Answer
$$\frac{1}{2}{\sin ^2}x + C$$
Work Step by Step
$$\eqalign{
& \int {\sin x\cos x} dx \cr
& {\text{Integrate by parts,}} \cr
& {\text{Let }}u = \sin x,{\text{ }}du = \cos xdx, \cr
& dv = \cos xdx,{\text{ }}v = \sin x \cr
& \int {\sin x\cos x} dx = \left( {\sin x} \right)\left( {\sin x} \right) - \int {\sin x\left( {\cos x} \right)} dx \cr
& \int {\sin x\cos x} dx = {\sin ^2}x - \int {\sin x\cos x} dx \cr
& 2\int {\sin x\cos x} dx = {\sin ^2}x \cr
& {\text{Solve for }}\int {\sin x\cos x} dx \cr
& \int {\sin x\cos x} dx = \frac{1}{2}{\sin ^2}x + C \cr
& \cr
& {\text{Integrate by substitution}} \cr
& {\text{Let }}u = \sin x,{\text{ }}du = \cos xdx, \cr
& \int {\sin x\cos x} dx = \int {udu} \cr
& = \frac{1}{2}{u^2} + C \cr
& {\text{Back - substitute }}u = \sin x \cr
& = \frac{1}{2}{\sin ^2}x + C \cr
& \cr
& {\text{The results are the same}} \cr} $$
