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

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Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Indefinite Integration

Question:

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

Options:

$log|\cos x+\sin x|+c$

$log|\cos x-\sin x|+c$

$log|\sin x|+c$

$-log|\cos x|+c$

Correct Answer:

$log|\cos x+\sin x|+c$

Explanation:

$I=\int\frac{1-tanx}{1+tanx}dx$

$=\int\frac{1-\frac{sinx}{cosx}}{1+\frac{sinx}{cosx}}dx=\int\frac{cosx-sinx}{cosx+sinx}dx$

Let $cosx+sinx=t$

$⇒dt=(-sinx+cosx)dx$

$⇒I=\int\frac{dt}{t}=\log|t|+C$

$=\log|cosx+sinx|+C$