Does the recurrence $I_n=\frac{n-1}nI_{n-2}$ for $I_n=\int_0^{\pi/2}\cos^nx\,dx$ have anything to do with Fourier series?

Question

I was finding the recurrence relation for the following integral: $$I_n = \int_{0}^{\pi/2}\cos^n(x)\,dx$$ I derived the following relation: $$I_{n} = \frac{n-1}{n}\cdot I_{n-2}$$

What I than did was to plot the $I_n$ as the function of $n$. At first I didn't restrict $n$ to be natural number, which is not what is assumed in those types recurrence relations, as far as I was taught. This resulted in the graph shown at this Desmos link, and a function $I(n)$ that looked like some kind of irrational function, at least from $x\in (-1,+\infty)$.

Immediately, this reminded me of the Fourier series, but that's as far as I went with my knowledge. I am wondering

Is my assumption that this has to do something with the Fourier series right, and if so, is there any method that could be used to get a more closed-form expression of the function that was transformed?

I wouldn't expect that to be true for every function, but having even a few examples of something like that could be interesting to me.

To recap, my question is:

Does this have anything to do with the Fourier transform, and if so, could the transform be reversed, in such a way that we get the original function back?

Thanks to everyone on their answers in advance!

Answer 1

Note that

$$I_n=\int\limits_0^{\frac{\pi}{2}}\cos^n(x)\,dx=\frac{\sqrt{\pi}}{2}\ \frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}+1\right)}\ ,\quad\Re(n)>-1.\tag{1}$$

So

$$I_n=\frac{n-1}{n}\ I_{n-2}\tag{2}$$

follows from the properties of the gamma function.