Find $\lim_{n\to \infty} \left [\tan \frac{\pi}{2n}\tan \frac{2\pi}{2n}..\tan \frac{n\pi}{2n}\right]^{1/n}$ $$\ln y = \lim _{n\to \infty} \frac 1n \left [\ln (\tan \frac{\pi}{2n})+....\ln (\tan \frac{n\pi}{2n}) \right]$$
$$\ln y = \int _0^1 \ln (\tan \frac{\pi x}{2}).dx$$
Now this integral is a pain to solve, and online calculators aren’t able to solve it, so is this method even correct? If yes, then how do I solve the integral, and if not, what is the correct way?
 A: $$I=\int_0^1\ln (\tan \frac{\pi (1-x)}{2})dx=\int_0^1\ln (\cot \frac{\pi x}{2})dx$$
$$I+I=\int_0^1\ln (\cot \frac{\pi x}{2})dx+\int_0^1\ln (\tan \frac{\pi x}{2})dx=\int_0^1\ln (1)dx=0$$
A: Hint: Assuming the final $\tan(\frac{n\pi}{2n})$ isn't there (that being $\tan\frac{\pi}{2}=\infty$), just use $\tan(\frac{\pi}{2}-\theta)=\cot\theta$
A: Too long for a comment.
SInce you already received good answer and comments, let me address the problem of the antiderivative about which you write
"this integral is a pain to solve, and online calculators aren’t able to solve it"
In fact, it is quite simple : let $\tan \left(\frac{\pi  x}{2}\right)=t$
$$I=\int \log (\tan \frac{\pi x}{2})\,dx=\frac 2\pi \int \frac{\log (t)}{t^2+1}\,dt$$
$$\frac{1}{t^2+1}=\frac{1}{(t+i)(t-i)}=\frac i 2\left(\frac{1}{t+i}-\frac{1}{t-i} \right)$$
$$I=\frac i \pi\left(\int \frac{\log(t)}{t+i}dt-\int \frac{\log(t)}{t-i}dt \right)$$
$$I=\frac i \pi\left(\Big[\text{Li}_2(i t)+\log (1-i t) \log (t)\Big]-\Big[\text{Li}_2(-i t)+\log (1+i t) \log (t)\Big]\right)$$
$$I=\frac 1 \pi\Big[2 \log (t) \tan ^{-1}(t)+i (\text{Li}_2(i t)-\text{Li}_2(-i t)) \Big]$$
