Answer
$$ = - \frac{1}{2}\cos \left( {2x} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\sin 2x} dx \cr
& {\text{set }}u = 2x{\text{ then }}\frac{{du}}{{dx}} = 2,\,\,\,\,\,\,\,\,\frac{{du}}{2} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\sin 2x} dx = \int {\sin u} \left( {\frac{{du}}{2}} \right) \cr
& {\text{use multiple constant rule}} \cr
& = \frac{1}{2}\int {\sin u} du \cr
& {\text{integrate by using the Basic Trigonometric integral }}\int {\sin x} dx = - \cos x + C \cr
& = \frac{1}{2}\left( { - \cos u} \right) + C \cr
& = - \frac{1}{2}\cos u + C \cr
& {\text{write in terms of }}x \cr
& = - \frac{1}{2}\cos \left( {2x} \right) + C \cr} $$
