∫ _ 3 2 4 ^ 3 3 4 48 9 - 4 x ^ 2 dx is equal to

Given integration is

${∫}_{\dfrac{3\sqrt{2}}{4}}^{\dfrac{3\sqrt{3}}{4}}\dfrac{48}{\sqrt{9 - 4{x}^{2}}}dx$

We know that

$∫ \dfrac{dx}{\sqrt{{a}^{2} - {x}^{2}}} = \sin^{- 1}\dfrac{x}{a} + C$

So, the equation can be re-written as

${∫}_{\dfrac{3\sqrt{2}}{4}}^{\dfrac{3\sqrt{3}}{4}}\dfrac{48}{2\sqrt{{\left(\dfrac{3}{2}\right)}^{2} - {x}^{2}}}dx = \dfrac{48}{2} \times {\left[\sin^{- 1}\dfrac{2x}{3}\right]}_{\dfrac{3\sqrt{2}}{4}}^{\dfrac{3\sqrt{3}}{4}}$

$= 24 \times \left[\sin^{- 1}\left(\dfrac{2}{3} \times \dfrac{3\sqrt{3}}{4}\right) - \sin^{- 1}\left(\dfrac{2}{3} \times \dfrac{3\sqrt{2}}{4}\right)\right]$

$= 24 \times \left[\sin^{- 1}\left(\dfrac{\sqrt{3}}{2}\right) - \sin^{- 1}\left(\dfrac{1}{\sqrt{2}}\right)\right]$

$= 24 \times \left[\dfrac{\pi}{3} - \dfrac{\pi}{4}\right] = 2\pi$