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.
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.
RE: Can you solve this integral from MIT's Integration Bee?