Problem 70 - Thomas Calculus in SI Units

Question

Evaluate the integrals in Exercises \(67-74\) in terms of a. inverse hyperbolic functions. b. natural logarithms. $$\int_{0}^{1 / 2} \frac{d x}{1-x^{2}}$$

Step-by-Step Solution

Answer

\( \frac{1}{2} \ln(3) \).

Step 1: Recognize the Form of the Integral

The integral \( \int \frac{1}{1-x^2} \, dx \) is of a standard form that can be evaluated using inverse hyperbolic functions as well as natural logarithms. Recognize it as resembling the derivative of the inverse hyperbolic tangent.

Step 2: Solution Using Inverse Hyperbolic Functions

The integral \( \int \frac{1}{1-x^2} \, dx \) corresponds to inverse hyperbolic tangent \( \text{artanh}(x) \). So the antiderivative of \( \frac{1}{1-x^2} \) is \( \text{artanh}(x) \). Thus, \( \int_{0}^{1/2} \frac{1}{1-x^2} \, dx = \text{artanh} \left( \frac{1}{2} \right) - \text{artanh}(0) \). Noting that \( \text{artanh}(0) = 0 \), evaluate \( \text{artanh}(\frac{1}{2}) \).

Step 3: Evaluate artanh(1/2)

The function \( \text{artanh}(x) \) is defined as \( \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) \). Substitute \( x = \frac{1}{2} \) to get \( \text{artanh}(1/2) = \frac{1}{2} \ln \left( \frac{1+1/2}{1-1/2} \right) = \frac{1}{2} \ln \left( \frac{3/2}{1/2} \right) = \frac{1}{2} \ln(3) \).

Step 4: Solution Using Natural Logarithms

The integral \( \int \frac{1}{1-x^2} \, dx \) can also be expressed in terms of natural logarithms using partial fraction decomposition: \( \frac{1}{1-x^2} = \frac{1}{2} \left( \frac{1}{1-x} + \frac{1}{1+x} \right) \). Integration yields: \( \frac{1}{2} (\ln|1-x| - \ln|1+x|) + C \). Thus, \( \int_{0}^{1/2} \frac{1}{1-x^2} \, dx = \frac{1}{2} \left[ \ln \left( \frac{3/2}{1/2} \right) \right] = \frac{1}{2} \ln(3) \).

Step 5: Final Evaluation

Both methods (inverse hyperbolic functions and natural logarithms) have confirmed the same result. The value of the integral \( \int_{0}^{1/2} \frac{d x}{1-x^{2}} \) is \( \frac{1}{2} \ln(3) \).

Key Concepts

Integral Calculus

Integral calculus deals with the concept of integrating functions to find areas under curves, solve differential equations, or compute quantities. The process of integration accumulates the values to find a total value, opposite to differentiation which finds a rate of change.
In this exercise, the integral of the function \( \int \frac{d x}{1-x^{2}} \) focuses on the area under the curve from 0 to 1/2. This involves the integration process, which may employ various methods.
The function \( \frac{1}{1-x^{2}} \) is special because it can be associated with inverse hyperbolic functions, allowing the use of specific integration techniques. Evaluating definite integrals requires computing the antiderivative, followed by the application of the limits; in this case, from 0 to 1/2. This exercise demonstrates how knowledge of integral calculus helps in finding precise values for these antiderivatives, which is often applicable in practical contexts.

Natural Logarithms

Natural logarithms, denoted as \( \ln(x) \), are logarithms with the base \( e \), where \( e \approx 2.71828 \) is the Euler's number. They have unique properties that make them crucial in calculus, particularly when integrating functions.
In the given exercise, natural logarithms come into play while using inverse hyperbolic functions and partial fraction decomposition. The connection arises because inverse hyperbolic functions, like the inverse hyperbolic tangent, can be expressed through natural logarithms.
For example, \( \text{artanh}(x) = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) \) directly relates the inverse hyperbolic tangent to natural logarithms. This relationship surfaces during solving the integral \( \int \frac{d x}{1-x^{2}} \), showing how logarithms can effectively find antiderivatives.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to simplify complex rational expressions into simpler fractions that are easier to integrate. This method is particularly useful when dealing with rational functions like \( \frac{1}{1-x^{2}} \).
The decomposition for \( \frac{1}{1-x^{2}} \) is \( \frac{1}{2} \left( \frac{1}{1-x} + \frac{1}{1+x} \right) \). By breaking it down into these simpler components, each part can be individually integrated using basic calculus techniques.

• Each fraction is simpler, reducing integration tasks to more manageable, basic natural logarithmic antiderivatives.
• The method highlights the usefulness and versatility of decomposition in tackling integral problems efficiently.

The decomposition technique simplifies the integration steps, making complex integrals like the one in this exercise more approachable, demonstrating the importance of partial fraction decomposition in integral calculus.