# CV7

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Evaluate $\int_0^{2\pi}\cos^8\theta\,d\theta\,$

Write $\cos\theta = \left(\frac{e^{i\theta}+e^{-i\theta}}{2}\right)\,$.

$\int_0^{2\pi} \left(\frac{e^{i\theta}+e^{-i\theta}}{2}\right)^8\,d\theta\,$

The binomial series is $(a+b)^n = \sum_{k=0}^n {n\choose k} a^k b^{n-k}\,$

$\frac{1}{2^8}\int_0^{2\pi}\sum_{k=0}^8{8\choose k}e^{ik\theta}e^{-i(8-k)\theta}\,d\theta = \frac{1}{2^8}\sum_{k=0}^8{8\choose k}\int_0^{2\pi}e^{i(2k-8)\theta}\,d\theta\,$

$\frac{1}{2^8} \sum_{k=0}^8 {8\choose k} \frac{e^{i(2k-8)\theta}}{i(2k-8)}\Bigg|_{\theta=0}^{2\pi}=\frac{1}{2^8}\sum_{k=0}^8 {8\choose k} \frac{1}{i(2k-8)}\left(e^{i(2k-8)2\pi}-1\right)\,$

The exponential is equal to $1\forall k\isin\mathbb{Z}\,$, so every term in the sum is zero for all valid $k\,$.

But when $k=4\,$ there is a zero in the denominator of the summand, so this term has to be evaluated seperately; let $k=4\,$ in the integral above.

$\frac{1}{2^8} {8\choose 4}\int_0^{2\pi}\,d\theta\,=\frac{70}{256}2\pi=\frac{35\pi}{64}$

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