Integrate $\int \sin (2x) \cos (2x)\,dx$ I have
$$\int \sin(2x) \cos (2x)\,dx = \frac12 \int \sin(4x)\,dx = -\frac18 \cos(4x),$$
but I also have
$$\int \sin(2x) \cos (2x)\,dx = \frac12 \int \sin 2x \cdot 2 \cos 2x \, dx = \frac14 \sin^2(2x).$$
Which one is correct, and why is the other method wrong?

To illustrate the phenomena in your calculation, I've chosen the number $\pi/8$ as the lower boundary.
\begin{align*}
  \int_{\frac{\pi}{8}}^{x} \sin(4t) \,\mathrm{d}t &= \int_{\frac{\pi}{8}}^{x} 2 \sin(2t) \cos(2t) \,\mathrm{d}t \\
  &= \int_{\frac{\pi}{8}}^{x} \sin(2t) \,\mathrm{d}(\sin(2t)) \\
  \left . -\frac{\cos(4t)}{4} \right\rvert_{\frac{\pi}{8}}^{x} &= \left . \frac{\sin^2(2t)}{2} \right\rvert_{\frac{\pi}{8}}^{x} \\
  -\frac{\cos(4x)}{4} + \color{red}{\frac{\cos(4\cdot\frac{\pi}{8})}{4}} &= 
  \frac{\sin^2(2x)}{2} - \color{red}{\frac{\sin^2(2\cdot\frac{\pi}{8})}{2}} \\
  -\frac{\cos(4x)}{4} + \color{red}{0} &= \frac{\sin^2(2x)}{2} + \color{red}{\frac{1}{4}}
\end{align*}
