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I having some problems with the integration of $\sqrt{2-2\cos t}$. I got it by calculating the length of the arc

\begin{align} \gamma (t) &= (t - \sin t,1 - \cos t), \quad t\in[0,2\pi] \\ \implies \dot{\gamma(t)} &= (1 - \cos t, \sin t) \\ \implies \vert\vert\dot{\gamma(t)}\vert\vert &= \sqrt{(1 - \cos t)^2+(\sin t)^2} \\ &= \sqrt{2-2\cos t} \end{align}

How can I do?

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    Hint: 1-cos(x)=2sin^2(x/2) – 2017-01-14
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    Hint: $\sqrt{\frac{1}{2}(1-\cos t)}=\sin \frac{t}{2}$ – 2017-01-14

1 Answers 1

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We have,

$\sqrt2 \sqrt{1 - \cos t }$

= $\sqrt2 \sqrt{2 \sin^2 \frac{t}{2}}$

= $ 2 \sin \frac{t}{2}$

Now integrate it.

= 2 $\int_0^{2\pi} \sin \frac{t}{2}$

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    May be with an absolute value ;o) – 2017-01-14
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    @Bernard can you please explain the meaning of your comment. – 2017-01-15
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    @Semiclassical Can you please explain taking my answer as refrence because most of the time I do questions like this. And always get correct result. – 2017-01-15
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    Sorry, your answer is fine. The issue I was concerned with was whether $\sqrt{\sin^2(t/2)}=\sin(t/2)$ was valid; it isn't true for $t\in(2\pi,4\pi),$ for instance. _But_ this objection is actually spurious: the equality is valid on the domain of integration. So there's no problem. – 2017-01-15
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    @Semiclassical then why someone downvoted my answer. – 2017-01-15
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    Yes, is corret, But why $1-cost$=$2sin^2\frac{t}{2}$ – 2017-01-15
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    This is a formula $1 - cos n\theta = 2 sin^2 \frac{n\theta}{2}$. Also $1 + cos n\theta = 2 cos^2 \frac{n\theta}{2}$ – 2017-01-15
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    If -ve sign before cos its change into sin and angle becomes half. – 2017-01-15
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    If still any doubt please ask. – 2017-01-15
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    I meant the same (general) objection as Semiclassical: there should be some words to justify why no absolute value. Let me add I didn't down vote – it was just a comment. Unfortunately there's a number of temperamental persons on this site :-( – 2017-01-15
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    @Bernard ok. But semiclassical says its fine. Is it not? – 2017-01-15
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    For *me*, it's fine. Actually, it depends on whom it is addressed to: for some people, it goes without saying, for other people, it might let them think that in all cases $\sqrt{A^2}=A$. That's what I wanted to point out. – 2017-01-15