Question Statement

Evaluate the integral:

$\int \frac{dx}{x^2+9}$

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Background and Explanation

This integral follows the standard form $\int \frac{dx}{x^2+a^2}$, which typically requires trigonometric substitution to derive the inverse tangent solution. The key technique involves substituting $x=a\tan \theta$ to exploit the identity $1+{\tan }^2\theta ={\sec }^2\theta$.

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Solution

Let $I$ denote the integral we wish to evaluate:

$I=\int \frac{dx}{x^2+9}$

First, recognize that the denominator is a sum of squares, which can be written as:

$I=\int \frac{1}{x^2+3^2}dx$

To simplify this, we apply the trigonometric substitution $x=3\tan \theta$. This substitution is chosen because it will allow us to factor out $9$ and use the Pythagorean identity to simplify the denominator.

Differentiating $x=3\tan \theta$ with respect to $\theta$:

$dx=3{\sec }^2\theta d\theta$

Substituting both $x$ and $dx$ into the integral:

$I=\int \frac{1}{(3\tan \theta {)}^2+3^2}\cdot 3{\sec }^2\theta d\theta$

Expand the squared term in the denominator:

$I=\int \frac{3{\sec }^2\theta }{9{\tan }^2\theta +9}d\theta$

Factor out $9$ from the denominator:

$I=\int \frac{3{\sec }^2\theta }{9({\tan }^2\theta +1)}d\theta$

Simplify the numerical coefficient $\frac{3}{9}=\frac{1}{3}$:

$I=\frac{1}{3}\int \frac{{\sec }^2\theta }{{\tan }^2\theta +1}d\theta$

Apply the Pythagorean identity ${\tan }^2\theta +1={\sec }^2\theta$:

$I=\frac{1}{3}\int \frac{{\sec }^2\theta }{{\sec }^2\theta }d\theta$

The ${\sec }^2\theta$ terms cancel, leaving:

$I=\frac{1}{3}\int 1d\theta$

Integrating with respect to $\theta$:

$I=\frac{1}{3}\theta +c$

Now we convert back to the original variable $x$. From our substitution $x=3\tan \theta$, we solve for $\theta$:

$\tan \theta =\frac{x}{3}$

Therefore:

$\theta ={\tan }^{-1}\left(\frac{x}{3}\right)$

Substituting this back into our expression for $I$:

$I=\frac{1}{3}{\tan }^{-1}\left(\frac{x}{3}\right)+c$

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Key Formulas or Methods Used

• Trigonometric substitution: $x=a\tan \theta$ for integrals containing $x^2+a^2$, with $dx=a{\sec }^2\theta d\theta$
• Pythagorean identity: ${\tan }^2\theta +1={\sec }^2\theta$
• Standard integral form: $\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+c$ (here $a=3$)
• Algebraic simplification: Factoring and cancellation of common terms

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Summary of Steps

1. Identify the integral as type $\frac{1}{x^2+a^2}$ with $a=3$, rewriting as $\int \frac{dx}{x^2+3^2}$
2. Apply trigonometric substitution: let $x=3\tan \theta$, so $dx=3{\sec }^2\theta d\theta$
3. Substitute into the integral and factor the denominator: $\frac{3{\sec }^2\theta }{9({\tan }^2\theta +1)}$
4. Simplify using ${\tan }^2\theta +1={\sec }^2\theta$ to reduce to $\frac{1}{3}\int d\theta$
5. Integrate to obtain $\frac{1}{3}\theta +c$
6. Back-substitute using $\theta ={\tan }^{-1}\left(\frac{x}{3}\right)$ to get the final answer $\frac{1}{3}{\tan }^{-1}\left(\frac{x}{3}\right)+c$