Question

If $f(x)$ is the antiderivative of $(1+2\tan x(\tan x+\sec x){)}^{\frac{1}{2}}$ and $f\left(\frac{\pi }{6}\right)=\log 2$, then the value of $f(0)$ is

Answer 1

Answer: 0

$f(x)=\int (1+2\tan x(\tan x+\sec x){)}^{\frac{1}{2}}dx$
$=\int {\left(1+2{\tan }^{2}x+2\tan x\sec x\right)}^{\frac{1}{2}}dx$
$=\int {\left(1+{\tan }^{2}x+{\tan }^{2}x+2\tan x\sec x\right)}^{\frac{1}{2}}dx$
$=\int {\left({\sec }^{2}x+{\tan }^{2}x+2\tan x\sec x\right)}^{\frac{1}{2}}dx$
$=\int {\left((\sec x+\tan x{)}^2\right)}^{\frac{1}{2}}dx$
$=\log (\sec x+\tan x)+\log \sec x+c$
$f=\left(\frac{\pi }{6}\right)=\log 2$
$\Rightarrow \log \frac{2}{\sqrt{3}}\left(\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}\right)+c=\log 2$
$\Rightarrow c=0$
$f(x)=\log (\sec x(\sec x+\tan x))$
$\Rightarrow f(0)=\log ((1)(1+0))=0$