[_3 2 4^3 3 4 48 9 - 4x^2 dx ] is equal to :

Question

$$\int\limits_{{{3\sqrt 2 } \over 4}}^{{{3\sqrt 3 } \over 4}} {{{48} \over {\sqrt {9 - 4{x^2}} }}dx} $$ is equal to :

Accepted Answer

$$
\int_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x
$$

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

Hence $\int_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x=\frac{48}{2} \times\left[\sin ^{-1} \frac{2 x}{3}\right]_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}}$

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

$=24 \times\left(\frac{\pi}{3}-\frac{\pi}{4}\right)$

$=24 \times \frac{\pi}{12}=2 \pi$

Solutions · 1