If $f\left(x\right)=\frac{1}{\left({\cos }^{2}x\right)\sqrt{1+\tan x}}$, then its anti-derivative $F\left(x\right)$ , where F(0) = 4.

$f\left(x\right)=\frac{1}{{\cos }^{2}x\sqrt{1+\tan x}}\Rightarrow$ Anti derivative of f(x) is $F\left(x\right)=\int \frac{1}{{\cos }^{2}x\sqrt{1+\tan x}}dx$

$=\int \frac{{\sec }^{2}xdx}{\sqrt{1+\tan x}}=2\sqrt{1+\tan x}+c.F\left(0\right)=4\Rightarrow 2+c=4\Rightarrow c=2$

$\therefore F\left(x\right)=2\sqrt{1+\tan x}+2=2[\sqrt{1+\tan x}+1].$