Does the integral of $\sqrt{1-\sin{t}}$ exist? 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.
 A: $1-\sin{t}$ = $(\sin{t/2} - \cos{t/2})^2$
A: $$\sqrt{1 - \sin{t}} = {\sqrt{1 - \sin{t}}\sqrt{1 + \sin{t}} \over \sqrt{1 + \sin{t}}}$$
$$ = {\sqrt{1 - \sin^2(t)} \over \sqrt{1 + \sin(t)}}$$
$$= {|\cos(t)| \over \sqrt{1 + \sin(t)}}$$
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. 
A: 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
