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Evaluate the following integral

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\[\int\limits_{0}^{2\pi} (\sqrt{\frac{1}{3\pi}}+ \sqrt{\frac{1}{6\pi}}*e^{ix}) (\sqrt{\frac{1}{3\pi}}+ \sqrt{\frac{1}{6\pi}}*e^{-ix}) dx\]
The integral should come out to be 1 because \[\sqrt{\frac{1}{6\pi}}e^{ix}*\sqrt{\frac{1}{6\pi}}e^{-ix}=\frac{1}{6\pi}\] and the ingetegral between 0 and 2pi will be 1/3 and if we do the same for \[\sqrt{\frac{1}{3\pi}}*\sqrt{\frac{1}{3\pi}}=\frac{1}{3\pi}\] ad the integral between 0 and 2pi for that is 2/3 I don't understand how the other stuff goes to zero when evaluating the integral.
The integrated result is: \[\frac{x}{3\pi}+\frac{x}{6\pi}+\sqrt{\frac{1}{18\pi ^{2}}}\times \frac{1}{i}e ^{ix}+\sqrt{\frac{1}{18\pi ^{2}}}\times-\frac{1}{i}e ^{-ix}\] The last two terms cancel the reason being that: \[e ^{x}=e ^{-x}\] By substitution the value of the definite integral = 1 unit

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Oh I see it now thanks!!
You're welcome :)

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