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How to calculate:

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

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You could use the Pythagorean identity to simplify your integrand... –  Guess who it is. May 17 '11 at 16:13
The identity $1 - \sin^2{\theta} = \cos^2{\theta}$ should prove useful =) –  Adrián Barquero May 17 '11 at 16:14
Additionally, you could reduce it to the integral $$2\int_{\pi/2}^{3\pi/2}\sqrt{1-\sin^2\theta}\mathrm d\theta$$ –  Guess who it is. May 17 '11 at 16:15
More elaborate hint: don't blindly use the Pythagorean identity. Plot the integrand and then apply the identity accordingly. –  Guess who it is. May 17 '11 at 16:20
@J.M.: Is the potential pitfall blindly using the trig identity, or thinking that $\sqrt{x^2}=x$ rather than $\sqrt{x^2}=|x|$? –  Isaac May 17 '11 at 22:22

4 Answers 4

up vote 11 down vote accepted

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

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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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The integral of $\cos{\theta}$ is $\sin{\theta}$ not $-\sin{\theta}$ :) –  Osama Gamal May 17 '11 at 16:33
@Osama: Sorry. I have rectified it now. Thanks for pointing out. –  user9413 May 17 '11 at 16:34

\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}

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


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Great tool, thanks for the tip –  bobobobo Jan 28 '12 at 17:23

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