Question

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

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

Answer 1

The correct option is A $\frac{1}{1-\tan x}.$

Let $I=\int \frac{1}{{\cos }^{2}x{(1-\tan x)}^2}dx=\int \frac{{\sec }^{2}xdx}{{(1-\tan x)}^2}$
Put $1-\tan x=t\Rightarrow -{\sec }^{2}xdx=dt$
Therefore
$I=-\int \frac{1}{t^2}dt=-(-\frac{1}{t})=\frac{1}{t}=\frac{1}{1-\tan x}$
Hence, option 'A' is correct.