Question

If $\int \sqrt{1+2\tan x(\tan x+\sec x)}dx=\ln |f\left(x\right)|+C,$ then the value of $f\left(0\right)$ is equal to $(0\le x<\frac{\pi }{2})$
(where $C$ is constant of integration )

Answer 1

$I=\int \sqrt{1+2\tan x(\tan x+\sec x)}dx=\int \sqrt{(\sec x+\tan x{)}^2}dx=\int (\sec x+\tan x)dx=\ln |\sec x+\tan x|+\ln |\sec x|+C\Rightarrow \ln |f\left(x\right)|=\ln |\sec x(\sec x+\tan x)|+C$
So, $f\left(x\right)=\sec x(\sec x+\tan x),f\left(0\right)=1$