$\displaystyle
\int x sin2x cos2x dx
$
i started of by making u=2x, du=x, which will cancel out the x and give me
$\displaystyle
\int sinucosu du
$
how would i go from here
if u=2x then du=2dx not x sorry
try this trig fomula
$\displaystyle 2\sin(\theta)\cos(\theta)=\sin(2\theta)$
so you would get
$\displaystyle \sin(2x)\cos(2x)=\frac{1}{2}\sin(4x)$
now we have the integral
$\displaystyle \frac{1}{2}\int x \sin(4x)dx$
by parts we get...
$\displaystyle \frac{1}{2}\int x \sin(4x)dx=\frac{1}{2}[-\frac{x}{4}\cos(4x)+\frac{1}{16}\sin(4x)]=-\frac{x}{8}\cos(4x)+\frac{1}{32}\sin(4x)+C$