The value of the integral: 3 24^ 3 34 48 9 - 4x^2 dx is equal to: — Solution | JEE Mathematics

MCQMediumJEE 2023Definite Integrals

The value of the integral: $\int_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}} \frac{48}{\sqrt{9 - 4x^2}} \, dx$ is equal to:

Answer

Correct answer:A

Standard Method

Given:

I = \int_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}} \frac{48}{\sqrt{9 - 4x^2}} \, dx

Find: The value of the integral and the correct option.

Use the substitution shown in the solution:

x = \frac{3}{2}\sin\theta, \qquad dx = \frac{3}{2}\cos\theta \, d\theta

Then

9 - 4x^2 = 9 - 4\left(\frac{3}{2}\sin\theta\right)^2 = 9 - 9\sin^2\theta = 9\cos^2\theta

\sqrt{9 - 4x^2} = 3\cos\theta

Detailed Evaluation

Transform the limits:

x = \frac{3\sqrt{2}}{4} \Rightarrow \sin\theta = \frac{\sqrt{2}}{2} \Rightarrow \theta = \frac{\pi}{4}

x = \frac{3\sqrt{3}}{4} \Rightarrow \sin\theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = \frac{\pi}{3}

Using the answer key instead of the worked solution. The solution explicitly states The Correct Option is A, so the answer must be mapped from the options accordingly.
Missing that the integrand in the question is $\frac{48}{\sqrt{9-4x^2}}$ , not $48\sqrt{9-4x^2}$ . Reading the integrand incorrectly changes the entire substitution outcome.

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.