Find the integral: $ \int{\frac{\sin x+2\cos x}{\sin^2x+3\sin x+2}\,dx}$ I would like some help with the following integral

$$ \int{\frac{\sin x+2\cos x}{\sin^2x+3\sin x+2}\,dx}$$

I tried do split the fraction to $\displaystyle\int\frac{\sin x}{\sin^2x+3\sin x+2}\,dx + \displaystyle\int\frac{2\cos x}{\sin^2x+3\sin x+2}\,dx$
and for $\displaystyle\int\frac{2\cos x}{\sin^2x+3\sin x+2}\,dx$, letting $[ \sin x=t,\frac{dt}{dx}=\cos x ]$ and we get $\displaystyle \int\frac{2dt}{t^2+3t+2}$
but what can I do with $\displaystyle \int\frac{\sin x}{\sin^2x+3\sin x+2}\,dx$?
thx

hint:$\dfrac{\sin x}{\sin^2x+3\sin x + 2} =\dfrac{2}{\sin x+2}-\dfrac{1}{\sin x+1}$, and for each of them use $t = \tan\left(\frac{x}{2}\right)$
