$\int_0^\pi \frac{x an x}{\sec x+\cos x} \mathrm{~d} x=$

Question

$\int_0^\pi \frac{x 	an x}{\sec x+\cos x} \mathrm{~d} x=$

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Let } \mathrm{I}=\int_0^\pi \frac{x 	an x}{\sec x+\cos x} \mathrm{~d} x \quad \ldots 	ext { (i) } \ & 	herefore \quad \mathrm{I}=\int_0^\pi \frac{(\pi-x) 	an x}{\sec x+\cos x} \mathrm{~d} x \ldots 	ext { (ii) } \ & \ldots\left[\because \int_0^{\mathrm{a}} \mathrm{f}(x) \mathrm{d} x=\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{a}-x) \mathrm{d} xight]\end{aligned}$ Adding (i) and (ii), we get $\begin{aligned} & 2 \mathrm{I}=\pi \int_0^\pi \frac{	an x}{\sec x+\cos x} \mathrm{~d} x \ & \Rightarrow \mathrm{I}=\frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\cos ^2 x} \mathrm{~d} x \end{aligned}$ Put $\cos x=\mathrm{t} \Rightarrow \sin x \mathrm{~d} x=-\mathrm{dt}$ $\begin{aligned} 	herefore \quad \mathrm{I} & =-\frac{\pi}{2} \int_1^{-1} \frac{\mathrm{dt}}{1+\mathrm{t}^2} \ & =-\frac{\pi}{2}\left[	an ^{-1} \mathrm{t}ight]_1^{-1} \ & =\left(-\frac{\pi}{2}ight)\left(-\frac{\pi}{2}ight)=\frac{\pi^2}{4} \end{aligned}$

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

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