So we are trying to integrate the following expression $~~~\rightarrow ~~~ \displaystyle\int \cos^{3} (2x)\ dx$.
To do the this, we will need to make an appropriate substitution inside of the integrand. Doing this leads us to the following:
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\displaystyle\int \cos^{3} (2x)\ dx$
Let: $~u =2x$
$du=2\ dx$
$dx=\dfrac{1}{2}\ du$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\dfrac{1}{2}\displaystyle\int \cos^{3} (u)\ du$
Using the reduction formula, $\int \cos^{m}(u) du = \dfrac{1}{m} \cos^{m-1}(u) \sin (u) + \dfrac{m-1}{m} \int \cos^{m-2}(u)\ du,~ \text{where }~ m = 3,~\text{gives}:$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{2}\Bigg[\dfrac{1}{3} \cos^{2}(u) \sin (u) + \dfrac{2}{3} \displaystyle\int \cos (u)\ du \Bigg]$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=~\dfrac{1}{6} \cos^{2} (2x) \sin (2x) + \dfrac{1}{3} \sin (2x) + C~~~~~~~~~~~~~~~~~~~~~~~~~~~~\blacksquare$
Which can be simplified further to this:
$\dfrac{1}{24}\Bigg(9 \sin (2x) + \sin (6x)\Bigg) + C.$
Okay, I hope that this has helped out and now you see where the $\dfrac{1}{2}$ came from. Let me know if there is any step covered that did not make much sense for doing so.
Thanks.
Good Luck.