Question

Evaluate $\int e^x\left[\tan x-\log \left(\cos x\right)\right]dx$

• A. $e^x\log \left(\sec x\right)+C$
• B. $e^x\log \left(\mathrm{cosec\; x}\right)+C$
• C. $e^x\log \left(\cos x\right)+C$
• D. $e^x\log \left(\sin x\right)+C$

Answer 1

The correct option is A $e^x\log \left(\sec x\right)+C$

Let say

$I=\int e^x[\tan x-\log (\cos x\left)\right]dx$

$I=\int e^x[\tan x+\log (\cos x{)}^{-1}]dx$

$I=\int e^x[\log \sec x+\tan x]dx$

Now if I check differentiation of '$\log \sec x$' then -

$=\frac{1}{\sec x}\times \sec x\tan x=\tan x$

So, given integral is in the form also -

$\int e^x\left(f\right(x_{+}{f}^{\prime }\left(x\right)dx=e^{x}f\left(x\right)+C$

So, $I=e^x\log \left(\sec x\right)+C$

option A