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How to calculate:

$ \int_0^{2\pi} \sqrt{1 - \sin^2 \theta}\;\mathrm d\theta $

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    @Isaac: Apparently I was not thinking straight writing that last comment; so yes, the second one.2011-06-02

4 Answers 4

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Hint: note that $\sqrt{1-\sin^2\theta}=\sqrt{\cos^2\theta}=|\cos\theta|.$

2

Use the fact that $\cos^{2}\theta = 1-\sin^{2}\theta$ and the fact that integral of $\cos\theta$ is $\sin\theta$. Also $\sqrt{1-\sin^{2}\theta} = |\cos{\theta}|$. And note that $\cos\theta$ is positive in the first and the fourth quadrant.

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    @Osama: Sorry. I have rectified it now. Thanks for pointing out.2011-05-17
1

Use Wolfram Alpha! Plug in "integrate sqrt(1-sin^2(x))". Then press "show steps". You can enter the bounds by hand...

http://www.wolframalpha.com/input/?i=integrate+sqrt%281-sin%5E2%28x%29%29

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    Great tool, thanks for the tip2012-01-28
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\begin{align} \int_0^{2\pi} \sqrt{1 - \sin^2 \theta} d\theta &= \int_0^{2\pi} \sqrt{\cos^2 \theta} d\theta \\ &= \int_0^{2\pi} | \cos \theta | d\theta \\ &= 4 \int_0^{\frac{pi}{4}} \cos \theta d\theta \\ &= 4 \end{align}