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Integrate the function
$\frac{1}{\cos ^2 x(1-\tan x)^2}$
$\frac{1}{\cos ^2 x(1-\tan x)^2}$
Solution:
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Verified Answer
Put $1-\tan x=t$, so that $-\sec ^2 x d x=d t$
$\begin{aligned}
&\therefore \int \frac{1}{\cos ^2 x(1-\tan x)^2} d x=\int \frac{\sec ^2 x}{(1-\tan x)^2} d x \\
&=-\int \frac{d t}{t^2}=\frac{1}{t}+\mathrm{C}=\frac{1}{(1-\tan x)}+\mathrm{C}
\end{aligned}$
$\begin{aligned}
&\therefore \int \frac{1}{\cos ^2 x(1-\tan x)^2} d x=\int \frac{\sec ^2 x}{(1-\tan x)^2} d x \\
&=-\int \frac{d t}{t^2}=\frac{1}{t}+\mathrm{C}=\frac{1}{(1-\tan x)}+\mathrm{C}
\end{aligned}$
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