TITLE: Putnam Exam Integral: $\lim_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \cos^2\big(\frac{\pi}{2n}(x_1+x_2+...x_n)\big)dx_1 dx_2...dx_n$
QUESTION [18 upvotes]: I am trying to evaluate$$
\lim_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \cos^2\big(\frac{\pi}{2n}(x_1+x_2+...x_n)\big)dx_1 dx_2...dx_n.
$$
This is from an old Putnam mathematics competition. Either 1965 or 1987 I forget. Should we re-write the $\cos^2$ term first or how should we approach it?  Thanks

REPLY [4 votes]: $\displaystyle \lim_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \cos^2\big(\frac{\pi}{2n}(x_1+x_2+...x_n)\big)dx_1 dx_2...dx_n $
$\displaystyle = \lim_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \cos^2\big(\frac{\pi}{2n}(1-x_1+1-x_2+...+1-x_n)\big)dx_1 dx_2...dx_n $
$\displaystyle = \lim_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \sin^2\big(\frac{\pi}{2n}(x_1+x_2+...x_n)\big)dx_1 dx_2...dx_n$
$\displaystyle = \frac12 \lim_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \sin^2\big(\frac{\pi}{2n}(x_1+x_2+...x_n)\big) + \cos^2\big(\frac{\pi}{2n}(x_1+x_2+...+x_n)\big)dx_1 dx_2...dx_n$
$=\dfrac12$