Use de Moivre's Theorem and the result that integral (with limits pi, 0) of cos[nx]dx = 0 for all non-zero integers of n. when n does equal 0 the answer is 1.
[this i proved in the first section of the qu using logic and limits(x->0).
so we need to prove that
integral[with limits pi, 0] [(cos(x))2n]dx = [pi / (2)2n] * [(2n)! / (n!)2]
de moivre -> (cos(nx) + i*sin(nx)) = (cos(x) + i*sin(x))n
eulers -> (cos(x) + i*sin(x)) = eix
The Attempt at a Solution
okay I've tried not using de moivres and trying to find a pattern by doing the integral a few times, and I've also thought that using eulers formula might help, but i seriously just can't get started...i know that i need to use the real part of the formula but i don't know what to do first.
this is a past exam paper by the way and my uni doesn't issue mark schemes, plus it being bank hols weekend no one will be there until wednesday
i appreciate any help or even just ideas on how to get started