If \(\int \frac{\mathrm{d} x}{1+3 \sin ^2 x}=\frac{1}{2} \tan ^{-1}(\mathrm{f}(x))+\mathrm{c}\), where c is -a constant of integration, then \(\mathrm{f}(x)\) is equal to

\(\int \frac{\mathrm{d} x}{1+3 \sin ^2 x}=\int \frac{\mathrm{d} x}{\sin ^2 x+\cos ^2 x+3 \sin ^2 x} \)
\( =\int \frac{\mathrm{d} x}{4 \sin ^2 x+\cos ^2 x} \)
\( =\int \frac{\sec ^2 x \mathrm{~d} x}{4 \tan ^2 x+1} \)
\( =\frac{1}{4} \int \frac{\sec ^2 x \mathrm{~d} x}{\tan ^2 x+\frac{1}{4}} \)
\( \text { Put } \mathrm{t}=\tan x \Rightarrow \mathrm{dt}=\sec ^2 x \mathrm{~d} x \)
\( \therefore \int \frac{\mathrm{~d} x}{1+3 \sin ^2 x}=\frac{1}{4} \int \frac{\mathrm{dt}}{\mathrm{t}^2+\left(\frac{1}{2}\right)^2}=\frac{1}{4}\) \(\cdot 2 \tan ^{-1}(2 \mathrm{t})+\mathrm{c} \)
\( =\frac{1}{2} \tan ^{-1}(2 \mathrm{t})+\mathrm{c} \)
\( =\frac{1}{2} \tan ^{-1}(2 \tan x)+\mathrm{c} \)
\( \therefore \mathrm{f}(x)=2 \tan x\)