Answer
$$\frac{1}{{\sqrt 5 }}\ln \left| {\frac{{2\tan \left( {\theta /2} \right) - 3 - \sqrt 5 }}{{2\tan \left( {\theta /2} \right) - 3 + \sqrt 5 }}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{2 - 3\sin \theta }}d\theta } \cr
& {\text{Using Substitution for Rational Functions of Sine and Cosine}} \cr
& \sin x = \frac{{2u}}{{1 + {u^2}}},{\text{ }}dx = \frac{{2du}}{{1 + {u^2}}},{\text{ }}u = \tan \left( {\frac{\theta }{2}} \right) \cr
& {\text{Substituting}} \cr
& \int {\frac{1}{{2 - 3\sin \theta }}d\theta } = \int {\frac{1}{{2 - 3\left( {\frac{{2u}}{{1 + {u^2}}}} \right)}}\left( {\frac{2}{{1 + {u^2}}}} \right)du} \cr
& = \int {\frac{{1 + {u^2}}}{{2 + 2{u^2} - 6u}}\left( {\frac{2}{{1 + {u^2}}}} \right)du} \cr
& = \int {\frac{2}{{2 + 2{u^2} - 6u}}du} \cr
& = \int {\frac{1}{{{u^2} - 3u + 1}}du} \cr
& {\text{Completing the square}} \cr
& = \int {\frac{1}{{{u^2} - 3u + \frac{9}{4} - \frac{9}{4} + 1}}du} \cr
& = \int {\frac{1}{{{{\left( {u - \frac{3}{2}} \right)}^2} - \frac{5}{4}}}du} \cr
& {\text{Integrate by tables }}\int {\frac{{dz}}{{{z^2} - {a^2}}} = \frac{1}{{2a}}\ln \left| {\frac{{z - a}}{{z + a}}} \right| + C} \cr
& = \frac{1}{{2\left( {\sqrt 5 /2} \right)}}\ln \left| {\frac{{u - \frac{3}{2} - \frac{{\sqrt 5 }}{2}}}{{u - \frac{3}{2} + \frac{{\sqrt 5 }}{2}}}} \right| + C \cr
& = \frac{1}{{\sqrt 5 }}\ln \left| {\frac{{2u - 3 - \sqrt 5 }}{{2u - 3 + \sqrt 5 }}} \right| + C \cr
& {\text{Write in terms of }}\theta {\text{, let }}u = \tan \left( {\theta /2} \right) \cr
& = \frac{1}{{\sqrt 5 }}\ln \left| {\frac{{2\tan \left( {\theta /2} \right) - 3 - \sqrt 5 }}{{2\tan \left( {\theta /2} \right) - 3 + \sqrt 5 }}} \right| + C \cr} $$
