3.6 Q-10

Question Statement

Evaluate the integral: $\int_{0}^{3} \frac{dx}{x^2 + 9}$

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Background and Explanation

To solve this integral, you need to recognize the standard form of an integral involving a sum of squares. Specifically, this is a form of the arctangent function integral. The general formula for integrals of this type is: $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C$

Here, $a = 3$, so we will use this formula to solve the problem. The limits of integration are from 0 to 3, which we will apply after integrating.

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Solution

1. Recognize the Integral Form:

   The integral can be written as: $\int_{0}^{3} \frac{dx}{x^2 + 9}$

   Notice that $9 = 3^2$, so the integral has the form $\frac{1}{x^2 + a^2}$ where $a = 3$.

2. Apply the Formula for Standard Integrals:

   Using the formula for the integral of $\frac{1}{x^2 + a^2}$: $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right)$

   In this case, $a = 3$, so the integral becomes: $\frac{1}{3} \tan^{-1} \left( \frac{x}{3} \right)$

3. Evaluate the Integral at the Limits:

   Now, apply the limits of integration from 0 to 3: $\int_{0}^{3} \frac{dx}{x^2 + 9} = \frac{1}{3} \left[ \tan^{-1} \left( \frac{3}{3} \right) - \tan^{-1} \left( \frac{0}{3} \right) \right]$

4. Simplify the Expression:

   We know that: $\tan^{-1} (1) = \frac{\pi}{4} \quad \text{and} \quad \tan^{-1} (0) = 0$

   Substituting these values into the equation gives: $\frac{1}{3} \left[ \frac{\pi}{4} - 0 \right] = \frac{1}{3} \times \frac{\pi}{4} = \frac{\pi}{12}$

5. Final Result:

   Therefore, the value of the integral is: $\boxed{\frac{\pi}{12}}$

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Key Formulas or Methods Used

• Integral Formula for Arctangent:
  $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right)$

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Summary of Steps

1. Recognize the standard form $\frac{dx}{x^2 + 9}$, where $a = 3$.
2. Apply the arctangent integral formula:
   $\frac{1}{3} \tan^{-1} \left( \frac{x}{3} \right)$
3. Evaluate the integral from 0 to 3: $\frac{1}{3} \left[ \tan^{-1}(1) - \tan^{-1}(0) \right]$
4. Simplify and compute: $\frac{1}{3} \times \frac{\pi}{4} = \frac{\pi}{12}$
5. The final result is: $\boxed{\frac{\pi}{12}}$