\int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x$ is equal to
$\int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x$ is equal to
$\frac{\pi(\pi-2)}{2}$
$\frac{\pi+2}{2}$
$\frac{\pi(\pi+2)}{2}$
$\frac{\pi-2}{2}$
Solution
$\int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x$
Let $I=\int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x...(i)$
$\begin{aligned} \because \quad \int_a^b f(x) d x & =\int_a^b f(a+b-x) d x \\ \Rightarrow \quad I & =\int_0^\pi \frac{(\pi-x) \tan (\pi-x)}{\sec (\pi-x)+\tan (\pi-x)} d x \\ I & =\int_0^\pi \frac{(\pi-x)(-\tan x)}{-\sec x-\tan x} d x \\ I & =\int_0^\pi \frac{(\pi-x) \tan x}{\sec x+\tan x} d x...(ii)\end{aligned}$
Adding Eqs. (i) and (ii), we get
$
\begin{aligned}
2 I & =\int_0^\pi \frac{x \tan x+(\pi-x) \tan x}{\sec x+\tan x} d x \\
2 I & =\int_0^\pi \frac{\pi \tan x}{\sec x+\tan x} d x \\
I & =\frac{\pi}{2} \int_0^\pi \frac{\tan x}{\sec x+\tan x} d x=\frac{\pi}{2} \int_0^\pi \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}} \\
& =\frac{\pi}{2} \int_0^\pi\left(\frac{\sin x}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}\right) d x \\
& =\frac{\pi}{2} \int_0^\pi \frac{\sin x(1-\sin x)}{1-\sin ^2 x} d x \\
& =\frac{\pi}{2} \int_0^\pi \frac{\sin x-\sin ^2 x}{\cos ^2 x} d x
\end{aligned}
$
$
\begin{aligned}
I= & \frac{\pi}{2} \int_0^\pi\left(\frac{\sin x}{\cos ^2 x}-\frac{\sin ^2 x}{\cos ^2 x}\right) d x \\
& \frac{\pi}{2} \int_0^\pi\left(\sec x \tan x-\tan ^2 x\right) d x \\
& \frac{\pi}{2} \int_0^\pi\left\{\sec x \tan x-\left(\sec ^2 x-1\right)\right\} d x
\end{aligned}
$
$\begin{aligned} I & =\frac{\pi}{2} \int_0^\pi\left(\sec x \tan x-\sec ^2 x+1\right) d x \\ & =\frac{\pi}{2}[\sec x-\tan x+x]_0^\pi \\ & =\frac{\pi}{2}\left\{(\sec \pi-\tan \pi+\pi)-\frac{\pi}{2}\right. \\ & =\frac{\pi}{2}\{(-1-0+\pi)-(1-0+0)\} \\ & =\frac{\pi}{2}(\pi-1-1)=\frac{\pi}{2}(\pi-2) \\ I & =\frac{\pi(\pi-2)}{2}\end{aligned}$