Answer
$\frac{1}{4}\sin^2{2x}+C$
Work Step by Step
$\int\sin{2x}\cos{2x}dx$
Method 1:
Let $u=\sin{2x}\hspace{8mm}du=2\cos{2x}dx\hspace{8mm}dx=\frac{du}{2\cos{2x}}$
Therefore
$\int\sin{2x}\cos{2x}dx=\int u\cos{2x}\frac{du}{2\cos{2x}}=\frac{1}{2}\int udu$
$=\frac{1}{2}\frac{u^2}{2}+C=\frac{1}{4}\sin^2{2x}+C$
