The correct option is C tan x x+ c ∫1 cos2x1+cos2xdx =∫2sin^2x2cos^2xdx =∫tan^2xdx =∫ (^2x 1)dx =tan x x+c ...

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Integration by Parts

∫1-cos2x/1+co...

Question

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

tan x + x + c

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tan x - x + c

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x tan x + c

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x - tan x + c

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Solution

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

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is equal to

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