MCQ
$\int_0^\pi {\frac{{xdx}}{{1 + \sin x}}} $ is equal to
  • A
    $ - \pi $
  • B
    $\frac{\pi }{2}$
  • $\pi $
  • D
    None of these

Answer

Correct option: C.
$\pi $
c
(c) Let, $I = \int_0^\pi {\frac{{xdx}}{{1 + \sin x}}} $ ....$(i)$

$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 $.

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