Question

The value of \(\displaystyle\int_{0}^{\sqrt3}\frac{6}{9+x^2}dx\) is equal to

• A. \(\frac{\pi}{3}\)
• B. \(\frac{\pi}{6}\)
• C. \(\frac{\pi}{4}\)
• D. \(\frac{2\pi}{3}\)
• E. 1

Answer 1

The Correct Option is A

We are given the integral \( \int_{0}^{\sqrt{3}} \frac{6}{9 + x^2} \, dx \) and are asked to find the value of the integral.

We can solve this by recognizing the form of the integral. The standard integral for \( \int \frac{dx}{a^2 + x^2} \) is \( \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) \).

Here, we have \( \frac{6}{9 + x^2} \), which is of the form \( \frac{6}{3^2 + x^2} \). So, we apply the formula with \( a = 3 \).

The integral becomes: 

\( \int_{0}^{\sqrt{3}} \frac{6}{9 + x^2} \, dx = \int_{0}^{\sqrt{3}} \frac{2}{3} \cdot \frac{1}{1 + \left( \frac{x}{3} \right)^2} \, dx \).

This simplifies to:

\( \frac{2}{3} \int_{0}^{\sqrt{3}} \frac{dx}{1 + \left( \frac{x}{3} \right)^2} \).

Now, applying the standard formula \( \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) \) with \( a = 3 \), we get:

\( \frac{2}{3} \left[ \tan^{-1} \left( \frac{x}{3} \right) \right]_{0}^{\sqrt{3}} \).

Evaluating the limits:

\( \frac{2}{3} \left( \tan^{-1} \left( \frac{\sqrt{3}}{3} \right) - \tan^{-1}(0) \right) \).

Since \( \tan^{-1} \left( \frac{\sqrt{3}}{3} \right) = \frac{\pi}{6} \), we get:

\( \frac{2}{3} \cdot \frac{\pi}{6} = \frac{\pi}{9} \).

The correct answer is \( \frac{\pi}{6} \).