Answer
$$\frac{1}{{\sqrt 3 }}\arctan \left( {\frac{{\sin \theta - 3}}{{\sqrt 3 }}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\cos \theta }}{{{{\sin }^2}\theta - 6\sin \theta + 12}}} d\theta \cr
& \cr
& {\text{write the integrand in terms of }}u \cr
& u = \sin \theta ,\,\,\,\,du = \cos \theta d\theta \cr
& {\text{then}} \cr
& \int {\frac{{\cos \theta }}{{{{\sin }^2}\theta - 6\sin \theta + 12}}} d\theta = \int {\frac{{du}}{{{u^2} - 6u + 12}}} \cr
& \cr
& {\text{completing the square}} \cr
& \int {\frac{{du}}{{{u^2} - 6u + 12}}} = \int {\frac{{du}}{{{u^2} - 6u + 9 + 3}}} \cr
& = \int {\frac{{du}}{{{{\left( {u - 3} \right)}^2} + 3}}} \cr
& = \int {\frac{{du}}{{{{\left( {u - 3} \right)}^2} + {{\left( {\sqrt 3 } \right)}^2}}}} \cr
& \cr
& {\text{Integrating by the formula }}\int {\frac{{dz}}{{{z^2} + {a^2}}}} = \frac{1}{a}\arctan \left( {\frac{z}{a}} \right) + C \cr
& \int {\frac{{du}}{{{{\left( {u - 3} \right)}^2} + 3}}} = \frac{1}{{\sqrt 3 }}\arctan \left( {\frac{{u - 3}}{{\sqrt 3 }}} \right) + C \cr
& \cr
& {\text{write the integrand in terms of }}\theta ,{\text{ replace sin}}\theta {\text{ for }}u \cr
& \int {\frac{{du}}{{{{\left( {u - 3} \right)}^2} + 3}}} = \frac{1}{{\sqrt 3 }}\arctan \left( {\frac{{\sin \theta - 3}}{{\sqrt 3 }}} \right) + C \cr} $$
