Question
Integrate the functions in Exercises:
$\frac{1}{\cos^2\text{x}(1-\tan\text{x})^2}$

Answer

$\text{Let I}=\int\frac{1}{\cos^2\text{x}(1-\tan\text{x})^2}\text{ dx}=\int\frac{\sec^2\text{x}}{(1 - \tan\text{x})^2}\text{ dx}$
$=-\int\frac{-\sec^2\text{x}}{(1 - \tan\text{x})^2}\text{ dx}\ \ \ \ \ \ \ \ \ \ ...\text{(i)} $
Putting $1-\tan\text{x = t}\ \ \Rightarrow\ \ \ \ -\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \Rightarrow\ \ \ \ -\sec^2\text{x}\text{ dx}=\text{dt} $
$\therefore \ \ \ \ $From eq. (i), $\text{I}=-\int\frac{\text{dt}}{\text{t}^2}=-\int\text{t}^{-2}\text{ dt}=\frac{-\text{t}^{-1}}{-1}+\text{c}=\frac{1}{\text{t}}+\text{c}$
$=\frac{1}{1 - \tan \text{x}}+\text{c} $

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