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