Question

If $\int e^x\left(\tan x-x-2 \tan x \sec ^2 x\right) d x=e^x f(x)+C$ where $f(0)=0$, then the value of $f\left(\frac{\pi}{4}\right)$ equals (where $C$ is the constant of integration)

• A. $\frac{\pi}{4}$
• B. $1-\frac{\pi}{4}$
• C. $\frac{-\pi}{4}$
• D. $\frac{\pi}{2}$

Answer 1

Answer: (C)

$ \int e ^{ x }\left(\tan x - x +\tan ^2 x -\tan ^2 x -2 \tan x \sec ^2 x \right) dx $
$=\int e ^{ x }\left(\tan x - x +\tan ^2 x \right) dx -\int e ^{ x }\left(\tan ^2 x +2 \tan x \sec ^2 x \right) dx$
$= e ^{ x }\left(\tan x - x -\tan ^2 x \right)+ C$
$f ( x )=\tan x - x -\tan ^2 x$
$f \left(\frac{\pi}{4}\right)=\frac{-\pi}{4}$