_ ^ dx 1 - x = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{1 - \sin x}}} = $

A
$x + \cos x + c$
B
$1 + \sin x + c$
C
$\sec x - \tan x + c$
✓
$\sec x + \tan x + c$

✓

Answer

Correct option: D.

$\sec x + \tan x + c$

d
(d) $\int_{}^{} {\frac{{dx}}{{1 - \sin x}} = \int_{}^{} {\frac{{(1 + \sin x)}}{{1 - {{\sin }^2}x}}\,dx} } $

$ = \int_{}^{} {{{\sec }^2}x\,dx + \int_{}^{} {\tan x\,.\,\sec x\,dx} } $$ = \tan x + \sec x + c$.

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