Integral of $\sqrt{1-\sin{x}}$: Wrong Result?

[catellanos]

Homework Statement

I need to calculate \int_0^{2\pi} \sqrt{1-\sin{x}}\, dx when I compute the indefinite integral I get 2\sqrt{\sin{x}+1}+C but when I evaluate it into the integration limits it turns out to be 0 which doesn't make sense since the original function is clearly positive...I already tested this result in wolfram alpha and my calculation of the definite integral is perfect but when I tested the definite integral wolfram computed its value to be 4\sqrt{2}. I can't find out what's wrong with my procedure. Please help!

Homework Equations

The Attempt at a Solution

 

[Hernaner28]

The indefinite integral you computed is not right I guess. Wolfram Alpha shows that but when I do it in Derive I get another thing which verifies the result 4sqrt2...

This is what I get (no idea what floor is, guess it's the function?)

$$\frac{{2\cos x}}{{\sqrt {1 - \sin x} }} - 4\sqrt 2 \cdot FLOOR\left( {\frac{1}{4} - \frac{x}{{2\pi }}} \right)$$

 

[SrEstroncio]

@Hernaner28
Floor's not even a continuous function, much less differentiable that can't possibly be a primitive for \sqrt{1-\sin{x}}.

 

[Dick]

When you computed the antiderivative you replaced \sqrt{\cos^2 x} with \cos x. That's only correct if cos(x) is positive. cos(x) isn't positive on [0,2pi]. I'd suggest you split the integral up.

 

[catellanos]

Thank you very much guys, your contributions were very valuable; particularly thank you Dick, with your observation I solved my trouble