By integration by substitution, we get $\frac{d}{dx} (\frac{1}{2} (x \sqrt{a^2 - x^2} + a^2 \arcsin (\frac{x}{a}))) = \sqrt{a^2 - x^2}$ on $[-a, a]$.
By integration by parts, we get $\frac{d}{dx} (\frac{1}{2} (x \sqrt{a^2 - x^2} + a^2 \arcsin (\frac{x}{a}))) = \sqrt{a^2 - x^2}$ on $(-a, a)$.
I wanna prove that $\frac{d}{dx} (\frac{1}{2} (x \sqrt{a^2 - x^2} + a^2 \arcsin (\frac{x}{a}))) = \sqrt{a^2 - x^2}$ on $[-a, a]$.
Is the following argument right?
(1)
Let $a > 0$.
Let $f : [-a, a] \to \mathbb{R}$ and $f(x) = \sqrt{a^2 - x^2}$.
Let $\phi : [-\frac{\pi}{2}, \frac{\pi}{2}] \to [-a, a]$ and $\phi(t) = a \sin(t)$.
$f$ is continuous on $[-a, a]$.
$\phi$ is differentiable on $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $\phi'$ is continuous on $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
$\phi$ is a bijective function and has the inverse function $\phi^{-1} : [-a, a] \to [-\frac{\pi}{2}, \frac{\pi}{2}]$ and $\phi^{-1}(s) = \arcsin(\frac{s}{a})$.
By integration by substitution, $\int \sqrt{a^2 - x^2} dx = \int \sqrt{a^2 - (a \sin(t))^2} a \cos(t) dt = \int \sqrt{a^2 - (a \sin(t))^2} a \cos(t) dt \\= \int a^2 \cos^2(t) dt = a^2 \int \cos^2(t) dt = a^2 \int \frac{\cos(2 t) + 1}{2} dt = \frac{a^2}{2} (\frac{1}{2} \sin(2 t) + t) \\= \frac{a^2}{2} (\sin(t) \cos(t) + t) = \frac{1}{2} (x \sqrt{a^2 - x^2} + a^2 \arcsin (\frac{x}{a})).$
This means $\frac{d}{dx} (\frac{1}{2} (x \sqrt{a^2 - x^2} + a^2 \arcsin (\frac{x}{a}))) = \sqrt{a^2 - x^2}$ on $[-a, a]$.
(2)
Let $a > 0$.
Let $f : [-a, a] \to \mathbb{R}$ and $f(x) = \sqrt{a^2 - x^2}$.
$f$ is differentiable on $(-a, a)$.
By integration by parts,
$F(x) = \int f(x) dx = x f(x) - \int x f'(x) dx = x \sqrt{a^2 - x^2} - \int x \frac{-x}{\sqrt{a^2 - x^2}} dx \\= x \sqrt{a^2 - x^2} - \int \frac{-a^2 + (a^2 - x^2)}{\sqrt{a^2 - x^2}} dx = x \sqrt{a^2 - x^2} + \int \frac{a^2}{\sqrt{a^2 - x^2}} dx - \int \sqrt{a^2 - x^2} dx \\= x \sqrt{a^2 - x^2} + a^2 \arcsin(\frac{x}{a}) - F(x).$
So, $F(x) = \frac{1}{2} (x \sqrt{a^2 - x^2} + a^2 \arcsin (\frac{x}{a}))$.
This means $\frac{d}{dx} (\frac{1}{2} (x \sqrt{a^2 - x^2} + a^2 \arcsin (\frac{x}{a}))) = \sqrt{a^2 - x^2}$ on $(-a, a)$.
We want to prove $\frac{d}{dx} (\frac{1}{2} (x \sqrt{a^2 - x^2} + a^2 \arcsin (\frac{x}{a}))) = \sqrt{a^2 - x^2}$ on $[-a, a].$
Let $G(x) = \int_{0}^{x} \sqrt{a^2 - t^2} dt$.
Then, by a famous theorem, $G(x)$ is differentiable on $[-a, a]$ and $G'(x) = \sqrt{a^2 - x^2}$ on $[-a, a]$.
By a famous theorem, $F(x) = G(x) + C$ on $(-a, a)$ for some real number $C$.
Since $G(x)$ is continuous on $[-a, a]$, $\lim_{x \to a} G(x) = G(a)$ and $\lim_{x \to -a} G(x) = G(-a)$.
And obviously $F(x)$ is continuous on $[-a, a]$.
So, $\lim_{x \to a} F(x) = F(a)$ and $\lim_{x \to -a} F(x) = F(-a)$.
So, $F(a) = \lim_{x \to a} F(x) = \lim_{x \to a} (G(x) + C) = G(a) + C$ and $F(-a) = \lim_{x \to -a} F(x) = \lim_{x \to -a} (G(x) + C) = G(-a) + C$
Then, $\lim_{x \to a} \frac{F(x) - F(a)}{x - a} = \lim_{x \to a} \frac{(G(x) + C)- (G(a) + C)}{x - a} = \lim_{x \to a} \frac{G(x) - G(a)}{x - a} = G'(a)$ and $\lim_{x \to -a} \frac{F(x) - F(-a)}{x - (-a)} = \lim_{x \to -a} \frac{(G(x) + C)- (G(-a) + C)}{x - (-a)} = \lim_{x \to -a} \frac{G(x) - G(-a)}{x - (-a)} = G'(-a)$.
So, $\frac{d}{dx} F(x) = \sqrt{a^2 - x^2}$ on $[-a, a].$
By the way, I noticed that $\frac{1}{2} (x \sqrt{a^2 - x^2} + a^2 \arcsin (\frac{x}{a}))$ is differentiable at $x = -a$ and $x = a$, but $x \sqrt{a^2 - x^2}$ is not differentiable at $x = -a$ and $x = a$ and $a^2 \arcsin (\frac{x}{a})$ is not differentiable at $x = -a$ and $x = a$.