Question

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

• A. $\frac{1}{1+\tan x}+c$
• B. $\frac{1}{1-\tan x}+c$
• C. $\frac{-1}{1+\tan x}+c$
• D. $\frac{-1}{1-\tan x}+c$

Answer 1

The correct option is B $\frac{1}{1-\tan x}+c$

$I=\int \frac{1}{{\cos }^{2}x(1-\tan x{)}^2}.dx$

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

Let $\tan x=t$

On differentiation, we have

${\sec }^{2}x.dx=dt$

Then, we have

$I=\int \frac{dt}{(1-t{)}^2}$

$=\int (1-t{)}^{-2}dt$

$=(-1)\frac{(1-t{)}^{-2+1}}{(-2+1)}$

$=(1-t{)}^{-1}+c$

$=\frac{1}{1-t}+c$

$=\frac{1}{1-\tan x}+c$