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I have searched for answers everywhere, but none of them seems to be correct. Since the area under $\sqrt{1-\sin{t}}$ is always positive, I imagine the integral to be a graph resembling an infinitely rising stairs, but the answers floating around the internet (including WolframAlpha) seem to be wrong.

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    $\begingroup$ How does it give zero? $\endgroup$ Commented Dec 4, 2012 at 13:55
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    $\begingroup$ The answer $F(t)$ from WolframAlpha is a function whose derivative is $\sqrt{1-\sin t}$ for all complex numbers $t$ except perhaps a few branch points. So, unless $F(t)$ happens to be continuous on an interval, you cannot get a definite integral by subtraction at the endpoints. In this case, the jumps in $F$ are where $\sin t = 1$. There is no good way to make $F$ work in the complex plane without having discontinuities at these branch points. $\endgroup$
    – GEdgar
    Commented Dec 4, 2012 at 15:40

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$1-\sin{t}$ = $(\sin{t/2} - \cos{t/2})^2$

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  • $\begingroup$ Definite integral from $0$ to $2\pi$ and from $0$ to $4\pi$ should not be equal. $\endgroup$ Commented Dec 4, 2012 at 13:58
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    $\begingroup$ @user803253 Why does this contradict that? This hint means you need to compute: $$\int |\sin t/2 -\cos t/2|dt$$ which is gonna be positive for $[0,2\pi]$ and bigger for $[0,4\pi]$ $\endgroup$ Commented Dec 4, 2012 at 14:02
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$$\sqrt{1 - \sin{t}} = {\sqrt{1 - \sin{t}}\sqrt{1 + \sin{t}} \over \sqrt{1 + \sin{t}}}$$ $$ = {\sqrt{1 - \sin^2(t)} \over \sqrt{1 + \sin(t)}}$$ $$= {|\cos(t)| \over \sqrt{1 + \sin(t)}}$$ Doing a $u$ substitution $u = 1 + \sin(t)$ shows that an indefinite integral of this is $2\sqrt{1 + \sin(t)}$ when $\cos(t) > 0$ and $-2\sqrt{1 + \sin(t)}$ when $\cos(t) < 0$. The former case corresponds to $-{\pi \over 2} < t < {\pi \over 2}$ and the latter to when ${\pi \over 2} < t < {3\pi \over 2}$.

If you want these to match up to a continuous function, you can use $2\sqrt{1 + \sin(t)}$ for the $-{\pi \over 2} < t < {\pi \over 2}$ range and $4\sqrt{2} - 2\sqrt{1 + \sin(t)}$ for the ${\pi \over 2} < t < {3\pi \over 2}$ range. If you want to keep going beyond $t = {3\pi \over 2}$ or below $t = -{\pi \over 2}$, you add further constants to make the indefinite integral you're using continuous.

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You are correct. The integral should not evaluate to 0, if we are working with real numbers. Sometimes a CAS needs a little help. Some creatively placed absolute value signs worked for Wolfram: http://www.wolframalpha.com/input/?i=integral%28|sqrt%281-sin%28t%29%29|+from+0+to+2pi%29

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  • $\begingroup$ And yet a numerical answer is hardly useful from a learning perspective. The actual answer has a closed form that WA does not give you. (Hint: divide the answer you get by $4$ and what do you get?) $\endgroup$ Commented Dec 4, 2012 at 14:27
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    $\begingroup$ Agreed wholeheartedly. However, I often feel people do not understand that computers sometimes need a bit of "coaxing" to produce the correct answer. I posted to share a tip that can help a computer produce an accurate answer. Probably should have done so as a comment, rather than an answer... $\endgroup$
    – apnorton
    Commented Dec 4, 2012 at 14:29

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