The integral $\int \cos^3(2x) \, \mathrm dx$ Given the following problem:

integrate $\cos^3(2x)$

I was given the solution 

\begin{align*}
  \int\cos^3(2x)\, \mathrm dx &= \int\cos^2(2x) \cos 2x\, \mathrm dx = \int(1-\sin^2 2x)\cos 2x\, \mathrm dx\\
  &= \frac12\int (1-u^2)\, \mathrm du = \cdots
  \end{align*}

but the problem is I am stuck on the 1/2. Where did it come from?
 A: When the substitution is made, $u=\sin 2x$, so $du=2\cos 2x\;dx$, or $\frac{1}{2}du=\cos 2x\;dx$.
A: 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.
