$\text{ If }\int \frac{{\log }_e\left(1+{\sin }^{2}x\right)}{{\cos }^{2}x}dx=f\left(x\right)+C,\text{ then the value of }f\left(0\right)\text{ is }$

$\begin{array}{l}\int \frac{{\log }_{\mathrm{e}}\left(1+{\sin }^2\mathrm{x}\right)\mathrm{dx}}{{\cos }^2\mathrm{x}}\\ ={\log }_{\mathrm{e}}\left(1+{\sin }^2\mathrm{x}\right)\cdot \tan \mathrm{x}-2\int \frac{{\sin }^2\mathrm{x}+1-1}{1+{\sin }^2\mathrm{x}}\mathrm{dx}\\ ={\log }_{\mathrm{e}}\left(1+{\sin }^2\mathrm{x}\right)\cdot \tan \mathrm{x}-2\mathrm{x}+2\int \frac{{\sec }^2\mathrm{xdx}}{1+2{\tan }^2\mathrm{x}}\\ ={\log }_{\mathrm{e}}\left(1+{\sin }^2\mathrm{x}\right)\cdot \tan \mathrm{x}-2\mathrm{x}+\sqrt{2}{\tan }^{-1}(\sqrt{2}\tan \mathrm{x})+\mathrm{C}\\ =\mathrm{f}\left(\mathrm{x}\right)+\mathrm{C}\\ \therefore \mathrm{f}\left(0\right)=0\end{array}$