Question

$\int \frac{Secx}{Secx+Tanx}dx=$

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