Question

In the following exercises, evaluate each integral in terms of an inverse trigonometric function. \(\int_{1 / \sqrt{3}}^{\sqrt{3}} \frac{d x}{1+x^{2}}\)

Short answer

The integral evaluates to \( \frac{\pi}{6} \).

Step-by-step solution

Recognize the Formula

The integral \( \int \frac{dx}{1+x^2} \) is a standard form that corresponds to the derivative of the inverse trigonometric function \( \arctan(x) \). The formula for the derivative is \( \frac{d}{dx}(\arctan(x)) = \frac{1}{1+x^2} \). Therefore, \( \int \frac{dx}{1+x^2} = \arctan(x) + C \), where \( C \) is the constant of integration.

Apply Limits to the Antiderivative

As we are dealing with a definite integral from \( \frac{1}{\sqrt{3}} \) to \( \sqrt{3} \), we need to evaluate \( \arctan(x) \) over these limits. This means evaluating \( \arctan(x) \) at the upper limit \( \sqrt{3} \) and subtracting its evaluation at the lower limit \( \frac{1}{\sqrt{3}} \).

Evaluate at Upper Limit

Substitute \( x = \sqrt{3} \) into \( \arctan(x) \). Knowing that \( \arctan(\sqrt{3}) = \frac{\pi}{3} \), compute this result as the value of the antiderivative at the upper limit.

Evaluate at Lower Limit

Substitute \( x = \frac{1}{\sqrt{3}} \) into \( \arctan(x) \). Since \( \arctan(\frac{1}{\sqrt{3}}) = \frac{\pi}{6} \), calculate this result as the value of the antiderivative at the lower limit.

Calculate the Definite Integral

Subtract the result from the lower limit from the result at the upper limit: \( \frac{\pi}{3} - \frac{\pi}{6} \). Perform the subtraction by finding a common denominator: \( \frac{2\pi}{6} - \frac{\pi}{6} = \frac{\pi}{6} \). This is the value of the definite integral.

Key concepts

Definite Integrals

A definite integral is a fundamental concept in calculus that represents the area under a curve between two points on the x-axis. It is expressed as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration, and \( f(x) \) is the function being integrated.
The definite integral can be thought of as summing up an infinite number of infinitesimally small areas under the curve between \( a \) and \( b \). It provides a numerical value representing something physical, like distance, area, or volume.
Here are key points to remember about definite integrals:

• The limits \( a \) and \( b \) are crucial for determining the exact area under the curve between these points.
• The Fundamental Theorem of Calculus connects differentiation and integration, providing a method to compute definite integrals using antiderivatives.
• Definite integrals can have negative values if the curve lies below the x-axis between \( a \) and \( b \).

Overall, definite integrals are a core tool in calculus, enabling solutions to a wide range of real-world problems and mathematical contexts.

Arctan Function

The arctan function, also known as the inverse tangent function, is a trigonometric function denoting the angle whose tangent is a given number.
It is represented as \( \arctan(x) \), and is the inverse function of the tangent function, \( \tan(x) \). This means that if \( y = \arctan(x) \), then \( \tan(y) = x \).
Understanding the properties of the arctan function is essential for solving integrals involving it:

• The range of the arctan function is \( \frac{-\pi}{2} \) to \( \frac{\pi}{2} \) radians.
• Its domain extends from negative to positive infinity, \( \(-\infty, \infty\) \).
• The derivative of \( \arctan(x) \) is \( \frac{1}{1+x^2} \), which shapes how we handle integrals involving this function.

In calculus, recognizing how the arctan function interacts with other functions is vital, especially when encountered as an integral like \( \int \frac{dx}{1+x^2} \), which simplifies directly to \( \arctan(x) + C \). Functions and angles represented with the help of inverse trigonometric expressions like arctan reveal underlying geometric meanings that are invaluable in both pure and applied mathematics.

Calculus Problem Solving

Calculus problem solving covers a broad spectrum of techniques and strategies used to analyze and solve mathematical problems involving derivatives, integrals, and limits. It requires a solid understanding of fundamental concepts and the ability to apply them to various scenarios.
When confronted with a calculus problem, follow these basic steps to guide your approach:

• Identify the type of problem: Derivative, indefinite integral, definite integral, limit, etc.
• Recognize standard forms: Many calculus problems involve recognizing familiar patterns, such as inverse trigonometric integrals.
• Apply relevant theorems: Use the rules and theorems such as the Fundamental Theorem of Calculus to simplify or directly solve problems.
• Evaluate systematically: Break down the problem into smaller, manageable parts and solve each one step-by-step.
• Check your work: Verify answers by differentiating results (for integrals) or calculating limits directly (for derivative problems).

Effective problem solving in calculus often involves intuitive thinking coupled with methodical execution. Whether integrating an inverse trigonometric function or tackling more complex scenarios, the essential skills developed through calculus enhance mathematical reasoning and problem-solving abilities across various disciplines.