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

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Question

(A) ધારો કે $I = \int_0^\pi \frac{x 	an x}{\sec x \cdot \operatorname{cosec} x} d x$  $(i)$ ગુણધર્મ $\int_0^a f(x) d x = \int_0^a f(a-x) d x$ નો ઉપયો

Vedclass · Accepted Answer

$\pi^2 / 4$

Vedclass · Answer

(A) ધારો કે $I = \int_0^\pi \frac{x 	an x}{\sec x \cdot \operatorname{cosec} x} d x$  $(i)$ ગુણધર્મ $\int_0^a f(x) d x = \int_0^a f(a-x) d x$ નો ઉપયોગ કરતા: $I = \int_0^\pi \frac{(\pi-x) 	an(\pi-x)}{\sec(\pi-x) \operatorname{cosec}(\pi-x)} d x$ કારણ કે $	an(\pi-x) = -	an x$,$\sec(\pi-x) = -\sec x$,અને $\operatorname{cosec}(\pi-x) = \operatorname{cosec} x$,તેથી: $I = \int_0^\pi \frac{(\pi-x)(-	an x)}{(-\sec x)(\operatorname{cosec} x)} d x = \int_0^\pi \frac{(\pi-x) 	an x}{\sec x \operatorname{cosec} x} d x$  (ii) $(i)$ અને (ii) નો સરવાળો કરતા: $2I = \int_0^\pi \frac{x 	an x + (\pi-x) 	an x}{\sec x \operatorname{cosec} x} d x = \int_0^\pi \frac{\pi 	an x}{\sec x \operatorname{cosec} x} d x$ કારણ કે $\frac{	an x}{\sec x \operatorname{cosec} x} = \sin^2 x$ હોવાથી: $2I = \pi \int_0^\pi \sin^2 x d x = \pi \int_0^\pi \frac{1 - \cos 2x}{2} d x$ $2I = \frac{\pi}{2} \left[ x - \frac{\sin 2x}{2} ight]_0^\pi = \frac{\pi}{2} [(\pi - 0) - (0 - 0)] = \frac{\pi^2}{2}$ $I = \frac{\pi^2}{4}$

$\int_0^\pi \frac{x \tan x}{\sec x \cdot \operatorname{cosec} x} d x$ ની કિંમત શોધો.

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