Answer
$\left( {{{\sin }^2}\theta + 1} \right)\arctan \left( {\sin \theta } \right) - \sin \theta + C$
Work Step by Step
$$\eqalign{
& \int {\sin 2\theta } \arctan \left( {\sin \theta } \right)d\theta \cr
& {\text{Use the double angle formula sin2}}\theta = 2\sin \theta \cos \theta \cr
& = \int {2\sin \theta \cos \theta } \arctan \left( {\sin \theta } \right)d\theta \cr
& = 2\int {\sin \theta } \arctan \left( {\sin \theta } \right)\cos \theta d\theta \cr
& {\text{Let }}u = \sin \theta ,{\text{ }}du = \cos \theta d\theta ,{\text{ substituting we obtain}} \cr
& = 2\int u \arctan udu \cr
& \cr
& {\text{Using the entry 92 }}\left( {{\text{Integrals on Reference Pages 6}}--{\text{1}}0} \right) \cr
& \int {u{{\tan }^{ - 1}}} udu = \frac{{{u^2} + 1}}{2}{\tan ^{ - 1}}u - \frac{u}{2} + C \cr
& 2\int u \arctan udu = 2\left( {\frac{{{u^2} + 1}}{2}} \right)\arctan u - 2\left( {\frac{u}{2}} \right) + C \cr
& = \left( {{u^2} + 1} \right)\arctan u - u + C \cr
& {\text{Write in terms of }}\theta ,{\text{ }}u = \sin \theta ,{\text{ then}} \cr
& = \left( {{{\sin }^2}\theta + 1} \right)\arctan \left( {\sin \theta } \right) - \sin \theta + C \cr} $$
