${\displaystyle \int \frac{dx}{\sin x\left(3+2\cos x\right)}=}$

$={\displaystyle \int \frac{\sin x}{{\sin }^{2}x\left(3+2\cos x\right)}}dx$

$=-{\displaystyle \int \frac{-\sin x}{\left(1-{\cos }^{2}x\right)\left(3+2\cos x\right)}}dx$

Let $\cos x=t\Rightarrow -\sin xdx=dt$

$=-{\displaystyle \int \frac{1}{\left(1-t^2\right)\left(3+2t\right)}dt}$

Let $\frac{1}{\left(1-t^2\right)\left(3+2t\right)}=\frac{1}{\left(1+t\right)\left(1-t\right)\left(3+2t\right)}=\frac{A}{1+t}+\frac{B}{1-t}+\frac{C}{3+2t}$

$\Rightarrow 1=A\left(1-t\right)\left(3+2t\right)+B\left(1+t\right)\left(3+2t\right)+C\left(1-t^2\right)$

Put $t=1\Rightarrow A=\frac{1}{10}$

Put $t=-1\Rightarrow A=\frac{1}{2}$

Constant $1=3A+3B+C$

$C=1-3\left(\frac{1}{2}\right)-3\left(\frac{1}{10}\right)=\frac{10-15-3}{10}$

$C=\frac{-8}{10}=\frac{-4}{5}$

$\because -{\displaystyle \int \frac{1}{\left(1-t^2\right)\left(3+2t\right)}dt=-}{\displaystyle \int \frac{\frac{1}{2}}{1+t}+\frac{\frac{1}{10}}{1-t}}-\frac{\frac{4}{5}}{3+2t}dt$

$=-\frac{1}{2}\log \left(1+\cos x\right)+\frac{1}{10}\log \left(1-\cos x\right)+\frac{2}{5}\log \left(3+2\cos x\right)+c$