Explanation of this integral's solution

Question

While doodling around with circles and associated geometry, I've stumbled across this integral (I'm led to believe it is correct but I have not found or created any proof):

$$2\int_{-r}^{r}\sqrt{r^2-x^2}dx=\pi r^2;r \ge 0 \tag 1$$

(edit, changed $\sqrt{r-x^2}$ to $\sqrt{r^2-x^2}$ as the former is not correct geometrically)

If it isn't immediately clear, this resulted from using an infinite series of parallel chords to compute area. I'm aware there are simpler approaches, however this one is intriguing, knowing the result... I've tried integrating it by hand, couldn't figure out how to approach it.

Wolfram Alpha (for lack of human assistance at my disposal) gave me this fascinating indefinite integral for $r=1$:

$$F(x)=\int\sqrt{1-x^2}dx=\frac{1}2(x\sqrt{1-x^2}+\arcsin x) \tag 2$$

It is apparent that: $2(F(1)-F(-1))=(\arcsin{1}) - (\arcsin{-1}) = \pi$

I understand why to expect $\arcsin$ to appear in the result, after all $\sqrt{1-x^2}=\cos\arcsin x$, but I don't understand why it has appeared in the result or its significance. I can see already that any definite integral for (1) will cancel itself out leaving only the arcsin term...

A step-by-step solution to (1) or (2) would be most helpful but not required.

Answer 1

Defining $x=r\cos(\theta)$ we can find that $\text{d}x=-r\sin(\theta)\text{d}\theta$ and $\theta$ goes from $\pi$ to $0$ the integral becomes $$\int_{-r}^{r}\sqrt{r^2-x^2}\text{d}x=-r\int_{\pi}^{0}\sqrt{r^2-r^2\cos^2(\theta)}\sin(\theta)\text{d}\theta=-r^2\int_{\pi}^{0}\sin^2(\theta)\text{d}\theta$$ and you may continue.

Answer 2

By recalling $$\sin^2 x + \cos^2 x = 1,$$ we are tempted to make a substitution $x = \sin t$ in $(2)$. Then, the indefinite integral becomes $$\int \sqrt{1-\sin^2t} \cos t \,dt = \int \cos^2 t \, dt.$$ Using integration by parts you will get your desired result.

What similar substitution can you do in order to solve $(1)$?

Answer 3

By decomposition, then by parts, $$\int\sqrt{1-x^2}dx=\int\frac{dx}{\sqrt{1-x^2}}-\int\frac{x^2\,dx}{\sqrt{1-x^2}}=\arcsin(x)+x\sqrt{1-x^2}-\int\sqrt{1-x^2}\,dx.$$

You can conclude.

Answer 4

The integral that you want to compute is the area of a circle segment defined by intersecting the trigonometric circle with a vertical at $x$.

This area is that of the sector, equal to the subtended angle, minus the triangle.

$$A=2\int_{-1}^x\sqrt{1-x^2}\,dx=2\arcsin(x)+x\sqrt{1-x^2}.$$