$\begin{align}  & \int{\sin x\cos xdx}=\frac{1}{2}\int{\sin 2xdx}=\frac{1}{2}\left( \frac{-\cos 2x}{2} \right)+C \\ & \int{\sin x\cos xdx}=-\frac{1}{4}\cos 2x+C \\\end{align}$