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Obviously, <math>\frac{1-\cos{x}}{2} = \sin^2{\frac{x}{2}}</math>
 
Obviously, <math>\frac{1-\cos{x}}{2} = \sin^2{\frac{x}{2}}</math>
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 +
So, the integral should then look like:
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 +
<math>\int_{0}^{2Pi}\sin{\frac{x}{2}}dx</math>

Revision as of 12:17, 19 October 2008

This one looks pretty easy, but I keep getting the wrong answer. Here's the problem.

$ \int_{0}^{2Pi}\sqrt{\frac{1-\cos{x}}{2}}dx $

Obviously, $ \frac{1-\cos{x}}{2} = \sin^2{\frac{x}{2}} $

So, the integral should then look like:

$ \int_{0}^{2Pi}\sin{\frac{x}{2}}dx $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva