Integrate $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ How can I evaluate this integral $$\int\dfrac{2\cos x-\sin x}{3\sin x+5\cos x } \mathop{dx}$$
I tried using the half angle formula
$$\sin x=\dfrac{2\tan\dfrac x2}{1+\tan^2\dfrac x2}, \cos x=\dfrac{1-\tan^2\dfrac x2}{1+\tan^2\dfrac x2}$$
substituted and simplified I got
$$\int\dfrac{2-2\tan^2\dfrac{x}{2}-2\tan\dfrac{x}{2}}{5\tan^2\dfrac{x}{2}+6\tan\dfrac{x}{2}-5}dx$$
substituted $\tan^2\dfrac x2=\sec^2\dfrac x2-1$
$$\int\dfrac{4-2\sec^2\dfrac{x}{2}-2\tan\dfrac{x}{2}}{5\left(\tan\dfrac{x}{2}+\dfrac{3}{5}\right)^2-\dfrac{34}{5}}dx$$
I can't eliminate $\tan\frac x2$ term in numerator. I think I am not in right direction. your help to solve this integral is appreciated. thank in advance

hint
Dividing by $\cos(x)$ and putting $\tan(x)=t$, it becomes
$$\int \frac{(2-t)dt}{(5+3t)(1+t^2)}$$
$$=\frac{1}{34}\int \Bigl(\frac{33}{5+3t}+\frac{-11t+7}{t^2+1}\Bigr)dt$$
$$=\frac{11}{34}\int \frac{3dt}{5+3t}-\frac{11}{68}\int \frac{2tdt}{t^2+1}+$$
$$\frac{7}{34}\int \frac{dt}{t^2+1}$$
