Question

Prove that $\int \sqrt{a^2-x^2}dn=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)+c$.

Answer 1

$I=\int \sqrt{a^2-x^2}dx=\int a\sqrt{1-\frac{x^2}{a^2}}dx$

Let $\frac{c}{a}=\sin \theta ⟹dx=a\cos \theta d\theta$

$=\int a\sqrt{1-{\sin }^2\theta }.a\cos \theta d\theta$

$=a^2\int {\cos }^2\theta d\theta$

$=a^2[\frac{\theta }{2}+\frac{\sin 2\theta }{4}]+c$

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