Question
$\int\frac{1}{\cos^2\text{x}(1-\tan\text{x})^2}\text{dx}$

Answer

Let $\text{I}=\int\frac{1}{\cos^2\text{x}(1-\tan\text{x})^2}\text{dx}$
$=\int\frac{\sec^2\text{x}}{(1-\tan\text{x})^2}\text{dx}$
$=\int\frac{\sec^2\text{x dx}}{(1-\tan\text{x})^2}$
$-\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\sec^2\text{x dx}=-\text{dt}$
$\therefore\text{I}=\int\frac{-\text{dt}}{\text{t}^2}$
$=-\int\text{t}^{-2}\text{dt}$
$=-\bigg[\frac{\text{t}^{-2+1}}{-2+1}\bigg]+\text{c}$
$=\frac{1}{\text{t}}+\text{c}$
$=\frac{1}{1-\tan\text{x}}+\text{c}$

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