Problem 4 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Which of the two indefinite integrals $$ \int \frac{1}{1+t^{2}} t d t \quad \text { or } \int \frac{1}{1+t^{2}} d t $$ is \(\ln \left(1+t^{2}\right)^{0.5}+C ?\) Explain your answer. Note: The other indefinite integral is arctan \(t+C\).

Step-by-Step Solution

Answer

The integral \( \int \frac{1}{1+t^{2}} t \, dt \) equals \( \ln((1+t^2)^{0.5}) + C \).

Step 1: Analyze the first integral

Consider the first integral \( \int \frac{1}{1+t^{2}} t \, dt \). This can be evaluated using a substitution method. Let \( u = 1 + t^2 \), then \( du = 2t \, dt \), or equivalently, \( t \, dt = \frac{1}{2} \, du \). Substitute into the integral to get \( \frac{1}{2} \int \frac{1}{u} \, du \).

Step 2: Integrate after substitution

After the substitution, the integral becomes \( \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln|1+t^2| + C \). This simplifies to \( \ln((1+t^2)^{0.5}) + C \).

Step 3: Analyze the second integral

Consider the second integral \( \int \frac{1}{1+t^{2}} \, dt \). This is a standard integral that results in the expression \( \tan^{-1} t + C \), because the derivative of \( \tan^{-1} t \) is \( \frac{1}{1+t^{2}} \).

Step 4: Compare both integrals with given expressions

The expression \( \ln((1+t^2)^{0.5}) + C \) matches the result of the first integral \( \int \frac{1}{1+t^{2}} t \, dt \). The expression \( \tan^{-1} t + C \) matches the result of the second integral \( \int \frac{1}{1+t^{2}} \, dt \).

Key Concepts

Substitution Method

The substitution method is a powerful technique used in calculus to simplify complex integrals by changing variables. It is particularly useful when dealing with integrals where the integrand contains a composite function or its derivative. Here's how it works:

• Identify a portion of the integrand that can be substituted with a single variable. This part often looks like a derivative of another part of the expression.
• Assign this expression to a new variable, say \( u \), and find its derivative, \( du \), in terms of the original variable.
• Substitute \( u \) and \( du \) into the integral, transforming it into an easier integral to solve.
• After integration, substitute back the original variable to get the final result.

For example, in the integral \( \int \frac{1}{1 + t^2} t \, dt \), by setting \( u = 1 + t^2 \), we derived \( du = 2t \, dt \) leading to a transformed integral of \( \frac{1}{2} \int \frac{1}{u} \, du \). This simplification made it possible to integrate easily by recognizing it as a natural logarithmic form.

Integration Techniques

Integration techniques encompass various methods used to find antiderivatives, which are crucial in solving indefinite integrals. Here are some popular techniques:

• Basic Antiderivative Rules: These include power rule, exponential rule, and trigonometric rules, typically the foundation for integration.
• Substitution Method: As already discussed, this method simplifies the integral through variable transformation.
• Integration by Parts: Useful for products of functions, based on the product rule for derivatives.
• Partial Fraction Decomposition: Breaks down rational expressions into simpler fractions, useful for polynomials.

The specific technique employed depends on the structure of the integrand. Recognizing patterns in the integrand allows for the appropriate technique to be implemented. In our exercise, substitution was the key technique for simplifying the integral \( \int \frac{1}{1 + t^2} t \, dt \), transforming it using a logarithmic antiderivative.

Arctangent Function

The arctangent function, also known as the inverse tangent, is critical in integration problems involving the form \( \frac{1}{1+t^2} \). This is because the derivative of \( \arctan(t) \) is exactly \( \frac{1}{1+t^2} \).

This unique property allows for simple evaluation of integrals that match this pattern. When given \( \int \frac{1}{1+t^2} \, dt \), we can directly write down the antiderivative as \( \tan^{-1}(t) + C \), where \( C \) is the constant of integration.

Understanding the role of the arctangent function in calculus not only helps in integration but also provides insights into the behavior of functions, particularly in relation to angles and trigonometric identities. In our exercise, recognizing the form of the second integral allowed us to immediately identify it with \( \tan^{-1}(t) + C \), simplifying the solution process.