$\int \frac{cos 2 x - 1}{cos 2 x + 1} d x$ is equal to.

tan x - x + c

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x + tan x + c

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x - tan x + c

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- x - cot x + c

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Solution

The correct option is C $x - tan x + c$
Let $I = \int \frac{cos 2 x - 1}{cos 2 x + 1} d x$
$\Rightarrow I = - \int \frac{(1 - cos 2 x)}{(1 + cos 2 x)} d x$
$\Rightarrow I = - \int \frac{2 {sin}^{2} x}{2 {cos}^{2} x} d x$
$\Rightarrow I = - \int {tan}^{2} x d x$
$\Rightarrow I = - \int (s e c^{2} x - 1) x$
$\Rightarrow I = - \int {sec}^{2} x d x + \int 1 d x$
$\Rightarrow I = - tan x + x + c$
$\Rightarrow I = x - tan x + c$

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Similar questions

\int \frac{\cos 2 x - 1}{\cos 2 x + 1} d x =

(a) tan x − x + C
(b) x + tan x + C
(c) x − tan x + C
(d) − x − cot x + C

is equal to

A. tan x + cot x + C

B. tan x + cosec x + C

C. − tan x + cot x + C

D. tan x + sec x + C

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∫cos2x−1cos2x+1dx is equal to.

The correct option is C x−tanx+cLet I=∫cos2x−1cos2x+1dx⇒I=−∫(1−cos2x)(1+cos2x)dx⇒I=−∫2sin2x2cos2xdx⇒I=−∫tan2xdx⇒I=−∫(sec2x−1)x⇒I=−∫sec2xdx+∫1dx⇒I=−tanx+x+c⇒I=x−tanx+c

$\int \frac{cos 2 x - 1}{cos 2 x + 1} d x$ is equal to.