Question

Solve : $\underset{0}{\stackrel{1}{\int }}\sqrt{9-4x^2}dx$

Answer 1

$\int \sqrt{9-4x^2}dx$

$x=\frac{3}{2}\sin u\to dx=\frac{3}{2}\cos udu$

$=\int \left(\sqrt{9-4\frac{9}{4}{\sin }^{2}u}\right)\frac{3}{2}\cos udu$

$=\int \sqrt{9(1-{\sin }^{2}u)}\frac{3}{2}\cos udu$

$=\int \left(3\cos u\right)\frac{3}{2}\cos udu$

$=\frac{9}{2}\int {\cos }^{2}udu$

$=\frac{9}{2}\int \left(\frac{1+\cos 2u}{2}\right)du$

$=\frac{9}{2}\cdot \frac{1}{2}(u+\frac{1}{2}\sin u)$

$u={\sin }^{-1}\left(\frac{2}{3}x\right)$

$\int \sqrt{9-4x^2}dx=\frac{9}{2}\cdot \frac{1}{2}({\sin }^{-1}\left(\frac{2}{3}x\right)+\frac{1}{2}\sin \left({\sin }^{-1}\left(\frac{2}{3}x\right)\right))+C$

${\int }_0^1\sqrt{9-4x^2}dx=\frac{9}{4}{\sin }^{-1}\left(\frac{2}{3}\right)+\frac{\sqrt{5}}{2}$