Running through math confusion how to finish this integral using Beta function. I am working on this easy integral but I want to use Beta Trigonometric Function to solve it. I know the answer is $\pi$ by using the half angle formula and help from u-substitution.
$\int_{0}^{2\pi} \cos^{2}(2x) dx $
So what I did was again used u-substitution which makes my integral and limit bounds to change $ \frac{1}{2} \int_{0}^{4\pi} \cos^{2}(u) du $.
My confusing is where do I go from here so Beta trigonometric form can be used and evaluate. I know the formula of Beta function involves trigonometric function $B(x,y) = \int_{0}^{\frac{\pi}{2}} 2 \cos^{2x-1}(\theta)\sin^{2y-1} (\theta) d\theta $.
 A: Using the Beta Integral
$$\newcommand{\B}{\operatorname{B}}
\begin{align}
\int_0^{2\pi}\cos^2(2x)\,\mathrm{d}x
&=\frac12\int_0^{4\pi}\cos^2(x)\,\mathrm{d}x\tag{1a}\\
&=2\int_0^{\pi}\cos^2(x)\,\mathrm{d}x\tag{1b}\\
&=4\int_0^{\pi/2}\cos^2(x)\,\mathrm{d}x\tag{1c}\\[6pt]
&=2\B\left(\tfrac12,\tfrac32\right)\tag{1d}\\[12pt]
&=\pi\tag{1e}
\end{align}
$$
Explanation:
$\text{(1a)}$: substitute $x\mapsto x/2$
$\text{(1b)}$: $\cos^2(x)=\cos^2(x+\pi)$ (periodicity)
$\text{(1c)}$: $\cos^2(x)=\cos^2(\pi-x)$ (symmetry)
$\text{(1d)}$: $\B(x,y)=2\int_0^{\pi/2}\sin^{2x-1}(\theta)\cos^{2y-1}(\theta)\,\mathrm{d}\theta$
$\text{(1e)}$: evaluate

A Simpler Approach
$$
\begin{align}
\int_0^{2\pi}\cos^2(2x)\,\mathrm{d}x
&=\frac12\int_0^{4\pi}\cos^2(x)\,\mathrm{d}x\tag{2a}\\
&=\frac12\int_{\pi/2}^{9\pi/2}\sin^2(x)\,\mathrm{d}x\tag{2b}\\
&=\frac12\int_0^{4\pi}\sin^2(x)\,\mathrm{d}x\tag{2c}\\
&=\frac14\int_0^{4\pi}\left(\sin^2(x)+\cos^2(x)\right)\,\mathrm{d}x\tag{2d}\\[6pt]
&=\pi\tag{2e}
\end{align}
$$
Explanation:
$\text{(2a)}$: substitute $x\mapsto x/2$
$\text{(2b)}$: $\cos^2(x)=\sin^2\left(x+\frac\pi2\right)$
$\text{(2c)}$: $\sin^2(x+4\pi)=\sin^2(x)$
$\text{(2d)}$: average $\text{(2a)}$ and $\text{(2c)}$
$\text{(2e)}$: $\sin^2(x)+\cos^2(x)=1$
