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

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

Vedclass · Accepted Answer

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

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(D) Let $I = \int_0^\pi \frac{x 	an x}{\sec x+	an x} d x \quad \dots(i)$Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,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:$I = \int_0^\pi \frac{-(\pi-x) 	an x}{-\sec x - 	an x} d x = \int_0^\pi \frac{(\pi-x) 	an x}{\sec x + 	an x} d x \quad \dots(ii)$Adding $(i)$ and $(ii)$:$2I = \int_0^\pi \frac{\pi 	an x}{\sec x + 	an x} d x = \pi \int_0^\pi \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} d x$$2I = \pi \int_0^\pi \frac{\sin x}{1 + \sin x} d x = \pi \int_0^\pi \frac{1 + \sin x - 1}{1 + \sin x} d x$$2I = \pi \left[ \int_0^\pi 1 d x - \int_0^\pi \frac{1}{1 + \sin x} d x ight]$$2I = \pi \left[ [x]_0^\pi - \int_0^\pi \frac{1 - \sin x}{\cos^2 x} d x ight]$$2I = \pi \left[ \pi - \int_0^\pi (\sec^2 x - \sec x 	an x) d x ight]$$2I = \pi \left[ \pi - [	an x - \sec x]_0^\pi ight]$$2I = \pi \left[ \pi - ((	an \pi - \sec \pi) - (	an 0 - \sec 0)) ight]$$2I = \pi \left[ \pi - ((0 - (-1)) - (0 - 1)) ight] = \pi [\pi - (1 + 1)] = \pi(\pi - 2)$$I = \frac{\pi(\pi - 2)}{2}$

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

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