Question

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 1

Let $I=\int \sqrt{a^2-x^2}dx$
$=\int \sqrt{a^2-x^2}\cdot 1dx$
On integrating by parts, we get
$I=\sqrt{a^2-x^2}\int 1dx-\int \left[\frac{d}{dx}\right(\sqrt{a^2-x^2})\int 1dx]dx$
$=x\sqrt{a^2-x^2}-\int \frac{-2x}{2\sqrt{a^2-x^2}}x\cdot dx$
$=x\sqrt{a^2-x^2}-\int \frac{(a^2-x^2)-a^2}{\sqrt{a^2-x^2}}dx$
$=x\cdot \sqrt{a^2-x^2}-\int [\sqrt{a^2-x^2}-\frac{a^2}{\sqrt{a^2-x^2}}]dx$
$=x\cdot \sqrt{a^2-x^2}-\int \sqrt{a^2-x^2}dx+a^2\int \frac{1}{\sqrt{a^2-x^2}}dx$
$=x\cdot \sqrt{a^2-x^2}-I+a^2\cdot {\sin }^{-1}\left(\frac{x}{a}\right)+c_1$
$2I=x\sqrt{a^2-x^2}+a^2\cdot {\sin }^{-1}\left(\frac{x}{a}\right)+c_1$
$I=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)+c_1$
$\therefore \int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)+c$ where, $c=\frac{c_1}{2}$.