int_0^pi frac{x tan x}{sec x + cos x} ,dx = | vedclass

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(A) Let $I = \int_0^\pi \frac{x 	an x}{\sec x + \cos x} \,dx$. Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$,we get: $I = \int_0^\pi

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$\frac{\pi^2}{4}$

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(A) Let $I = \int_0^\pi \frac{x 	an x}{\sec x + \cos x} \,dx$. Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$,we get: $I = \int_0^\pi \frac{(\pi - x) 	an(\pi - x)}{\sec(\pi - x) + \cos(\pi - x)} \,dx = \int_0^\pi \frac{(\pi - x)(-	an x)}{-\sec x - \cos x} \,dx = \int_0^\pi \frac{(\pi - x) 	an x}{\sec x + \cos x} \,dx$. Adding the two expressions for $I$: $2I = \int_0^\pi \frac{\pi 	an x}{\sec x + \cos x} \,dx = \pi \int_0^\pi \frac{\sin x / \cos x}{1/\cos x + \cos x} \,dx = \pi \int_0^\pi \frac{\sin x}{1 + \cos^2 x} \,dx$. Let $t = \cos x$,then $dt = -\sin x \,dx$. When $x=0, t=1$ and when $x=\pi, t=-1$. $2I = \pi \int_1^{-1} \frac{-dt}{1 + t^2} = \pi \int_{-1}^1 \frac{dt}{1 + t^2} = \pi [	an^{-1} t]_{-1}^1$. $2I = \pi (	an^{-1}(1) - 	an^{-1}(-1)) = \pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = \pi (\frac{\pi}{2}) = \frac{\pi^2}{2}$. Therefore,$I = \frac{\pi^2}{4}$.

$\int_0^\pi \frac{x \tan x}{\sec x + \cos x} \,dx = $

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