Question

The value of \(\displaystyle\int_0^{2/3} \frac{dx}{4+9x^2}\) is:

• A. \(\dfrac{\pi}{6}\)
• B. \(\dfrac{\pi}{12}\)
• C. \(\dfrac{\pi}{24}\)
• D. \(\dfrac{\pi}{4}\)

Hint:

Write 4+9x²=9(x²+(2/3)²) and use ∫dx/(x²+a²)=(1/a)tan⁻¹(x/a).

Answer 1

The Correct Option is C

Step 1: Key Formula or Approach:
Use the standard result \(\displaystyle\int \frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C\). Here write \(4+9x^2 = 9\left(x^2 + \dfrac49\right)\), so \(a = \dfrac23\) after rewriting in the form \(9(x^2+a^2)\).

Step 2: Rewriting the integrand:
\[ \frac{1}{4+9x^2} = \frac{1}{9}\cdot\frac{1}{x^2+(2/3)^2} \]
So \(\displaystyle\int \frac{dx}{4+9x^2} = \frac19 \cdot \frac{1}{2/3}\tan^{-1}\left(\frac{x}{2/3}\right) + C = \frac16 \tan^{-1}\left(\frac{3x}{2}\right) + C\).

Step 3: Applying the limits:
\[ \left[\frac16\tan^{-1}\left(\frac{3x}{2}\right)\right]_0^{2/3} = \frac16\left(\tan^{-1}(1) - \tan^{-1}(0)\right) = \frac16\left(\frac{\pi}{4} - 0\right) = \frac{\pi}{24} \]

Final Answer:
The value of the definite integral is \(\pi/24\). \[ \boxed{\dfrac{\pi}{24}} \]