Let $f(x)=\frac{\sqrt{\tan x}}{\sin x \cos x}$ and F(x) is its anti derivative, if $F(\frac{π}{4})$, then F(x) is equal to

$2(\sqrt{\tan x}+1)$ $2(\sqrt{\tan x}+3)$ $2(\sqrt{\tan x}+2)$ none of these

$2(\sqrt{\tan x}+2)$

$F(x)=\int\frac{\sqrt{\tan x}}{\sin x\cos x}dx=\int\frac{\sqrt{\tan x}}{\tan x}\sec^2x\,dx\,\,\,(t=\tan x)$  $=\int\frac{1}{\sqrt{t}}dt+c=2\sqrt{t}+c=2\sqrt{\tan x}+c$ Since, $6=F(\frac{π}{4})=2+c$; so c = 4 Hence $F(x)=2(\sqrt{\tan x}+2)$