Question

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

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