Question

The value of $\int e^{-x}(1-\tan x)\sec xdx$ is equal to

• A. $e^{-x}\sec x+c$
• B. $e^{-x}\tan x+c$
• C. $-e^{-x}\tan x+c$
• D. $-e^{-x}\sec x+c$

Answer 1

The correct option is D $-e^{-x}\sec x+c$

$\int e^{-x}(1-\tan x)\sec xdx$
$\int e^{-x}(\sec x-\sec x\tan x)dx$
$-x=t$
$-\int e^t(\sec t+\sec t\tan t)dt$
It is in the form of $\int e^x\left(f\right(x\left)\right)+f^1\left(x\right))dx=e^{x}f(x)+c$
$-e^t\left(\sec t\right)+c$
$t=-x$
$=-e^{-x}\sec x+c$