_ ^ 1 x(1 + x) ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{1}{{\cos x(1 + \cos x)}}} \;dx = $

A
$\log (\sec x + \tan x) + 2\tan \frac{x}{2} + c$
B
$\log (\sec x + \tan x) - 2\tan \frac{x}{2} + c$
C
$\log (\sec x + \tan x) + \tan \frac{x}{2} + c$
✓
$\log (\sec x + \tan x) - \tan \frac{x}{2} + c$

✓

Answer

Correct option: D.

$\log (\sec x + \tan x) - \tan \frac{x}{2} + c$

d
(d)$\int_{}^{} {\frac{1}{{\cos x(1 + \cos x)}}} dx = \int_{}^{} {\frac{{dx}}{{\cos x}} - \int_{}^{} {\frac{{dx}}{{1 + \cos x}}} } $
$ = \int_{}^{} {\sec x\;dx - \frac{1}{2}\int_{}^{} {{{\sec }^2}\frac{x}{2}dx} } $
$ = \log (\sec x + \tan x) - \tan \frac{x}{2} + c.$

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