You are viewing a single comment's thread from:
RE: Can you solve this integral from MIT's Integration Bee?
The key step is using
sin(2x) = 2.sinx.cosx
so that
cos(x) = sin(2x)/2sin(x)
and by extension
cos(nx) = sin(2nx)/2sin(nx)
when plugged into the integral product, all the sines cancel out apart from the first and last terms.
Beautiful