7.2 definite integral — Maths STD 12 Science — Question

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

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

$\left\{ \because \,\int_{0}^{a}{f(x)\,dx}=\int_{0}^{a}{f(a-x)\,dx} \right\}\,$

Adding $(i)$ and $(ii),$ we get 

$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 $.