Question

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

Answer 1

$\frac{1}{\cos^2x(1-\tan x)^2}=\frac{\sec^2x}{(1-\tan x)^2}$

Let (1-tanx) = t

∴ -sec2 xdx=dt

$\Rightarrow \int\frac{\sec^2x}{(1-\tan x)^2}dx=\int\frac{-dt}{t^2}$

= $-\int t^2dt$

= $+\frac{1}{t}+C$

=$\frac{1}{(1-\tan x)}+C$