STD 12 - 7.2 definite integral — JEE कक्षा 11 साइन्स — Question

$I = \int_0^\pi {\frac{{(\pi - x)dx}}{{1 + \sin (\pi - x)}}} $

$I = \int_0^\pi {\frac{{(\pi - x)dx}}{{1 + \sin x}}} $….. $(ii),$

$\left\{ {\because \,\int_0^a {f(x)\,dx}  = \int_0^a {f(a - x)\,dx} } \right\}\,$

$(i)$ व $(ii)$ को जोड़ने पर,

$2I = \int_0^\pi {\frac{{\pi \,dx}}{{1 + \sin x}}} $

$2I = \pi \int_0^\pi {\frac{{1 - \sin x}}{{(1 + \sin x)(1 - \sin x)}}dx} $

$2I = \pi \int_0^\pi {\frac{{1 - \sin x}}{{{{\cos }^2}x}}} dx = \pi \int_0^\pi {({{\sec }^2}x - \sec x\tan x)dx} $

$2I = \pi [\tan x - \sec x]_0^\pi = \pi [0 - ( - 1) - (0 - 1)]$,$2I = 2\pi $

$\therefore$  $I = \pi $.