General

[quote=gibbsfree]@AnasF: check it on the integrator you're wrong

here i'll show you

integral of (-f(x))=-integral(f(x))

so integral of (-1/sqrt[25-x^2])=-integral(1/sqrt[25-x^2]

since integral of (1/sqrt[a^2-x^2])=arcsin(x/a)+c

then integral of (-1/sqrt[25-x^2])=-arcsin(x/5) +c

check it on the integrator i can't post links but search integrator in google and type in -1/Sqrt[25-x^2] and see what you get.[/quote]
All calculators take negative constants out first. But it is a mathematical identity that Integral of -1/Sqrt[25-x^2] is also cos^-1(x/5).

@Jake305: If I wanted to prove my superiority, I'd post something more puzzling, and something I understand perfectly.

[quote=gibbsfree]Integral of -1/Sqrt[25 - x^2] = cos^-1(x/5)
^^
thats wrong
the integral of 1/sqrt[a^2-x^2]=arcsin(x/a)+c[/quote]
http://www.wolframalpha.com/input/?i=differentiate+cos^-1%28x%2F5%29

[quote=Jake305]Congratulations, you have succeeded in proving your mathematical superiority over a community of [assumed] 14 year olds. You may now proceed in reading a long and exciting book.[/quote]

lol you have no idea right?

OT: you're crazy.

Going back to your original question, evaluating the definite integral with the integrand -1/Sqrt[1 - x^2], bounded from x=0 to x=1, results in -pi/2 regardless of whether arcsin or arccos is used. arccos(x) + arcsin(x) =/= 0. Could you have mixed up the signs?

[quote=ShockPenguin]@AnasF:

What? arccos[1] - arccos[0] = 0 - pi/2 = -pi/2. ._.[/quote]
Yes, I realise that.
My point is that, through legitimate methods, I've produced two answers for the integral of 1/sqrt[1-x^2] from 0 to 1.
Also, what piques my interest further is that neither of this are equivalent to "What is the area enclosed between y = 1/Sqrt[1-x^2], y = 0, x = 0, x = 1".

[quote=ShockPenguin]As long as your bounds are the same, definite integrals would give you zero.[/quote]

Let's say, integral of -1/[1 - x^2] bounds 0 to 1.
If you let the integral be cos^-1(x)... The definite integral results in 0.
If you let the integral be -sin^-1(x)... The definite integral becomes -pi/2.

[quote=ShockPenguin]Oh, sorry. I just woke up and your wording confused me a little.

The problem here is that you're forgetting your constant of integration. That little "+C" at the end of your indefinite integral.

If you kind of take a step back from calculus and reinterpret what arcsin[x] and arccos[x] are, you'd find that adding them gives you pi/2 (the constant you're looking for).
Draw a right triangle, label one of legs "x" and the hypotenuse "1", and you'll see that the two angles are always complementary given a suitable range.[/quote]
Ok sure.
But what about definite integrals?

[quote=ShockPenguin]If you take the negative out as a constant,
Integral of -1/Sqrt[25 - x^2] = -1 * sin^-1(x/5)

^ What?[/quote]
Let's say you have
"Find the integral of -f'(x)." It is legitimate to rephrase that as, "Find the integral of f'(x) and multiply it by -1."
However, this causes slight discrepancies with inverse trigonometric integrals.