Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-12, evaluate the given integral. $$ \int_{0}^{1 / 2} \frac{1}{1-t^{2}} d t $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{2} \ln(3) \).

Step 1: Identifying the Integral Type

The given integral \( \int_{0}^{1/2} \frac{1}{1-t^2} \, dt \) is a definite integral with a rational function in the integrand. Specifically, it resembles the form of a partial fraction decomposition problem if we were to evaluate it without any simplifications.

Step 2: Using an Algebraic Identity

Notice that \( \frac{1}{1-t^2} \) can be rewritten using the identity for a hyperbolic tangent inverse. The formula \( \frac{1}{1-t^2} = \sum \), typically needs expansion or hyperbolic functions, but in this case, we recognize it as a standard integral value.

Step 3: Recognizing a Standard Integral

The integral \( \int \frac{1}{1-x^2} \, dx \) is the derivative of the inverse hyperbolic tangent function. In simpler terms, this integral evaluates to directly \( \text{artanh}(x) + C \) in an indefinite form. For definite form, the boundaries apply.

Step 4: Evaluating the Definite Integral

Use the formula found: \( \int_{0}^{1/2} \frac{1}{1-t^2} \, dt = \text{artanh}(t) \bigg|_{0}^{1/2} = \text{artanh}(1/2) - \text{artanh}(0) \).\text{artanh}(0) = 0, so we only need to find \text{artanh}(1/2).

Step 5: Calculating \(\text{artanh}(1/2)\)

The inverse hyperbolic tangent function, \( \text{artanh}(x) \), can be computed as \( \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) \). Substitute \( x = 1/2 \) to get:\[\text{artanh}(1/2) = \frac{1}{2} \ln \left( \frac{1+1/2}{1-1/2} \right) = \frac{1}{2} \ln(3) \].

Key Concepts

Partial Fraction Decomposition

When faced with a rational function integral, like \( \frac{1}{1-t^2} \), partial fraction decomposition can be a handy tool. This technique helps break down complex fractions into simpler ones, making integration more manageable. Unfortunately, this specific integral does not require the traditional path of partial fraction decomposition, as it naturally aligns with a well-known identity from hyperbolic trigonometry.
This approach involves expressing a rational function as the sum of simpler fractions, which are easier to integrate individually.
In many cases, partial fraction decomposition is necessary to evaluate integrals. This integral, however, fits another category: standard integral values that greatly simplify our work.

Inverse Hyperbolic Tangent

The inverse hyperbolic tangent function, known as \( \text{artanh}(x) \), is significant in calculus, particularly when calculating integrals involving the fraction \( \frac{1}{1-t^2} \).
This function is the integral of the rational function \( \frac{1}{1-t^2} \), and appears frequently in these calculations. The inverse hyperbolic tangent provides a direct pathway to resolving certain types of integrals.

• The calculation of \( \text{artanh}(x) \) is represented by \( \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) \).
• Knowing this allows us to find the indefinite integral, \( \int \frac{1}{1-t^2} \, dt \).

This formula simplifies the evaluation process, especially when the definite integral involves specific limits like in this problem. Evaluating at these limits gives us the answer without additional algebraic manipulations.

Rational Functions

Rational functions, like \( \frac{1}{1-t^2} \), offer a world of integration challenges and possibilities. These are functions expressed as a fraction of two polynomials.
Learning how to handle rational functions is a crucial skill in calculus. It involves:

• Identifying patterns such as those that match standard integral forms.
• Breaking down complex expressions if necessary, using tools like partial fractions.

Understanding and identifying these functions in integrals allows solving them efficiently. While some require splitting and simplifying, others, like our example, match a standard integral value for a more straightforward computation.

Standard Integral Values

Standard integral values are pre-determined solutions for common integral forms which simplify the process of solving integrals. Recognizing these can save a great deal of time.
In this exercise, the integral \( \int \frac{1}{1-t^2} \, dt \) aligns with the standard integral value associated with the inverse hyperbolic tangent. We use this pre-existing knowledge to compute the definite integral rapidly.

• Instead of complex algebra, you can directly use the known results, such as \( \text{artanh}(t) \).
• This saves effort and reduces the chance for calculation errors during integration.

Grasping these standard values is beneficial for anyone tackling calculus problems, speeding up evaluation and understanding of integrals.