Closed form for $\int_0^{\pi/2}(\cos^2{\theta})^{1/n}d\theta$

Question

Is there a closed form for $f(n)=\int_0^{\pi/2}(\cos^2{\theta})^{1/n}d\theta$ for every $n\in\mathbb{N}$ ?

I suspect there may be, because of the following apparent equalities.

$f(1)=\frac{\pi}{4}$

$f(2)=1$

$f(3)\overset{?}{=}\dfrac{2^{1/3}\sqrt3 \pi^2}{\left(\left(\Gamma(\frac13\right)\right)^3}$

$f(4)\overset{?}{=}\dfrac{(2\pi)^{3/2}}{\left(\Gamma\left(\frac14\right)\right)^2}$

$f(5)= ?$ (I haven't found a possible closed form yet.)

$f(6)\overset{?}{=}\dfrac{3 \Gamma\left(\frac13\right)\Gamma\left(\frac56\right)}{2^{1/3}\Gamma\left(\frac16\right)}$

For $f(3), f(4), f(6)$, the two sides are equal to at least $11$ decimal places. I found the rightside expressions using desmos, wolfram and OEIS.

Context

I wanted to find the locus of point $P$ such that $\prod\limits_{k=1}^n PV_k=1$, where $V_1,\dots,V_n$ are the vertices of a regular $n$-gon of radius $1$. Assume that the centre of the $n$-gon is $(0,0)$, and one vertex is $(1,0)$.

The locus turns out to be the polar curve $r=(2\cos{n\theta})^{1/n}, r\ge 0$. (proof below)

The graph of $r=(2\cos{n\theta})^{1/n}$ is a flower with $n$ petals. For example, here is the graph with $n=5$, with the vertices of the $n$-gon in red.

The total area of the flower can be expressed as $A(n)=\int_0^{\pi/2}(4\cos^2{n\theta})^{1/n}d\theta=4^{1/n}f(n)$.

Does this area have a closed form for every $n\in\mathbb{N}$ ?

Proof that the locus is $r=(2\cos{n\theta})^{1/n}$:

$\prod\limits_{k=1}^n \left(\left(x-\cos{\left(\frac{2k\pi}{n}\right)}\right)^2+\left(y-\sin{\left(\frac{2k\pi}{n}\right)}\right)^2\right)=1$

$\prod\limits_{k=1}^n \left(\left(r\cos{\theta}-\cos{\left(\frac{2k\pi}{n}\right)}\right)^2+\left(r\sin{\theta}-\sin{\left(\frac{2k\pi}{n}\right)}\right)^2\right)=1$

$\prod\limits_{k=1}^n \left(r^2-2r\cos{\left(\theta-\frac{2k\pi}{n}\right)}+1\right)=1$

$\prod\limits_{k=1}^n \left(r-e^{\left(\theta-\frac{2k\pi}{n}\right)i}\right)\left(r-e^{-\left(\theta-\frac{2k\pi}{n}\right)i}\right)=1$

$(r^n-(\cos{n\theta}+i\sin{n\theta}))(r^n-(\cos{(-n\theta)}+i\sin{(-n\theta})))=1$

$r^{2n}-2r^n \cos{n\theta}=0$

$\therefore r=(2\cos{n\theta})^{1/n}$

Answer 1

Use beta function:

$$B(a,b)=2\int_0^{\pi/2} (\sin\theta)^{2a-1}(\cos\theta)^{2b-1}d\theta$$

So you get:

$$\int_0^{\pi/2}(\cos^2{\theta})^{1/n}d\theta=\frac{1}2 B\left(\frac{1}2,\frac{1}2+\frac{1}n\right)=\frac{n\sqrt\pi}{2}\cdot\frac{\Gamma\left(\frac{1}2+\frac{1}n\right)}{\Gamma\left(\frac{1}n\right)}$$