Question

Let $f\left(x\right)=\underset{0}{\overset{x}{\int }}{e}^t(\ln \sec t-{\sec }^{2}t)dt$ and $g\left(x\right)=-2e^x\tan x$. Then, the area bounded by the curves $y=f\left(x\right)$ and $y=g\left(x\right)$ between the ordinates $x=0$ and $x=\frac{\pi }{3}$ is

• A. $\frac{1}{2}{e}^{\pi /3}\ln 2$
• B. $\frac{1}{2}{e}^{\pi /3}\ln 4$
• C. $e^{\pi /3}\ln 2$
• D. None of these

Answer 1

The correct option is C $e^{\pi /3}\ln 2$

$f\left(x\right)=\underset{0}{\overset{x}{\int }}{e}^t(\ln \sec t-{\sec }^{2}t)dt$
$=\underset{0}{\overset{x}{\int }}{e}^t(\ln \sec t-{\sec }^{2}t+\tan t-\tan t)dt$
$\left[\frac{d}{dt}\right(\ln \sec t)=\tan t]$
$=\underset{0}{\overset{x}{\int }}{e}^t(\ln \sec t+\tan t)dt-\underset{0}{\overset{x}{\int }}{e}^t(\tan t+{\sec }^{2}t)dt$
$=[e^t\left(\ln \sec t\right){]}_0^x-[e^t\tan t{]}_0^x$

$\therefore f\left(x\right)=e^x(\ln \sec x-\tan x)$


Required area $=\underset{0}{\overset{\pi /3}{\int }}\left(f\right(x)-g(x\left)\right)dx$
$=\underset{0}{\overset{\pi /3}{\int }}\left(e^x\right(\ln \sec x-\tan x)+2e^x\tan x)dx$
$=\underset{0}{\overset{\pi /3}{\int }}{e}^x(\ln \sec x+\tan x)dx$
$=[e^x\ln \sec x{]}_0^{\pi /3}=e^{\pi /3}\ln 2$