$\int e^{-x}(1-\tan x) \sec x d x$ is equal to

$e^{-x} \sec x+C$ $e^{-x} \tan x+C$ $-e^{-x} \tan x+C$ none of these

none of these

We have, $I =\int e^{-x}(1-\tan x) \sec x d x$ $\Rightarrow I =-\int e^{-x}(-\sec x+\sec x \tan x) d x$ $\Rightarrow I =-e^{-x} \sec (x)+C~~~~~~\left[∵ \int e^{k x}\left\{k f(x)+f'(x)\right\} d x=e^{k x} f(x)\right]$ $\Rightarrow I =-e^{-x} \sec x+C$