_0^ dx 1 + x = - Vidyadip

MCQ

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

A
$0$
B
$\frac{1}{2}$
✓
$2$
D
$\frac{3}{2}$

✓

Answer

Correct option: C.

$2$

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

$ = [\tan x - \sec x]_0^\pi = [\tan \pi - \sec \pi + 1] $

$= [0 + 1 + 1] = 2$.

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