Answer
$\frac{1}{2}\ln \left| {1 + {{\sin }^2}\theta } \right| + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sin \theta \cos \theta }}{{1 + {{\sin }^2}\theta }}} d\theta \cr
& {\text{Let }}u = 1 + {\sin ^2}\theta ,{\text{ then by the chain rule }}du = 2\sin \theta \cos \theta d\theta \cr
& \frac{1}{2}du = \sin \theta \cos \theta d\theta \cr
& {\text{Applying the substitution}} \cr
& \int {\frac{{\sin \theta \cos \theta }}{{1 + {{\sin }^2}\theta }}} d\theta = \int {\frac{1}{u}\left( {\frac{1}{2}} \right)} du \cr
& = \frac{1}{2}\int {\frac{1}{u}} du \cr
& {\text{Integrate }} \cr
& = \frac{1}{2}\ln \left| u \right| + C \cr
& {\text{Write in terms of }}t,{\text{ substitute }}u = 1 + {\sin ^2}\theta \cr
& = \frac{1}{2}\ln \left| {1 + {{\sin }^2}\theta } \right| + C \cr} $$
