Different methods of evaluating $\int\sqrt{a^2-x^2}dx$:

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.