Integrals — Class 12 Maths Solution

: Let $I = \int {\cfrac{1}{{{{\cos }^2}x{{\left( {1 - \tan x} \right)}^2}}}dx = I = \int {\cfrac{{{{\sec }^2}x}}{{{{\left( {1 - \tan x} \right)}^2}}}dx} }$

Put $1 - \tan x = t$ $\Rightarrow$ $- {\sec ^2}xdx = dt$

$\therefore$ $I = - \int {\cfrac{{dt}}{{{t^2}}}} = - \cfrac{{{t^{ - 2 + 1}}}}{{ - 2 + 1}} + C = \cfrac{1}{t} + C = \cfrac{1}{{1 - \tan x}} + C$