If the antiderivative of $f\left(x\right) =$

$\frac{\sin x}{{\sin }^{2}x+4{\cos }^{2}x}\text{ is }(1/\sqrt{3}){\tan }^{-1}\left(\right(1/\sqrt{3}\left)g\right(x\left)\right)+C$ then

$g\left(x\right)$ is equal to

$\begin{array}{l}f\left(x\right)=\frac{\sin x}{{\sin }^{2}x+4{\cos }^{2}x}=\frac{\tan x\sec x}{{\tan }^{2}x+4}\\ =\frac{\tan x\sec x}{{\sec }^{2}x+3}\end{array}$

Putting $\sec x=t,\mathrm{d}x\sec x\tan x=\mathrm{d}t$ so

$\int f\left(x\right)dx=\int \frac{\mathrm{d}t}{t^2+3}=\frac{1}{\sqrt{3}}{\tan }^{-1}\frac{\sec x}{\sqrt{3}}+C$