2x - 1 , , 2x + 1 dx = - Vidyadip

MCQ

$\int {\frac{{\cos 2x - 1\,\,}}{{\cos 2x + 1}}dx = } $

A
$\tan x - x + c$
B
$x + \tan x + c$
✓
$x - \tan x + c$
D
$ - x - \cot x + c$

✓

Answer

Correct option: C.

$x - \tan x + c$

c
(c)$\int {\frac{{\cos 2x - 1}}{{\cos 2x + 1}}\,} dx$
$ \Rightarrow \,I = - \int {\frac{{(1 - \cos 2x)}}{{(1 + \cos 2x)}}} \,dx$$ = - \int {\frac{{2{{\sin }^2}x}}{{2{{\cos }^2}x}}\,dx} $
$ \Rightarrow I = - \int {{{\tan }^2}x\,dx} $$ = - \int {({{\sec }^2}x - 1)\,dx} $
$ \Rightarrow I = - \int {{{\sec }^2}x\,dx + \int {1\,dx} } $$ = - \tan x + x + c$
==> $I = x - \tan x + c$

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