Answer
$$ = - ta{n^{ - 1}}\left( {\cos \theta } \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sin \theta }}{{{{\cos }^2}\theta + 1}}} d\theta \cr
& or \cr
& \int {\frac{{\sin \theta }}{{{{\left( {\cos \theta } \right)}^2} + 1}}} d\theta \cr
& {\text{substitutue }}u = \cos \theta ,{\text{ }}du = - {\text{sin}}\theta d\theta \cr
& \int {\frac{{\sin \theta }}{{{{\left( {\cos \theta } \right)}^2} + 1}}} d\theta = \int {\frac{{ - du}}{{{u^2} + 1}}} \cr
& = - \int {\frac{{du}}{{{u^2} + 1}}} \cr
& {\text{ use the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}}} = \frac{1}{a}ta{n^{ - 1}}\left( {\frac{u}{a}} \right) + C,{\text{ }}\left( {{\text{see page 468}}} \right) \cr
& = - \left( {\frac{1}{1}ta{n^{ - 1}}\left( {\frac{u}{1}} \right)} \right) + C \cr
& = - ta{n^{ - 1}}\left( u \right) + C \cr
& {\text{write in terms of }}x \cr
& = - ta{n^{ - 1}}\left( {\cos \theta } \right) + C \cr} $$
