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?
 A: They are differing by an integration constant, because of $\cos(2 y) = \cos^2(y) - \sin^2(y) = 1 - 2 \sin^2(y)$, and hence are the same as indefinite integration produces an anti-derivative up to a constant
A: 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*}
A: In second method lost step it is not
$$\int \sin(2x) \cos (2x)\,dx = \frac12 \int \sin 2x \cdot 2 \cos 2x \, dx = \frac14 \sin^2(2x).$$
It has to be
(since 2sin(2x). cos(2x)=sin(4x)) 
$$\int \sin(2x) \cos (2x)\,dx = \frac12 \int \sin 2x \cdot 2 \cos 2x \,dx = \frac12 \int \sin(4x) dx=-\frac18\cos(4x)$$
(Since $$\int \sin(4x)dx=-\frac14 \cos(4x).$$) 
