Integration/Fundamental Calculus/Transcendental Numbers

Question

Given $\int\sqrt{1 + \frac{-x}{\sqrt{4-x^2}}} dx$, how would this be integrated? On a universally-calculus side of things, do all functions have integrals? And is it easier to evaluate definite integrals as opposed to indefinite integrals? I would be led to believe that if there are, indeed, integral-less functions, this would be one. This is the case because the length along a curve formula follows $\int\sqrt{1 + f'(x)} dx$ and here $f(x) = \sqrt{4-x^2}$, a (semi) circle equation. So when this is evaluated using the proper limits (2 and -2), the yields should be half of pi, and because of the fundamental laws of transcendental numbers, this cannot yield an algebraic function. Insight?

Answer 1

Here are the questions I've identified:
1. How to integrate $\int\sqrt{1 + \frac{-x}{\sqrt{4-x^2}}} dx$
2. Do all functions have integrals
3. Definite vs. indefinite integrals
4. What are elementary functions

1: You don't. This is an impossible function to integrate with only elementary functions. See number 2.

2: There are functions that don't have integrals that can be expressed in terms of elementary functions (logarithmic, trigonometric, algebraic, etc.). For example, $\int{e^{x^2}}\,dx$ cannot be integrated - if you type it into Mathematica, for example, you'll get an answer involving the imaginary error function.

3: There's no good answer to this. There's not much issue with indefinite integrals; you just add a $C$ or $C(x, y)$ or something along those lines. But definite integrals give you answers, which is nice. It's really an opinion-based question.

4: See the wikipedia page for more: Elementary Functions

If you want to find the arc length of that semicircle equation, you can do it as follows quite easily. It involves a little multivariable calculus, but not much. And it makes life easier because you don't need to do square root integrals. This semicircle has radius 2. In polar coordinates, the equation for this is $C=\big<2\cos{\theta},2\sin{\theta}\big>$ for $\theta\in[0,\pi]$. Arc length can be found with a line integral:
$$\int_C \,dC$$ $$dC=\sqrt{\bigg(\frac{dx}{d\theta}\bigg)^2+\bigg(\frac{dy}{d\theta}\bigg)^2}\,d\theta$$ So our integral becomes:
$$\int_0^\pi \sqrt{4\sin^2{\theta}+4\cos^2{\theta}}=2\int_0^\pi \,d\theta=2\pi.$$ $2\pi$ makes sense according to the formula $C=2\pi r$ - the radius is 2, yielding $4\pi$ but we're only traversing the top half of the circle, so $2\pi$ is right!

EDIT:
You do not always need a numerical answer: in the case of the arc length of a circle, if you integrate with some arbitrary constant radius $R$, you will get circumference as a function of radius ($2\pi R$).