Question

$\int e^x[\log \cos x+{\sec }^{2}x]dx=$

• A. $e^x[\log \cos x+{\sec }^{2}x]+c$
• B. $e^x[\log \cos x+\tan x]+c$
• C. $e^x\left(\cos x\right)+c$
• D. $e^x\left[\log \right(\tan x\left)\right]+c$

Answer 1

Answer: (B)

$e^x[\log \cos x+\tan x]+c$

$\int e^x[logcosx+se{c}^{2}x]dx$
$=\int e^x\left(logcosx\right)dx+\int e^{x}se{c}^{2}xdx$
Integrating by parts,
$=log\left(\cos x\right)e^x+\int \tan x{e}^{x}dx+e^x\tan x-\int e^x\tan xdx$
$=e^x\left[log\right(\cos x)+\tan x]+c$