Question Number 78352 by Khyati last updated on 16/Jan/20
$${Find}\:\int\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}\left(\mathrm{1}−{tanx}\right)^{\mathrm{2}} }\:{dx} \\ $$
Commented by jagoll last updated on 17/Jan/20
$$\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} {x}\:{dx}}{\left(\mathrm{1}−\mathrm{tan}\:{x}\right)^{\mathrm{2}} }\:=\:−\int\:\frac{{d}\left(\mathrm{1}−\mathrm{tan}\:{x}\right)}{\left(\mathrm{1}−\mathrm{tan}\:{x}\right)^{\mathrm{2}} } \\ $$$$=\:−\int\:{u}^{−\mathrm{2}} \:{du}\:=\:\frac{\mathrm{1}}{{u}}+{c} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{tan}\:{x}}+{c} \\ $$
Answered by MJS last updated on 16/Jan/20
![∫(dx/(cos^2 x (1−tan x)^2 ))=∫(dx/(1−sin 2x))= [t=tan x → dx=cos^2 x dt] =∫(dt/((t−1)^2 ))=(1/(1−t))=((cos x)/(cos x −sin x))+C](https://www.tinkutara.com/question/Q78354.png)
$$\int\frac{{dx}}{\mathrm{cos}^{\mathrm{2}} \:{x}\:\left(\mathrm{1}−\mathrm{tan}\:{x}\right)^{\mathrm{2}} }=\int\frac{{dx}}{\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\mathrm{cos}^{\mathrm{2}} \:{x}\:{dt}\right] \\ $$$$=\int\frac{{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{1}−{t}}=\frac{\mathrm{cos}\:{x}}{\mathrm{cos}\:{x}\:−\mathrm{sin}\:{x}}+{C} \\ $$
Commented by peter frank last updated on 17/Jan/20
$${thank}\:{you}\: \\ $$