Question Number 21679 by Arnab Maiti last updated on 30/Sep/17
$$\int\frac{\mathrm{sec}\theta\:\mathrm{d}\theta}{\mathrm{1}−\mathrm{sec}\theta} \\ $$
Answered by alex041103 last updated on 30/Sep/17
$${First}\:{we}\:{make}\:{the}\:{following}\:{transformations}: \\ $$$$\int\frac{\mathrm{sec}\theta\:\mathrm{d}\theta}{\mathrm{1}−\mathrm{sec}\theta}=\int\frac{\frac{\mathrm{1}}{{cos}\theta}}{\mathrm{1}−\frac{\mathrm{1}}{{cos}\theta}}{d}\theta= \\ $$$$=\int\frac{\frac{\mathrm{1}}{{cos}\theta}}{\mathrm{1}−\frac{\mathrm{1}}{{cos}\theta}}\:\frac{{cos}\theta}{{cos}\theta}{d}\theta= \\ $$$$=\int\frac{\mathrm{1}}{{cos}\theta−\mathrm{1}}{d}\theta \\ $$$${Also}\:{by}\:{the}\:{double}\:{angle}\:{indetity} \\ $$$${we}\:{have}\:{cos}\theta=\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right) \\ $$$$\Rightarrow\int\frac{\mathrm{1}}{{cos}\theta−\mathrm{1}}{d}\theta=−\int\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}\:\frac{{d}\theta}{\mathrm{2}} \\ $$$${Let}\:{x}=\frac{\theta}{\mathrm{2}}\Rightarrow{dx}=\frac{{d}\theta}{\mathrm{2}} \\ $$$$\int\frac{\mathrm{1}}{{cos}\theta−\mathrm{1}}{d}\theta=\int−{csc}^{\mathrm{2}} {x}\:{dx}={I} \\ $$$${We}\:{know}\:{that}\:\frac{{d}}{{dx}}\left({ctg}\:{x}\right)=−{csc}^{\mathrm{2}} {x} \\ $$$$\Rightarrow\int\frac{{sec}\theta}{\mathrm{1}−{sec}\theta}{d}\theta={ctg}\left(\frac{\theta}{\mathrm{2}}\right)+{C} \\ $$