Problem 84 - Calculus for Biology and Medicine

Question

Evaluate each definite integral. $$ \int_{2}^{3} \frac{1}{1-x^{2}} d x $$

Step-by-Step Solution

Answer

The integral diverges as it involves a hyperbolic function outside its domain.

Step 1: Recognize the Form

The integrand is \( \frac{1}{1-x^2} \). Notice that this resembles the form of the derivative of \( \text{arctanh}(x) \), which is \( \frac{1}{1-x^2} \) when \( |x| < 1 \). Since \( x = 2 \text{ and } 3 \) do not lie in the domain of the arctanh function, check alternatives such as partial fraction decomposition.

Step 2: Review Trigonometric Substitutions

Recognize that the function \( \frac{1}{1-x^2} \) can be related to trigonometric identities. Particularly, the integrand resembles that of a hyperbolic or circular trigonometric function. For a bounded \(|x| < 1\), \( \frac{1}{1-x^2} \) needs partial fraction decomposition for \( x \geq 1 \), integrate using limits.

Step 3: Partial Fractions Decomposition

The expression \( \frac{1}{1-x^2} \) can be written as \( \frac{1}{(1+x)(1-x)} \). This can be decomposed into partial fractions as \( \frac{1}{2} \left[ \frac{1}{1-x} - \frac{1}{1+x} \right] \).

Step 4: Integrate Each Fraction

Integrate each term separately: For \( \int \frac{1}{1-x} \, dx \), the antiderivative is \(-\ln|1-x|\), and for \( \int \frac{1}{1+x} \, dx \), the antiderivative is \(\ln|1+x|\).

Step 5: Evaluate the Definite Integral

Calculate the definite integral \( \int_{2}^{3} \frac{1}{2} \left( - \ln|1-x| + \ln|1+x| \right) \, dx \). Note the antiderivatives are \(-\frac{1}{2} \ln|1-x| + \frac{1}{2} \ln|1+x|\). Compute the value as \( \left[ \frac{1}{2} \ln|4| - \frac{1}{2} \ln|3| \right]_{2}^{3} \).

Step 6: Final Calculation

Substitute \( x=3 \) and \( x=2 \) into the antiderivatives and evaluate: \[ \left. \frac{1}{2} \left( \ln{4} - \ln{3} \right) \right|_{2}^{3} = \left( \ln(\frac{4}{3}) \right) \]. Cancel terms if possible, and check bounds for errors, thus solving gives; \( \infty \) due to undefined behaviour in bounds of integration.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions. This technique is particularly helpful when dealing with integrals of rational functions, where the integrand is a fraction whose numerator and denominator are polynomials.
When the denominator can be factored into linear factors, we can decompose the original function into a sum of simpler fractions, each with one of these linear factors in the denominator. For instance, the expression \( \frac{1}{1-x^2} \) can be written as \( \frac{1}{(1-x)(1+x)} \).

• Step 1: Factor the Denominator - Write \( 1-x^2 \) as \( (1-x)(1+x) \).
• Step 2: Set up the Partial Fractions - Assume \( \frac{1}{1-x^2} = \frac{A}{1-x} + \frac{B}{1+x} \).
• Step 3: Solve for Constants - Use algebraic techniques to solve for \( A \) and \( B \).
• Step 4: Integrate Each Fraction - Find the antiderivative of each simpler fraction separately.

Partial fraction decomposition is especially useful when the function has singularities (points where it is not defined), as it allows us to integrate without those complications.

Trigonometric Substitution

Trigonometric substitution is a technique used to evaluate integrals where a direct substitution can transform the integrand into a simpler form using trigonometric identities. This method is useful when dealing with expressions that involve square roots or quadratic polynomials.
In the context of the exercise \( \int \frac{1}{1-x^2} dx \), a trigonometric substitution is not directly applied because the form can be better managed through partial fraction decomposition. However, understanding the related trigonometric and hyperbolic identities helps recognize potential substitutions in similar problems.

• When the integral involves expressions like \( 1-x^2 \), consider substitutions related to \( \sin(x) \) or \( \tan(x) \).
• Learning these methods enhances one's ability to solve various integral types, though in this specific case, alternatives are preferred because of bounds and singularities.

While trigonometric substitution wasn't used in this solution, it's a powerful tool for integrals where simplification can be achieved.

Antiderivative

An antiderivative, or an indefinite integral, of a function is another function whose derivative is the original function. In the context of definite integrals, finding the antiderivative is often a step towards evaluating the integral over a given interval.
In the problem at hand, the antiderivatives for the partial fractions derived from \( \frac{1}{1-x^2} \) are:

• For \( \int \frac{1}{1-x} \, dx \): The antiderivative is \( -\ln|1-x| \).
• For \( \int \frac{1}{1+x} \, dx \): The antiderivative is \( \ln|1+x| \).

By combining these results, we can evaluate the original integral using the limits to find the exact value over the interval from 2 to 3.
Antiderivatives are crucial in calculus as they enable the transition from the rate of change (derivative) back to the original quantity.

Hyperbolic Trigonometric Functions

Hyperbolic trigonometric functions, such as \( \sinh \), \( \cosh \), and \( \tanh \), are analogs to the ordinary trigonometric functions. These functions often surface in calculus, particularly when dealing with integrals involving expressions like \( 1-x^2 \).
In the exercise related to \( \int \frac{1}{1-x^2} dx \), the derivative of \( \text{arctanh}(x) \), given by \( \frac{1}{1-x^2} \) for \(|x|<1\), initially suggested hyperbolic functions might play a role. However, due to the domain restrictions of \( \text{arctanh}(x) \), partial fraction decomposition was more suitable here.

• \( \text{arctanh}(x) \) is related to the logarithmic form tied to the antiderivatives found in the exercise.
• These functions expand the toolkit for solving integrals that are not easily simplified by polynomial-based methods alone.

Understanding when and how to apply hyperbolic trigonometric functions can simplify many calculus problems and connect different mathematical concepts.