Different methods of evaluating $\int\sqrt{a^2-x^2}dx$: Is there a simple and nice way to solve $\int\sqrt{a^2-x^2}dx$:
PS:I am not looking for a substitution like $x=a\sin p$,
 A: Firstly, I would like to point out that the simplest way to solve this problem is to use trigonometric substitution.
In addition, if you insist on do not using that method, it is surely have some other methods. Let $I$ denote the result of the indefinite integral. We have:
\begin{equation}
I=\int \frac{a^2-x^2}{\sqrt{a^2-x^2}}dx=a^2\arcsin\left(\frac{x}{a}\right)-\int\frac{x^2}{\sqrt{a^2-x^2}}dx\\[10mm]
I=x\sqrt{a^2-x^2}+\int\frac{x^2}{\sqrt{a^2-x^2}}dx
\end{equation}
The 2nd equation is given by integral by parts. Then add them together, we have
\begin{equation}
I=\frac{a^2}{2}\arcsin\left(\frac{x}{a}\right)+\frac{x}{2}\sqrt{a^2-x^2}
\end{equation}
I ignore the constant $C$ in the result.
A: Another way to attack such kinds of integrals is using Chebyshev's substitution. Consider the integral:
$$
I_{m,p} = \int x^m (\alpha x^n + \beta)^pdx
$$
Chebyshev has shown that $I_{m, p}$ is expressed in elementary functions if and only if one of the following cases is satisfied:
$$\begin{align*}
&1)\ p \in \Bbb Z \\
&2)\ {m+1 \over n} \in \Bbb Z \\
&3)\ {m+1 \over n} + {1\over p} \in \Bbb Z
\end{align*}
$$
For the first case substitute:
$$
x = t^N,\ \text{where}\ N\ \text{is common denominator of fractions}\ m\ \text{and}\ n
$$
Second case, substitute:
$$
\alpha x^n + \beta = t^M\ \text{where}\ M\ \text{is the denominator of}\ p
$$
Third case substitute:
$$
\alpha + \beta x^{-n} = t^M\ \text{where}\ M\ \text{is the denominator of}\ p
$$
If you check the third case is satisfied, hence the integral exists in elementary functions.
$$
(a^2 - x^2)^{1\over 2} = x^m(\alpha x^n + \beta)^p
$$
Therefore:
$$
p = {1\over 2}\\
m = 0\\
n = 2\\
\alpha = -1\\
\beta = a^2
$$
Let:
$$
\alpha + \beta x^{-n} = t^M
$$
So:
$$
t^2 = {a^2\over x^2} - 1\\
x^2 = {a^2\over t^2 + 1}\\
x^3 = \left({a^2\over t^2 + 1}\right)^{3\over 2}\\ 
tdt = -{a^2 \over x^3}dx \implies dx = -{t\over a^2}\left({a^2\over t^2 + 1}\right)^{3\over 2}dt
$$
The integral becomes:
$$\begin{align}
I &= -\int \sqrt{a^2 - {a^2\over t^2 + 1}}{t\over a^2}\left({a^2\over t^2 + 1}\right)^{3\over 2}dt \\
&= -\int \sqrt{a^2 t^2 \over t^2 + 1}{t\over a^2}\left({a^2\over t^2 + 1}\right)^{3\over 2}dt\\
&= -\int {|t|t\over a^2}\left({a^2\over t^2 + 1}\right)^2dt 
\end{align}
$$
And this is a regular rational function.
