int_0^pi frac{x tan x}{sec x cdot operatornam | vedclass

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(A) Let $I = \int_0^\pi \frac{x 	an x}{\sec x \cdot \operatorname{cosec} x} d x$  $(i)$ Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,

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$\pi^2 / 4$

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(A) Let $I = \int_0^\pi \frac{x 	an x}{\sec x \cdot \operatorname{cosec} x} d x$  $(i)$ Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we have: $I = \int_0^\pi \frac{(\pi-x) 	an(\pi-x)}{\sec(\pi-x) \operatorname{cosec}(\pi-x)} d x$ Since $	an(\pi-x) = -	an x$,$\sec(\pi-x) = -\sec x$,and $\operatorname{cosec}(\pi-x) = \operatorname{cosec} x$,we get: $I = \int_0^\pi \frac{(\pi-x)(-	an x)}{(-\sec x)(\operatorname{cosec} x)} d x = \int_0^\pi \frac{(\pi-x) 	an x}{\sec x \operatorname{cosec} x} d x$  (ii) Adding $(i)$ and (ii): $2I = \int_0^\pi \frac{x 	an x + (\pi-x) 	an x}{\sec x \operatorname{cosec} x} d x = \int_0^\pi \frac{\pi 	an x}{\sec x \operatorname{cosec} x} d x$ Since $\frac{	an x}{\sec x \operatorname{cosec} x} = \frac{\sin x / \cos x}{(1 / \cos x)(1 / \sin x)} = \sin^2 x$: $2I = \pi \int_0^\pi \sin^2 x d x = \pi \int_0^\pi \frac{1 - \cos 2x}{2} d x$ $2I = \frac{\pi}{2} \left[ x - \frac{\sin 2x}{2} ight]_0^\pi = \frac{\pi}{2} [(\pi - 0) - (0 - 0)] = \frac{\pi^2}{2}$ $I = \frac{\pi^2}{4}$

$\int_0^\pi \frac{x \tan x}{\sec x \cdot \operatorname{cosec} x} d x$ is equal to

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