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Integrate the function
$\frac{1}{1-\tan x}$
$\frac{1}{1-\tan x}$
Solution:
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Verified Answer
$\int \frac{1}{1-\tan x} d x=\frac{1}{2} \int \frac{2 \cos x d x}{\cos x-\sin x}$
$=\frac{1}{2} \int \frac{\cos x+\sin x+\cos x-\sin x}{\cos x-\sin x} d x$
$=\frac{1}{2} \int d x+\frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} d x$
Put $\cos x-\sin x=t \Rightarrow(\sin x+\cos x) d x=-d t$
$\Rightarrow \mathrm{I}=\frac{1}{2} \int d x+\frac{1}{2} \int \frac{-d t}{t}$
$=\frac{x}{2}-\frac{1}{2} \log (\cos x-\sin x)+\mathrm{C}$
$=\frac{1}{2} \int \frac{\cos x+\sin x+\cos x-\sin x}{\cos x-\sin x} d x$
$=\frac{1}{2} \int d x+\frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} d x$
Put $\cos x-\sin x=t \Rightarrow(\sin x+\cos x) d x=-d t$
$\Rightarrow \mathrm{I}=\frac{1}{2} \int d x+\frac{1}{2} \int \frac{-d t}{t}$
$=\frac{x}{2}-\frac{1}{2} \log (\cos x-\sin x)+\mathrm{C}$
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