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 $.
Best Answer
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 $\text{(2c)}$
$\text{(2e)}$: $\sin^2(x)+\cos^2(x)=1$