∫ 3 2 4 3 3 4 48 9 - 4 x 2 d x is equal to

Given integration is

${\int }_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}}\frac{48}{\sqrt{9-4x^2}}dx$

We know that

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

So, the equation can be re-written as

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

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

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

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