Question

$\int \frac{\sec x}{\sec x+\tan x}dx=$

• A. $\tan x-\sec x+C$
• B. $\log (1+\sin x)+C$
• C. $\sec x+\tan x+C$
• D. $\log \sin x+\log \cos x+C$

Answer 1

The correct option is A $\tan x-\sec x+C$

$=\int \frac{\sec x}{\sec x+\tan x}dx$
$=\int \frac{\sec x}{\sec x+\tan x}\cdot \frac{\sec x-\tan x}{\sec x-\tan x}dx$
$=\int \frac{\sec x(\sec x-\tan x)}{{\sec }^{2}x-{\tan }^{2}x}dx$
$=\int ({\sec }^{2}x-\sec x\cdot \tan x)dx$ $[\because {\sec }^{2}x-{\tan }^{2}x=1]$
$\therefore \tan x-\sec x+C$