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I was trying to determine the arc length of the polar curve $r = f(\theta) = a(1 - \cos \theta)$, and it was going well until I got to the definite integral. I know that $f'(\theta) = a \sin \theta$, so the integral would be

$$\int_0^{2 \pi} \sqrt{(a(1-\cos \theta))^2+(a(\sin \theta))^2} d \theta$$

I can simplify until I get to the point

$$a\int_0^{2 \pi} \sqrt{1 - 2 \cos \theta} d \theta$$

At which point I get stuck. Any hints/suggestions?

Thanks!

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Calculate carefully :

$$\int_0^{2 \pi} \sqrt{(a(1-\cos \theta))^2+(a(\sin \theta))^2} d \theta$$

$$=\sqrt 2a\int_0^{2\pi}\sqrt{1-\cos \theta}\,d\theta$$

$$= 2a \int_0^{2\pi} \ \ \sin \ \frac{\theta}{2} \ \ d\theta=8a$$

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