int_0^pi frac{x tan x}{sec x+tan x} d x is eq | vedclass

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

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$\frac{\pi(\pi-2)}{2}$

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(A) Let $I = \int_0^\pi \frac{x 	an x}{\sec x+	an x} d x \quad ...(i)$ Using the property $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$,we get: $I = \int_0^\pi \frac{(\pi-x) 	an(\pi-x)}{\sec(\pi-x)+	an(\pi-x)} d x$ Since $	an(\pi-x) = -	an x$ and $\sec(\pi-x) = -\sec x$,we have: $I = \int_0^\pi \frac{(\pi-x)(-	an x)}{-\sec x-	an x} d x = \int_0^\pi \frac{(\pi-x) 	an x}{\sec x+	an x} d x \quad ...(ii)$ Adding $(i)$ and $(ii)$: $2I = \int_0^\pi \frac{x 	an x + (\pi-x) 	an x}{\sec x+	an x} d x = \int_0^\pi \frac{\pi 	an x}{\sec x+	an x} d x$ $I = \frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\sin x} d x$ Multiply numerator and denominator by $(1-\sin x)$: $I = \frac{\pi}{2} \int_0^\pi \frac{\sin x(1-\sin x)}{1-\sin^2 x} d x = \frac{\pi}{2} \int_0^\pi \frac{\sin x - \sin^2 x}{\cos^2 x} d x$ $I = \frac{\pi}{2} \int_0^\pi (\sec x 	an x - 	an^2 x) d x = \frac{\pi}{2} \int_0^\pi (\sec x 	an x - (\sec^2 x - 1)) d x$ $I = \frac{\pi}{2} [\sec x - 	an x + x]_0^\pi$ $I = \frac{\pi}{2} [(\sec \pi - 	an \pi + \pi) - (\sec 0 - 	an 0 + 0)]$ $I = \frac{\pi}{2} [(-1 - 0 + \pi) - (1 - 0 + 0)] = \frac{\pi}{2} (\pi - 2)$

$\int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x$ is equal to

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