The correct option is D log|sin x+cos x|+c ∫1 tan x1+tan xdx =∫cos x sin xcos x+sin xdx Let cos x+sin x=t Then (cos x sin x)dx=dt Hence ...

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Formation of a Differential Equation from a General Solution

∫1-tan x/1+ta...

Question

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

log | 1 + tan x | + c

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log | 1 - tan x | + c

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log | sin x + cos x | + c

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log | sin x - cos x | + c

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Solution

The correct option is D $log | sin x + cos x | + c$
$\int \frac{1 - tan x}{1 + tan x} d x$
$= \int \frac{cos x - sin x}{cos x + sin x} d x$
Let
$cos x + sin x = t$
Then
$(cos x - sin x) d x = d t$
Hence the above integral transforms to
$\int \frac{d t}{t}$
$= l n | t | + c$
$= l n | sin x + cos x | + c$

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