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

Answer: (A)

Let $I=\int \frac{\sec x}{\sec x+\tan x} d x$
$=\int \frac{\sec x(\sec x-\tan x)}{\sec ^{2} x-\tan ^{2} x} d x$
$=\int\left(\sec ^{2} x-\sec x \tan x\right) d x$
$=\tan x-\sec x+C$