Question

\[\int\frac{1}{\text{cos}^2\text{ x }\left( 1 - \text{tan x} \right)^2} dx\]

Answer 1

\[\text{Let  I } = \int\frac{1}{\cos^2 x \left( 1 - \tan x \right)^2}dx\]
\[ = \int\frac{\sec^2 x}{\left( 1 - \tan x \right)^2} \text{dx} \]
\[ = \int\frac{\sec^2 \text{x dx}}{\left( 1 - \tan x \right)^2}\]

Let 1-  tan x = t

\[- \sec^2 \text{x dx} = dt\]
\[ \Rightarrow \sec^2\text{ x dx} = - dt\]

\[\therefore I = \int\frac{- dt}{t^2}\]
\[ = - \int t^{- 2} dt\]
\[ = - \left[ \frac{t^{- 2 + 1}}{- 2 + 1} \right] + C\]
\[ = \frac{1}{t} + C\]
\[ = \frac{1}{1 - \tan x} + C\]