Integration using complex numbers (Euler's theorem)

[zzinfinity]

Hi,
I'm stuck on a homework question.
[FONT=Verdana, Arial, Tahoma, Calibri, Geneva, sans-serif]Calculate Integral of Cos(x)^8 from 0 to 2pi by writing cos(x) as .5(e^ix+e^-1x)

No idea where to start after the substitution. Any input would be greatly appreciated. Thanks!
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[galactus]

You could rewrite as:

\(\displaystyle \frac{1}{256}\int_{0}^{2\pi}(e^{ix}+e^{-ix})^{8}dx\)

Let \(\displaystyle t=ix, \;\ dt=idx, \;\ -idt=dx\).

Expand out using the binomial expansion:

\(\displaystyle \displaystyle \frac{-i}{256}\int_{0}^{2\pi i}\left[e^{8t}+8e^{6t}+28e^{4t}+56e^{2t}+56e^{-2t}+28e^{-4t}+8e^{-6t}+e^{-8t}+70\right]dt\)

Integrate term-by-term.

i.e. \(\displaystyle \displaystyle \frac{-i}{256}\int_{0}^{2\pi i}e^{8t}dt=0\)

Note that all evaluate to 0 except the end term, \(\displaystyle \displaystyle\frac{-i}{256}\int_{0}^{2\pi i}70dt\).

This is because \(\displaystyle e^{2k\pi i}=1, \;\ e^{0}=1\)

Therein lies your solution.