- #1

- 124

- 0

## Homework Statement

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}]

## Homework Equations

de moivre -> (cos(nx) + i*sin(nx)) = (cos(x) + i*sin(x))

^{n}

eulers -> (cos(x) + i*sin(x)) = e

^{ix}

## The Attempt at a Solution

okay ive tried not using de moivres and trying to find a pattern by doing the integral a few times, and ive also thought that using eulers formula might help, but i seriously just cant get started...i know that i need to use the real part of the formula but i dont know what to do first.

this is a past exam paper by the way and my uni doesnt 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

thanks