Answer
$$\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{\sin \theta + 1}}{{\sqrt 2 }}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\cos \theta }}{{3 + 2\sin \theta + {{\sin }^2}\theta }}d\theta } \cr
& {\text{Let }}t = \sin \theta .{\text{ Then }}dt = \cos \theta d\theta ,{\text{}}{\text{ substituting you have}} \cr
& \int {\frac{{\cos \theta }}{{3 + 2\sin \theta + {{\sin }^2}\theta }}d\theta } = \int {\frac{{dt}}{{3 + 2t + {t^2}}}} \cr
& {\text{Completing the square}} \cr
& \int {\frac{{dt}}{{3 + 2t + {t^2}}}} = \int {\frac{{dt}}{{\left( {{t^2} + 2t + 1} \right) + 2}}} = \int {\frac{{dt}}{{{{\left( {t + 1} \right)}^2} + 2}}} \cr
& {\text{Let }}u = t + 1.{\text{ }}du = dt,{\text{ }}du{\text{ substituting you have}} \cr
& = \int {\frac{{du}}{{{u^2} + 2}}} \cr
& {\text{From the table of integrals in the back of the book}} \cr
& \int {\frac{1}{{{a^2} + {u^2}}}} du = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr
& \int {\frac{1}{{{u^2} + 2}}} du = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{u}{{\sqrt 2 }}} \right) + C \cr
& {\text{Write in terms of }}\theta ,{\text{ }}\theta = t + 1 \cr
& = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{t + 1}}{{\sqrt 2 }}} \right) + C \cr
& {\text{Write in terms of }}t,{\text{ }}\theta = \sin \theta \cr
& = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{\sin \theta + 1}}{{\sqrt 2 }}} \right) + C \cr} $$
