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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\in {\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}}

Complex Variables

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