# 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 $$B(x,y) = \int_{0}^{\frac{\pi}{2}} 2 \cos^{2x-1}(\theta)\sin^{2y-1} (\theta) d\theta$$.

• Do you know the formula for the Beta function that involves trigonometric functions?
– robjohn
Jan 14 at 0:11
• Could you remind us what your definition of $\beta$ is? Jan 14 at 0:19
• From symmetry and periodicity conditions on the function $\cos^2(u)$ you should be able to get $\int_0^{4\pi} \cos^2(u) du = 8 \int_0^{\pi/2} \cos^2(u) du$. Jan 14 at 0:23
• @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.
– EM4
Jan 14 at 0:31
• What do you get for $\operatorname{B}\left(\frac12,\frac32\right)$?
– robjohn
Jan 14 at 1:55

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