Question

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

Answer 1

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

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

$=\int \frac{co{s}^{2}x+sinxcosx}{co{s}^{2}x-si{n}^{2}x}dx$

$=\frac{1}{2}\int \frac{2co{s}^{2}x+sin2x}{cos2x}dx$

$=\frac{1}{2}\int \frac{(1+cos2x)+sin2x}{cos2x}dx$

$=\frac{1}{2}\int sec2xdx+\frac{1}{2}\int dx+\frac{1}{2}\int tan2xdx$

$=\frac{1}{4}log(sec2x+tan2x)+\frac{1}{2}x+\frac{1}{4}logsec2x+c$

$=\frac{1}{4}log(se{c}^{2}2x+\sec 2xtan2x)+\frac{x}{2}+c$