48 ∫_(3/√2)^(3√3/2) dx / √(9 - 4x²) is equal to (1) π/3, (2) π/2, (3) π/6, (4) 2π

Q: Wait, why can't I just use the power rule or treat $\sqrt{9-4x^2}$ as $(9-4x^2)^{-1/2}$?

That is a classic trap! The power rule $\int u^n du$ only works if the derivative of the inner function is present as a multiplier. Here, the derivative of $9-4x^2$ is $-8x$, which is not present. We must recognize this as a standard inverse trigonometric integral form.

Q: When I factor out the 4, does it become 2 or 4 outside the root?

Great question! Remember the property $\sqrt{A \cdot B} = \sqrt{A} \cdot \sqrt{B}$. So, $\sqrt{4(\frac{9}{4} - x^2)} = \sqrt{4} \cdot \sqrt{(\frac{3}{2})^2 - x^2} = 2 \cdot \sqrt{(\frac{3}{2})^2 - x^2}$. The 4 becomes a 2 because it is inside the square root.

Q: Why do we get $\sin^{-1}(\frac{2x}{3})$ instead of just $\sin^{-1}(\frac{x}{3/2})$?

Mathematically, they are identical! $\frac{x}{3/2}$ is the same as $\frac{2x}{3}$. Writing it as $\frac{2x}{3}$ is just cleaner and makes the substitution of the limits much easier to handle without dealing with complex fractions.

Question

$48 \int_{3/\sqrt{2}}^{3\sqrt{3}/2} \frac{dx}{\sqrt{9-4x^2}}$ is equal to

Options:

(A) $\frac{\pi}{3}$

(B) $\frac{\pi}{2}$

(C) $\frac{\pi}{6}$

(D) $2\pi$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, warriors of JEE! Today, we are not just solving an integral; we are peeling back the layers of a mathematical onion to reveal the elegant core hidden within.

Consider the problem:

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

At first glance, it looks like a daunting expression. Imagine you are standing on the edge of a curve, and this integral is simply the area beneath it. Our goal is to find the exact value of that area.

### The Algebraic Surgery

The first thing that should catch your eye is the denominator: $\sqrt{9-4x^2}$. It screams for the standard inverse sine integral form:

$$\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C$$

However, our formula demands that the coefficient of $x^2$ be exactly $1$. Currently, it is $4$. We must perform some algebraic surgery.

We factor out the $4$ from inside the square root:

$$\sqrt{9 - 4x^2} = \sqrt{4\left(\frac{9}{4} - x^2\right)}$$

When we pull that $4$ out of the square root, it becomes a $2$. Now, our integral transforms into:

$$48 \int \frac{dx}{2\sqrt{\left(\frac{3}{2}\right)^2 - x^2}}$$

Since $48$ divided by $2$ is $24$, we have successfully tamed the expression, leaving us with:

$$24 \int \frac{dx}{\sqrt{\left(\frac{3}{2}\right)^2 - x^2}}$$

### The Transformation

Now, we are in the home stretch. We apply the standard formula where $a = \frac{3}{2}$.

The integration yields:

$$24 \left[ \sin^{-1}\left(\frac{x}{3/2}\right) \right] = 24 \left[ \sin^{-1}\left(\frac{2x}{3}\right) \right]$$

This is the moment where the complexity vanishes. We are left with a simple inverse sine function evaluated over our specific limits.

### The Final Act

Now, we substitute our limits. For the upper limit, $x = \frac{3\sqrt{3}}{4}$, the expression $\frac{2x}{3}$ becomes:

$$\frac{2}{3} \cdot \frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$$

We know that $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$.

For the lower limit, $x = \frac{3}{2\sqrt{2}}$, the expression $\frac{2x}{3}$ becomes:

$$\frac{2}{3} \cdot \frac{3}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$$

We know that $\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$.

Finally, we calculate the difference:

$$24 \left( \frac{\pi}{3} - \frac{\pi}{4} \right) = 24 \left( \frac{4\pi - 3\pi}{12} \right) = 24 \left( \frac{\pi}{12} \right)$$

The final result is a clean, elegant **$2\pi$**. By staying calm and applying the standard forms, we arrived at a beautiful, simple answer.

Animated Solution for Mathematics - Definite Integration: $48 \int_{3/ 2}^{33 /2} \frac{d x}{9 - 4 x ^{2}}$ is equal to

Select Answer:

Visualized Solution

Analyze the Integral

Standard Integral Form

Factoring the Denominator

Simplifying the Integral

Applying the Formula

Upper Limit Substitution

Lower Limit Substitution

Evaluating Inverse Trigonometry

Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Algebraic Surgery

The Transformation

The Final Act

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