Question

Prove that $\int \sqrt{(a^2-x^2)}dx=\frac{x\sqrt{(a^2-x^2)}}{2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}$

Answer 1

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

Integration by parts differentiate $\sqrt{a^2-x^2}$ and integrate $1$

$\Rightarrow I=\sqrt{a^2-x^2}\int dx-\int \frac{d\sqrt{a^2-x^2}}{dx}(\int dx)dx$

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

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

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

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

$I=x\sqrt{a^2-x^2}-I+a^2{\sin }^{-1}\frac{x}{a}$

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

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

Hence, solved.