Question

$\underset{\frac{3\sqrt{2}}{4}}{\overset{\frac{3\sqrt{3}}{4}}{\int }}\frac{48}{\sqrt{9-4x^2}}dx$ is equal to

• A. $2\pi$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (A)

$\underset{\frac{3\sqrt{2}}{4}}{\overset{\frac{3\sqrt{3}}{4}}{\int }}\frac{48}{\sqrt{9-4x^2}}dx$
We have $\int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\frac{x}{a}+C$
Hence $\underset{\frac{3\sqrt{2}}{4}}{\overset{\frac{3\sqrt{3}}{4}}{\int }}\frac{48}{\sqrt{9-4x^2}}dx=\frac{48}{2}\times {\left[{\sin }^{-1}\frac{2x}{3}\right]}_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}}$
$=24\times \left[{\sin }^{-1}\left(\frac{2}{3}\times \frac{3\sqrt{3}}{4}\right)-{\sin }^{-1}\left(\frac{2}{3}\times \frac{3\sqrt{2}}{4}\right)\right]$
$=24\times \left[{\sin }^{-1}\frac{\sqrt{3}}{2}-{\sin }^{-1}\frac{1}{\sqrt{2}}\right]$
$=24\times \left(\frac{\pi }{3}-\frac{\pi }{4}\right)$
$=24\times \frac{\pi }{12}=2\pi$