I don't understand the circled part. Since U uses up the 2x in front of the cos2x how come a DU is still present. Also why is the 1/2 there?
$\displaystyle \int \textcolor{red}{\sin^2(2x)} \textcolor{blue}{\cos(2x) \, dx}$
$\displaystyle \textcolor{red}{u = \sin(2x)}$
$\displaystyle \textcolor{blue}{du = 2\cos(2x) \, dx}$
$\displaystyle \textcolor{blue}{\frac{1}{2}} \int \textcolor{red}{\sin^2(2x)} \textcolor{blue}{2\cos(2x) \, dx}$
note that the $\displaystyle \frac{1}{2}$ on the outside of the integral compensates for the 2 inserted in the integrand
$\displaystyle \textcolor{blue}{\frac{1}{2}} \int \textcolor{red}{u^2} \textcolor{blue}{du}$
see why now?