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 $.

  • 1
    $\begingroup$ Do you know the formula for the Beta function that involves trigonometric functions? $\endgroup$
    – robjohn
    Jan 14 at 0:11
  • $\begingroup$ Could you remind us what your definition of $\beta$ is? $\endgroup$
    – Alex Ortiz
    Jan 14 at 0:19
  • $\begingroup$ From symmetry and periodicity conditions on the function $\cos^2(u)$ you should be able to get $\int_0^{4\pi} \cos^2(u) du = 8 \int_0^{\pi/2} \cos^2(u) du$. $\endgroup$ Jan 14 at 0:23
  • $\begingroup$ @DanielSchepler this is what I had on my scrap paper but it gave me $2\pi$ not $\pi$, will this work? $\frac{1}{2}*4B(\frac{3}{2} , \frac{1}{2})$ because I factored out 8 for 2 times 4. $\endgroup$
    – EM4
    Jan 14 at 0:31
  • $\begingroup$ What do you get for $\operatorname{B}\left(\frac12,\frac32\right)$? $\endgroup$
    – robjohn
    Jan 14 at 1:55

1 Answer 1


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


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