Put $t=cosx$

$\frac{dt}{-\sin x}=dx$

$I=\int \frac{dt}{-\frac{\sin x}{\sin x(3+2t)}}$

$=-\int \frac{dt}{{{\sin }^{2}}x(3+2t)}=-\int \frac{dt}{(1-{{\cos }^{2}}x)(3+2t)}$

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

$\frac{1}{(1-{{t}^{2}})(3+2t)}=\frac{1}{(1-t)(1+t)(3+2t)}$

Putting

$\frac{1}{(1-t)(1+t)(3+2t)}=\frac{A}{1-t}+\frac{B}{1+t}+\frac{C}{3+2t}……(1)$

$A(1+t)(3+2t)+B(1-t)(3+2t)+C(1+t)(1-t)=1$

Now putting $1+t=0$

$t=-1$

$A(0)+B(2)(3-2)+C(0)=1$

$B=\frac{1}{2}$

Now putting $1-t=0$

$t=1$

$A(2)(5)+B(0)+C(0)=1$

$A=\frac{1}{10}$

Now putting $3+2t=0$

$t=-\frac{3}{2}$

$A\left( 0 \right)+B(0)+C\left( 1-\frac{9}{4} \right)=1$

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

$\frac{1}{(1-t)(1+t)(3+2t)}=\frac{1}{10}\times \frac{1}{1-t}+\frac{1}{2}\times \frac{1}{1+t}-\frac{4}{5}\times \frac{1}{3+2t}\int \frac{1}{(1-t)(1+t)(3+2t)}dt$

$=\frac{1}{10}\int \frac{1}{1-t}dt+\frac{1}{2}\int \frac{1}{t+1}dt-\frac{4}{5}\int \frac{1}{3+2t}dt$

$=-\frac{1}{10}\log \left| 1-t \right|+\frac{1}{2}\log \pi \left| 1+t \right|-\frac{4}{5}\times \frac{\log \left| 3+2t \right|}{3+2t}dt$

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