\(\int_0^\pi \frac{x \tan x}{\sec x+\cos x} \mathrm{~d} x=\)

\( \text { Let } \mathrm{I}=\int_0^\pi \frac{x \tan x}{\sec x+\cos x} \mathrm{~d} x \quad \ldots \text { (i) } \)
\( \therefore \mathrm{I}=\int_0^\pi \frac{(\pi-x) \tan x}{\sec x+\cos x} \mathrm{~d} x \ldots \text { (ii) } \)
\( \ldots\left[\because \int_0^{\mathrm{a}} \mathrm{f}(x) \mathrm{d} x=\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{a}-x) \mathrm{d} x\right]\)
Adding (i) and (ii), we get
\( 2 \mathrm{I}=\pi \int_0^\pi \frac{\tan x}{\sec x+\cos x} \mathrm{~d} x \)
\( \Rightarrow \mathrm{I}=\frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\cos ^2 x} \mathrm{~d} x\)
Put \(\cos x=\mathrm{t} \Rightarrow \sin x \mathrm{~d} x=-\mathrm{dt}\)
\(\therefore \mathrm{I} =-\frac{\pi}{2} \int_1^{-1} \frac{\mathrm{dt}}{1+\mathrm{t}^2} \)
\( =-\frac{\pi}{2}\left[\tan ^{-1} \mathrm{t}\right]_1^{-1} \)
\( =\left(-\frac{\pi}{2}\right)\left(-\frac{\pi}{2}\right)=\frac{\pi^2}{4}\)