Problem 30 - Calculus for Biology and Medicine

Question

$$ \text { In Problems } , \text { evaluate each integral. } $$ $$ \int \frac{1}{x^{2}+9} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{3} \tan^{-1} \left( \frac{x}{3} \right) + C \).

Step 1: Recognize the Integral Form

The integral \( \int \frac{1}{x^{2}+9} \, dx \) can be recognized as a standard form related to arctangent functions. Specifically, it resembles \( \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \).

Step 2: Identify Constants

In the integral \( \int \frac{1}{x^{2}+9} \, dx \), the expression \( x^{2} + 9 \) matches the form \( x^2 + a^2 \) where \( a^2 = 9 \). Identify the constant \( a \) such that \( a = 3 \).

Step 3: Substitute into Arctangent Formula

Apply the arctangent integral formula \( \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \) with \( a = 3 \). Therefore, the solution is \( \frac{1}{3} \tan^{-1} \left( \frac{x}{3} \right) + C \).

Step 4: Simplify and Conclude

The integral evaluates to \( \frac{1}{3} \tan^{-1} \left( \frac{x}{3} \right) + C \). This is the simplified form of the integral, where \( C \) is the constant of integration.

Key Concepts

Arctangent Function

The arctangent function, often represented as \( \tan^{-1}(x) \) or \( \arctan(x) \), is the inverse of the tangent function. In trigonometry, it is used to find an angle whose tangent is a given number. This function has a distinct property of mapping every real number to angles between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).
When dealing with integrals, the arctangent function becomes crucial in evaluating certain integrals, especially of the form \( \int \frac{1}{x^2 + a^2} \, dx \). The result of such an integral is \( \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \), where:

• \(a\) is a constant, derived from the expression under the integral.
• \(C\) is the integration constant, representing all vertical translations of the function.

Understanding the role of the arctangent function in integration helps bridge trigonometry and calculus, providing a method to solve integrals involving quadratic terms.

Definite Integrals

Definite integrals are a tool in calculus used to calculate the net area under a curve within a specified interval. Unlike indefinite integrals, which result in a family of functions, definite integrals represent a specific numerical value. This value corresponds to the total accumulation of quantities represented by the function over the interval.
When working with expressions like \( \int \frac{1}{x^{2} + 9} \, dx \), it is important to properly evaluate it over given limits if required. The evaluation would involve:

• Finding the antiderivative: Here, it involves using the arctangent formula for integration.
• Applying boundary limits: Substitute the upper and lower limits into the antiderivative and calculate the difference.

This computation helps determine how the function accumulates over a particular range, and it is particularly useful for finding physical quantities like area and volume.

Integration Techniques

In integral calculus, various integration techniques are employed to find antiderivatives of complex functions. The choice of technique largely depends on the form of the function being integrated. Here are some common techniques:

• Substitution: This involves changing variables to simplify the integral, often using a new variable \(u\) that relates to \(x\).
• Integration by Parts: Based on the product rule for differentiation, this technique integrates products of functions.
• Trigonometric Substitution: Used when an integral contains a square root of a quadratic expression, allowing the use of trigonometric identities.
• Partial Fractions: Useful when integrating rational functions by breaking them down into simpler fractions.

For the integral \( \int \frac{1}{x^2 + 9} \, dx \), the arctangent formula was the chosen technique. This highlights how certain forms, like \( \frac{1}{x^2 + a^2} \), align with known antiderivative results, simplifying the process.
Mastering these techniques allows simplification of complex integrals and builds a strong foundation for tackling diverse problems in calculus.