How to calculate:
$ \int_0^{2\pi} \sqrt{1 - \sin^2 \theta}\;\mathrm d\theta $
How to calculate:
$ \int_0^{2\pi} \sqrt{1 - \sin^2 \theta}\;\mathrm d\theta $
Hint: note that $\sqrt{1-\sin^2\theta}=\sqrt{\cos^2\theta}=|\cos\theta|.$
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.
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
\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}