$\int\limits_0^\pi {\cfrac{{x\tan x}}{{\sec x + {\mathop{\rm tanx}\nolimits} }}dx}$

: Let $I = \int\limits_0^\pi {\cfrac{{x\tan x}}{{\sec x + {\mathop{\rm tanx}\nolimits} }}dx}$

…(i)
Also, $I = \int\limits_0^\pi {\cfrac{{\left( {\pi - x} \right)\tan \left( {\pi - x} \right)}}{{\sec \left( {\pi - x} \right) + \tan \left( {\pi - x} \right)}}dx}$

$= \int\limits_0^\pi {\cfrac{{\left( {\pi - x} \right)\left( { - \tan x} \right)}}{{\left( { - \sec x} \right) + \left( { - \tan x} \right)}}dx} = \int\limits_0^\pi {\cfrac{{\left( {\pi - x} \right)\tan \,x}}{{\sec x + \tan x}}} dx$

….(ii)
Adding (i) and (ii),

we get
$2I = \int\limits_{0\,}^\pi {\cfrac{{\left( {x + \pi - x} \right)\tan x}}{{\sec \,x + tan\,x}}} = \pi \int\limits_{0\,}^\pi {\cfrac{{\tan x}}{{\sec \,x + tan\,x}}} dx$

$= \pi \int\limits_0^\pi {\cfrac{{\cfrac{{\sin x}}{{\cos x}}}}{{\cfrac{1}{{\cos x}} + \cfrac{{\sin x}}{{\cos x}}}}dx} = \pi \int\limits_0^\pi {\cfrac{{\sin x}}{{1 + \sin x}}dx}$

$= \pi \int\limits_0^\pi {\cfrac{{\left( {1 + \sin x} \right) - 1}}{{1 + \sin x}}} dx = \pi \int\limits_0^\pi {\left( {1 - \cfrac{1}{{1 + \sin x}}} \right)} dx$

$= \pi \int\limits_0^\pi {\left( 1 \right)dx} - \pi \int\limits_0^\pi {\cfrac{{1 - \sin x}}{{1 - {{\sin }^2}x}}dx} = \pi \left[ x \right]_0^\pi - \pi \int\limits_0^\pi {\cfrac{{1 - \sin x}}{{{{\cos }^2}x}}} dx$

$= \pi \left( {\pi - 0} \right) - \pi \int\limits_0^\pi {\left( {{{\sec }^2}x - \sec x\tan x} \right)dx}$

$= {\pi ^2} - \pi \left[ {\tan x - \sec x} \right]_0^\pi = {\pi ^2} - \pi \left[ {\left( {\tan \pi - \tan 0} \right) - \left( {\sec \pi - \sec 0} \right)} \right]$

$= {\pi ^2} - \pi \left( {0 - 0} \right) + \pi \left( { - 1 - 1} \right) = {\pi ^2} - 2\pi = \pi \left( {\pi - 2} \right)$

$\therefore$ $I = \cfrac{\pi }{2}\left( {\pi - 2} \right)$