$\int e^{-x}(1-\tan x) \sec x d x$ is equal to
Answer & explanation
Correct answer: option 4
We have,
$I =\int e^{-x}(1-\tan x) \sec x d x$
$\Rightarrow I =-\int e^{-x}(-\sec x+\sec x \tan x) d x$
$\Rightarrow I =-e^{-x} \sec (x)+C~~~~~~\left[∵ \int e^{k x}\left\{k f(x)+f'(x)\right\} d x=e^{k x} f(x)\right]$
$\Rightarrow I =-e^{-x} \sec x+C$