If f ( x ) = 1 cos 2 x 1 + tan x , then its anti-derivative F ( x ) =          …

$\Rightarrow f\left(x\right)=\frac{1}{\left({\cos }^{2}x\right)\sqrt{1+\tan x}}$

$put \sqrt{1+\tan x}=t$

$\Rightarrow \frac{1·{\sec }^{2}x}{2·\sqrt{1+\tan x}}=\frac{dt}{dx}$

$\Rightarrow dx=\frac{2\sqrt{1+\tan x}}{{\sec }^{2}x}dt$

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

$\Rightarrow \int f\left(x\right)dx=\int \frac{2\sqrt{1+\tan x}}{{\sec }^{2}x·{\cos }^{2}x·\sqrt{1+\tan x}}dt$

$\Rightarrow F\left(x\right)=\int 2dt$

$\Rightarrow F\left(x\right)=2t+C$

$\Rightarrow F\left(x\right)=2\sqrt{1+\tan x}+C$

$\Rightarrow F\left(0\right)=4=2\sqrt{1+\tan 0}+C$

$\Rightarrow C=2$

$\Rightarrow F\left(x\right)=2\sqrt{1+\tan x}+2$

$\Rightarrow F\left(x\right)=2\left(\sqrt{1+\tan x}+1\right)$