Prove that : \(\int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + c\)

Answer: Proof using integration by parts.

Let \(I = \int \sqrt{a^2 - x^2}\cdot 1\,dx\). Integrating by parts with \(u = \sqrt{a^2 - x^2}\) and \(v = 1\):

\(I = \sqrt{a^2 - x^2}\cdot x - \int \frac{-2x}{2\sqrt{a^2 - x^2}}\cdot x\,dx = x\sqrt{a^2 - x^2} + \int \frac{x^2}{\sqrt{a^2 - x^2}}dx\)

\(= x\sqrt{a^2 - x^2} + \int \frac{a^2 - (a^2 - x^2)}{\sqrt{a^2 - x^2}}dx = x\sqrt{a^2 - x^2} + a^2\int \frac{dx}{\sqrt{a^2 - x^2}} - \int \sqrt{a^2 - x^2}\,dx\)

\(\therefore I = x\sqrt{a^2 - x^2} + a^2\sin^{-1}\left(\frac{x}{a}\right) - I + c_1\)

\(2I = x\sqrt{a^2 - x^2} + a^2\sin^{-1}\left(\frac{x}{a}\right) + c_1\)

\(\therefore I = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + c\), where \(c = \frac{c_1}{2}\).