If $\int \frac{1}{4+5\sin x}dx=\frac{1}{3}\log \left|f\left(x\right)\right|+c$ then f(x) =

$\int \frac{1}{4+5\sin x}dx$

$Put \tan \frac{x}{2}=t\Rightarrow {\sec }^2\frac{x}{2}\frac{1}{2}dx=dt$

$(1+t^2)\frac{dx}{2}=dt\Rightarrow dx=\frac{2dt}{1+t^2}$

$=\int \frac{1}{\left[4+5\left(\frac{2t}{1+t^2}\right)\right]}\frac{2dt}{(1+t^2)}$

$=\int \frac{2dt}{4+4t^2+5t}$

$=\int \frac{dt}{2t^2+5t+2}$

$=\int \frac{dt}{(2t+1)(t+2)}$

$=\frac{1}{2\left(2\right)-1\left(1\right)}\log \left|\frac{2t+1}{t+2}\right|+c$

$=\frac{1}{3}\log \left|\frac{{\displaystyle 2\tan \frac{{\displaystyle x}}{{\displaystyle 2}}+1}}{{\displaystyle \tan \frac{{\displaystyle x}}{{\displaystyle 2}}+2}}\right|+c$

$\therefore f\left(x\right)=\frac{2\tan \frac{x}{2}+1}{\tan \frac{x}{2}+2}$