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Integral of Sin


\int\limits_\0^\pi \sin\theta \; \mathrm{d}\theta

edited Jun 14 '19 at 05:38 Asked Jun 13 '19 at 15:52 by Bob


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\int\limits_\0^\pi sin\theta \; \mathrm{d}\theta \;=\; -\cos\theta|\limits_\0^\pi = \;-cos(\pi) - (-cos0)= 1- (-1) =2

asked Jun 14 '19 at 05:34 by Bob

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