Answer
$ - \ln \left( {1 + {{\cos }^2}x} \right) + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sin 2x}}{{1 + {{\cos }^2}x}}dx} \cr
& {\text{use }}\sin 2x = 2\sin x\cos x \cr
& = \int {\frac{{2\sin x\cos x}}{{1 + {{\cos }^2}x}}dx} \cr
& = 2\int {\frac{{cosx}}{{1 + {{\cos }^2}x}}\left( {\sin x} \right)dx} \cr
& {\text{set }}u = \cos x,\,\,\,\,\,then\,\,\,\,\,\,\,du = - \sin xdx \cr
& {\text{use the change of variable}} \cr
& 2\int {\frac{{cosx}}{{1 + {{\cos }^2}x}}\left( {\sin x} \right)dx} = 2\int {\frac{u}{{1 + {u^2}}}\left( { - du} \right)} \cr
& = - \int {\frac{{2u}}{{1 + {u^2}}}du} \cr
& {\text{integrate}} \cr
& = - \ln \left( {1 + {u^2}} \right) + C \cr
& {\text{replace }}u = \cos x \cr
& = - \ln \left( {1 + {{\cos }^2}x} \right) + C \cr} $$
