Question

$\int \frac{\sec 2x-1}{\sec 2x+1}dx$ is equal to

• A. ${\sec }^{2}x-x+C$
• B. $\tan x-x+C$
• C. ${\sec }^{2}x+x+C$
• D. $\tan x+x+C$

Answer 1

The correct option is B $\tan x-x+C$

$\int \frac{\sec 2x-1}{\sec 2x+1}dx$

$\int \frac{\frac{1}{cos2x}-1}{\frac{1}{cos2x}+1}dx$

Multiplying numerator and denominator with $\cos 2x$:
$\int \frac{1-\cos 2x}{1+\cos 2x}dx$
$=\int \frac{2{\sin }^{2}x}{2{\cos }^{2}x}dx$
$=\int {\tan }^{2}xdx$
$=\int (se{c}^{2}x-1)dx$
$\int \frac{\sec 2x-1}{\sec 2x+1}dx=\tan x-x+C$