Answer
$ - \ln \left( {{{\cos }^2}\theta + \sqrt {{{\cos }^4}\theta + 4} } \right) + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sin 2\theta }}{{\sqrt {{{\cos }^4}\theta + 4} }}} d\theta \cr
& {\text{Use the double angle formula sin2}}\theta = 2\sin \theta \cos \theta \cr
& = \int {\frac{{2\sin \theta \cos \theta }}{{\sqrt {{{\cos }^4}\theta + 4} }}} d\theta \cr
& {\text{Let }}u = {\cos ^2}\theta \Rightarrow du = - 2\cos \theta \sin \theta d\theta ,{\text{ then}} \cr
& \int {\frac{{2\sin \theta \cos \theta }}{{\sqrt {{{\cos }^4}\theta + 4} }}} d\theta = - \int {\frac{{du}}{{\sqrt {{u^2} + 4} }}} \cr
& {\text{Using the formula}} \cr
& \int {\frac{{du}}{{\sqrt {{u^2} + {a^2}} }}} = \ln \left( {u + \sqrt {{u^2} + {a^2}} } \right) + C \cr
& {\text{Therefore}}{\text{,}} \cr
& - \int {\frac{{du}}{{\sqrt {{u^2} + 4} }}} = - \ln \left( {u + \sqrt {{u^2} + 4} } \right) + C \cr
& {\text{Write in terms of }}\theta ,{\text{ }}u = {\cos ^2}\theta ,{\text{ then}} \cr
& \int {\frac{{\sin 2\theta }}{{\sqrt {{{\cos }^4}\theta + 4} }}} d\theta = - \ln \left( {{{\cos }^2}\theta + \sqrt {{{\cos }^4}\theta + 4} } \right) + C \cr} $$