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## Romero Group Title Evaluate the following integral 2 years ago 2 years ago

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1. Romero

$\int\limits_{0}^{2\pi} (\sqrt{\frac{1}{3\pi}}+ \sqrt{\frac{1}{6\pi}}*e^{ix}) (\sqrt{\frac{1}{3\pi}}+ \sqrt{\frac{1}{6\pi}}*e^{-ix}) dx$

2. Romero

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.

3. kropot72

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

4. Romero

Oh I see it now thanks!!

5. kropot72

You're welcome :)