JEE Main 2023 — Mathematics Calculus

Given integration is

$${\int }_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}}\frac{48}{\sqrt{9-4{x}^{2}}}dx$$∫432​​433​​​9−4x2​48​dx

We know that

$$\int \frac{dx}{\sqrt{{a}^{2}-{x}^{2}}}={\mathrm{sin}}^{-1}\frac{x}{a}+C$$∫a2−x2​dx​=sin−1ax​+C

So, the equation can be re-written as

$${\int }_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}}\frac{48}{2\sqrt{{(\frac{3}{2})}^{2}-{x}^{2}}}dx=\frac{48}{2}\times {[{\mathrm{sin}}^{-1}\frac{2x}{3}]}_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}}$$∫432​​433​​​2(23​)2−x2​48​dx=248​×[sin−132x​]432​​433​​​

$$=24\times [{\mathrm{sin}}^{-1}(\frac{2}{3}\times \frac{3\sqrt{3}}{4})-{\mathrm{sin}}^{-1}(\frac{2}{3}\times \frac{3\sqrt{2}}{4})]$$=24×[sin−1(32​×433​​)−sin−1(32​×432​​)]

$$=24\times [{\mathrm{sin}}^{-1}(\frac{\sqrt{3}}{2})-{\mathrm{sin}}^{-1}(\frac{1}{\sqrt{2}})]$$=24×[sin−1(23​​)−sin−1(2​1​)]

$$=24\times [\frac{\pi }{3}-\frac{\pi }{4}]=2\pi$$=24×[3π​−4π​]=2π