Question
Integrate the function: $\frac{1}{\cos ^{2} x(1-\tan x)^{2}}$

Answer

We have $\frac{1}{\cos ^{2} x(1-\tan x)^{2}}=\frac{\sec ^{2} x}{(1-\tan x)^{2}}$ 
Let (1 – tanx) = t
$\Rightarrow$ -sec2xdx = dt
$\therefore ~ \int \frac{\sec ^{2} x}{(1-\tan x)^{2}} d x=\int \frac{-d t}{t^2}=-\int t^{-2} d t$ 
$= \frac{1}{t}+C$ 
$= \frac{1}{(1-\tan x)}+C$

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