Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{1}{{\cos }^{2}x{\left(1-\tan x\right)}^2}dx \\ =\int \frac{{\sec }^{2}x}{{\left(1-\tan x\right)}^2}dx \\ =\int \frac{{\sec }^{2}x\mathit{}dx}{{\left(1-\tan x\right)}^2} \end{gathered}$$

Let 1 $-$ tan x = t
$$\begin{gathered} -{\sec }^{2}xdx=dt \\ \Rightarrow {\sec }^{2}xdx=-dt \end{gathered}$$

$$\begin{gathered} \therefore I=\int \frac{-dt}{t^2} \\ =-\int t^{-2}dt \\ =-\left[\frac{t^{-2+1}}{-2+1}\right]+\mathrm{C} \\ =\frac{1}{t}+\mathrm{C} \\ =\frac{1}{1-\tan x}+\mathrm{C} \end{gathered}$$