Question

$\int \frac{\sec x(1+\tan x)dx}{(e^{-x}+\sec x)}=f\left(x\right)+c$.
If $f\left(0\right)=\ln \left(2\right)$, then $f\left(\frac{\pi }{4}\right)$ is

• A. $\ln (1+\sqrt{2}{e}^{\pi /4})$
• B. $\ln \sqrt{2}$
• C. $\ln (2\sqrt{2}$)
• D. $\ln (\frac{e^{\pi /4}}{\sqrt{2}}+1)$

Answer 1

The correct option is A $\ln (1+\sqrt{2}{e}^{\pi /4})$

$\int \frac{\sec x(1+\tan x)dx}{(e^{-x}+\sec x)}$
$=\int \frac{e^x\sec x(1+\tan x)dx}{(1+e^x\sec x)}$
$Let1+e^x\sec x=t$
$\Rightarrow e^{x}secx(1+tanx)dx=dt$
$\int \frac{dt}{t}=\ln t+c=\ln (1+e^x\sec x)+c$
If $f\left(0\right)=\ln \left(2\right)$ then $c=0$
So $f\left(x\right)=\ln (1+e^x\sec x)$
$f\left(\frac{\pi }{4}\right)=\ln (1+\sqrt{2}{e}^{\pi /4})$