As in the title, I'm trying to find the exact length of the $r=1-cos(\theta)$ with $0 \le \theta \le 2 \pi$ and I have it set up as so: $$L={\int_0^{2 \pi} \sqrt{(1-cos \theta)^2+(-sin \theta)^2}}d \theta$$ But I can't figure out how to integrate this. Now, so you know what I understand, I'm familiar with trig identities and the basics of integration. Hopefully, that helps.
A u-substitution won't work here since there will be left-over non-constants, and I don't see any other way. I'm guessing that I will need to use some kind of trig-identity but I can't see it. I tried expanding $(1-cos \theta)^2$ but that still doesn't really help me.
Thank you to anyone who can help me.
Best Answer
Hint: $$(1-\cos\theta)^2+\sin^2\theta=2(1-\cos\theta)=2\cdot2\sin^2\dfrac{\theta}{2}$$