Question

Evaluate: $\int \frac{dx}{1-\tan x}$

Answer 1

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

$I=\int \frac{\cos x}{\cos x-\sin x}dx$

$I=\int \frac{2\cos x}{2(\cos x-\sin x)}dx$

$I=\frac{1}{2}\int \frac{\cos x+\cos x+\sin x-\sin x}{\cos x-\sin x}dx$

$I=\frac{1}{2}\int (\frac{\cos x-\sin x}{\cos x-\sin x}+\frac{\cos x+\sin x}{\cos x-\sin x})dx$

$I=\frac{1}{2}\int 1.dx+\frac{\cos x+\sin x}{\cos x-\sin x}dx$

Let $\cos x-\sin x=t$

$\Rightarrow (-\sin x-\cos x)dx=dt$

$\Rightarrow -(\sin x+\cos x)dx=dt$

$\Rightarrow d(\sin x+\cos x)dx=-dt$

$I=\frac{1}{2}[x+\int \frac{-dt}{t}]$

$I=\frac{1}{2}[x-\log |t|]+c$

$I=\frac{1}{2}[x-\log |\cos x-\sin x|]+c$