If $\int \frac{\sin 2x}{\sin 5x\sin 3x}dx=\frac{1}{3}\log \left|\sin 3x\right|-f\left(x\right)$, then $f\left(x\right)=$

$I=\int \frac{\sin 2x}{\sin 5x\sin 3x}dx$

$I=\int \frac{\sin (5x-3x)}{\sin 5x\sin 3x}dx$

We know, $\sin (A-B)=\sin A\cos B-\cos A\sin B$

$I=\int \frac{\sin 5x\cos 3x-\cos 5x\sin 3x}{\sin 5x\sin 3x}dx$

$=\int \frac{\sin 5x\cos 3x}{\sin 5x\sin 3x}-\int \frac{\cos 5x\sin 3x}{\sin 5x\sin 3x}dx$

$=\int \frac{\cos 3x}{\sin 3x}dx-\int \frac{\cos 5x}{\sin 5x}dx$

$=\int \cot 3xdx-\int \cot 5xdx$

$I=\frac{1}{3}\log |\sin 3x|-\frac{1}{5}\log |\sin 5x|+c$

$\therefore$ We can see value of $f\left(x\right)=\frac{1}{5}\log |\sin 5x|$