Calculate
$$\lim\limits_{n\to\infty}\left(\prod_{k=1}^{n-1}\cos\left(\frac{k\pi}{2n}\right)\right)^{\frac{1}{n}}$$
The following words are my answer:
$$=\exp\left(\lim{\sum{\ln\left(\cos\left(\frac{k\pi}{2n}\right)\right)^\frac{1}{n}}}\right)=\exp\left(\int_{0}^{1}{\ln\left(\cos\left(\frac{\pi}{2}x\right)\right)}\right)$$
But I can not solve this improper integral. Is there any idea about solving integral like this?
What is more, how to do it without using the definition of integral?
We have that
$$\log\left[\left(\prod_{k=1}^{n-1}\cos\left(\frac{k\pi}{2n}\right)\right)^{\frac{1}{n}}\right]=\frac1n\sum_{k=1}^{n-1}\log\left(\cos\left(\frac{\pi}{2}\frac{k}{n}\right)\right)=\int_0^1 \log\left(\cos\left(\frac \pi 2 x\right)\right)dx$$
which, as noticed in the comments, by $u=\frac{\pi}2x$ reduces to the well-known integral
$$\boxed{L=\frac 2\pi \int_0^{\frac \pi 2} \log\left(\cos u\right)du=\log \frac12}$$
As an alternative solution, we can avoid integrals using that by symmetry
$$C_n=\sum_{k=1}^{n-1}\log\left(\cos\left(\frac{\pi}{2}\frac{k}{n}\right)\right)=\sum_{k=1}^{n-1}\log\left(\sin\left(\frac{\pi}{2}\frac{k}{n}\right)\right)=S_n$$
and then
$$2\frac{C_n}n=\frac{C_n}n+\frac{S_n}n=\log\frac12+\frac1n\sum_{k=1}^{n-1}\log\left(\sin\left(\pi\frac{k}{n}\right)\right)$$
and assuming wlog $n=2m$
$$\frac1n\sum_{k=1}^{n-1}\log\left(\sin\left(\pi\frac{k}{n}\right)\right)=\frac1{2m}\sum_{k=1}^{2m-1}\log\left(\sin\left(\frac \pi 2\frac{k}{m}\right)\right)=\frac1m\sum_{k=1}^{m-1}\log\left(\sin\left(\frac \pi 2\frac{k}{m}\right)\right)=\frac{C_m}m$$
finally by $\frac{C_n}n \to L$ we have
$$2L=\log \frac12 + L \implies \boxed{L=\log \frac12}$$