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

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(D) Let $I = \int_0^\pi \frac{x 	an x}{\sec x + 	an x} \,dx$. 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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$\pi \left( \frac{\pi}{2} - 1 \right)$

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(D) Let $I = \int_0^\pi \frac{x 	an x}{\sec x + 	an x} \,dx$. Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$,we get: $I = \int_0^\pi \frac{(\pi - x) 	an(\pi - x)}{\sec(\pi - x) + 	an(\pi - x)} \,dx = \int_0^\pi \frac{(\pi - x)(-	an x)}{-\sec x - 	an x} \,dx = \int_0^\pi \frac{(\pi - x) 	an x}{\sec x + 	an x} \,dx$. Adding the two expressions for $I$: $2I = \int_0^\pi \frac{\pi 	an x}{\sec x + 	an x} \,dx = \pi \int_0^\pi \frac{\sin x}{1 + \sin x} \,dx$. $2I = \pi \int_0^\pi \frac{1 + \sin x - 1}{1 + \sin x} \,dx = \pi \left( \int_0^\pi 1 \,dx - \int_0^\pi \frac{1}{1 + \sin x} \,dx ight)$. Since $\int_0^\pi \frac{1}{1 + \sin x} \,dx = \int_0^\pi \frac{1 - \sin x}{\cos^2 x} \,dx = \int_0^\pi (\sec^2 x - \sec x 	an x) \,dx = [	an x - \sec x]_0^\pi = (0 - (-1)) - (0 - 1) = 2$. Thus,$2I = \pi (\pi - 2) = \pi^2 - 2\pi$. Therefore,$I = \frac{\pi^2}{2} - \pi = \pi \left( \frac{\pi}{2} - 1 ight)$.

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

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