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

MCQ

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

A
$2\tan \frac{x}{2} - x + c$
B
$\frac{1}{2}\tan \frac{x}{2} - x + c$
C
$x - \frac{1}{2}\tan \frac{x}{2} + c$
✓
$x - 2\tan \frac{x}{2} + c$

✓

Answer

Correct option: D.

$x - 2\tan \frac{x}{2} + c$

d
(d) $\int_{}^{} {\frac{{\cos x - 1}}{{\cos x + 1}}\,dx = - \int_{}^{} {{{\tan }^2}\frac{x}{2}\,dx} } $
$ = - \int {\left( {{{\sec }^2}\frac{x}{2} - 1} \right)\,dx} = \int {\left( {1 - {{\sec }^2}\frac{x}{2}} \right)dx = x - 2\tan \frac{x}{2} + c} $.

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