<s>[INST] 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 $x=a\sin p$,
 [/INST] 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.
</s>