QJEEM23Ma8aaa: JEE Main 2023 Mathematics - Class 12 JEEM23Ma8aaa Solution

Question

Verified Solution • 2025–26 Session moderate

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

Step-by-Step Verified Solution