Answer
$$\frac{1}{{2\left( {1 + \cos 2\theta } \right)}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sin 2\theta d\theta }}{{{{\left( {1 + \cos 2\theta } \right)}^2}}}} \cr
& \cr
& {\text{Integrate using the substitution method}}{\text{, }} \cr
& \,\,\,\,\,u = 1 + \cos 2\theta ,\,\,\,\,\,du = - 2\sin 2\theta d\theta ,\,\,\,\sin 2\theta d\theta = - \frac{1}{2}du \cr
& \int {\frac{{\sin 2\theta d\theta }}{{{{\left( {1 + \cos 2\theta } \right)}^2}}}} = \int {\frac{{\left( { - 1/2} \right)du}}{{{u^2}}}} \cr
& = - \frac{1}{2}\int {\frac{1}{{{u^2}}}} du \cr
& {\text{Integrate}} \cr
& = - \frac{1}{2}\left( { - \frac{1}{u}} \right) + C \cr
& = \frac{1}{{2u}} + C \cr
& \cr
& {\text{Write in terms of }}\theta ,{\text{ substitute }}1 + \cos 2\theta {\text{ for }}u \cr
& = \frac{1}{{2\left( {1 + \cos 2\theta } \right)}} + C \cr} $$
