I'm having a bit of an issue calculating the arc length of $r = a(1-\cos\theta)$.
I'll begin by listing the steps I made in my attempt to solve this exercise.
We know that the arc length formula for polar coordinates is $$L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \,d\theta$$
So with $r = a(1-\cos\theta)$ the arc length formula becomes $$L = \int_{0}^{2\pi} \sqrt{[a(1-\cos\theta)]^2 + (a\sin\theta)} \, d\theta.$$
Simplifying this further, we obtain $$L = a\sqrt{2}\int_{0}^{2\pi} \sqrt{1-\cos\theta} \, d\theta.$$
I'm stuck on solving the resultant integral. I've attempted u-substitution and integration by parts but to no avail. Is my resultant integral incorrect? How do I proceed?
Best Answer
Use $$\sin^2+\cos^2=1\implies \sqrt {1-\cos x}=\dfrac {\sin x}{\sqrt {1+\cos x}} $$ Let $ u=1-\cos x $, so $ du =\sin xdx $. Then, we get $$\int \dfrac {1}{\sqrt {1-u}}\cdot \dfrac {\sin x}{\sin x} du$$ Now, cancel out and do one more $u$ substitution, integrate and then plug back in to get $$2a\sqrt {2 (1-\cos x)}|^{2\pi}_{0}$$