Closed form of $\int\limits_0^{2\pi} \prod\limits_{j=1}^n \cos(jx)dx$ and combinatorial link

Hint. From $$\int\limits_{\gamma}f(z)dz=\int\limits_{a}^{b}f(\gamma(t))\gamma'(t)dt$$ with a transformation like $$\int\limits_0^{2\pi} \prod_{j=1}^n \cos(jx)dx= \int\limits_0^{2\pi} \prod_{j=1}^n \frac{e^{i j x}+e^{-i j x}}{2}dx=\\ \int\limits_0^{2\pi} \frac{1}{2^n} \cdot\prod_{j=1}^n \frac{1}{e^{i j x}} \cdot \prod_{j=1}^n \left(e^{2i j x}+1\right)dx=\\ \int\limits_0^{2\pi} \frac{1}{2^n} \cdot \frac{1}{e^{i \frac{n(n+1)}{2} x}} \cdot \prod_{j=1}^n \left(e^{2i j x}+1\right)dx=\\ \int\limits_0^{2\pi} \frac{1}{i2^n} \cdot \frac{1}{e^{i \frac{n(n+1)}{2} x+ix}} \cdot \left(\prod_{j=1}^n \left(e^{2i j x}+1\right)\right) \cdot ie^{i x}dx=\\ \int\limits_{|z|=1}\frac{1}{i2^n} \cdot \frac{1}{z^{\frac{n(n+1)}{2}+1}} \cdot \prod_{j=1}^n \left(z^{2 j }+1\right)dz=...$$ and noting $f(z)=\prod\limits_{j=1}^n \left(z^{2 j }+1\right)$, we have $$...=\frac{1}{i2^n} \int\limits_{|z|=1}\frac{f(z)}{z^{\frac{n(n+1)}{2}+1}} dz=...$$ recalling Cauchy's integral formula this is $$...=\frac{1}{i2^n}\cdot \frac{2 \pi i}{\left(\frac{n(n+1)}{2}\right)!}\cdot f^{\left(\frac{n(n+1)}{2}\right)}(0)= \frac{\pi}{2^{n-1}} \cdot \frac{1}{\left(\frac{n(n+1)}{2}\right)!} \cdot f^{\left(\frac{n(n+1)}{2}\right)}(0)$$

The integral is equal to $\frac{A_n\pi}{2^{n-1}}$ where $A_n$ is the number of subsets of $\{1,2,3,\dots,n\}$ whose sum is $\frac{n(n+1)}4.$

In particular, if $n\equiv 1,2\pmod 4,$ the $\frac{n(n+1)}4$ is not an integer, so there can be no such aubsets, so the integral is zero in that case.

You get this value by representing $\cos nx =\frac12 \left(e^{inx}+e^{-inx}\right)$ and realize the integral is zero for all terms expanding the produc, except for the constant term, which has coefficient $\frac{A_n}{2^n}.$

I don’t think there is an easy way to represent this term. There is an upper bound $\binom n{\lfloor n/2\rfloor}.$

See this answer for details. Compute $\lim\limits_{n\to\infty} \int_{0}^{2\pi} \cos x \cos 2x\cdots \cos nx \space{dx}$