Question

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

Answer 1

$\frac{1}{{\cos }^{2}x(1-\tan x{)}^2}=\frac{{\sec }^{2}x}{(1-\tan x{)}^2}$
Let $(1-\tan x)=t$
$\therefore -{\sec }^{2}xdx=dt$
$\Rightarrow \int \frac{{\sec }^{2}x}{(1-\tan x{)}^2}dx=\int \frac{-dt}{t^2}$
$=-\int t^{-2}dt$
$=\frac{1}{t}+C$
$=\frac{1}{(1-\tan x)}+C$