Let $f\left(x\right)=\frac{\sqrt{\tan x}}{\sin x\cos x}$ and $F\left(x\right)$ is its anti-derivative, if $F(\pi /4)=6$ then $F\left(x\right)$ is equal to

$F\left(x\right)=\int \frac{\sqrt{\tan x}}{\sin x\cos x}\mathrm{d}x=\int \frac{\sqrt{\tan x}}{\tan x}{\sec }^{2}x\mathrm{d}x$

$(t=\tan x)$

$=\int \frac{1}{\sqrt{t}}\mathrm{d}t+C=2\sqrt{t}+C=2\sqrt{\tan x}+C$

Since $6=F(\pi /4)=2+C\text{ so }C=4$ Hence 

$F\left(x\right)=2(\sqrt{\tan x}+2)$