Q. $\int_{}^{}{\frac{\sec x}{\sec{x + \tan x}}\ dx}$ is equal to
A) $\tan{x - \sec{x + c}}$
B) $\log{(1 + \sec{x)}} + c$
C) $\sec{x + \tan{x + c}}$
D) $\log{\sin{x + \log{\cos{x + c}}}}$
Explanation
(a)
Let $I = \int_{}^{}{\frac{\sec x}{\sec{x + \tan x}}\ dx}$
$= \int_{}^{}{\frac{\sec{x(\sec{x - \tan{x)}}}}{\sec^{2}{x - \tan^{2}x}}dx}$
$= \int_{}^{}{(\sec^{2}{x - \sec{x\tan{x)}}}}dx$
$= \tan{x - \sec{x + c}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.