Problem 49 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(39-56\) $$ \int \frac{3 \sec ^{2} t}{6+3 \tan t} d t $$

Step-by-Step Solution

Answer

The integral evaluates to \( \ln |2 + \tan t| + C \).

Step 1: Simplify the Integrand

First, observe that both the numerator and the denominator have a common factor of 3. Factor out the 3 from both the numerator and the denominator: \[ \int \frac{3 \sec^2 t}{6 + 3 \tan t} \, dt = \int \frac{\sec^2 t}{2 + \tan t} \, dt. \] This simplifies our integral and makes it easier to evaluate.

Step 2: Make a Substitution

Let \( u = 2 + \tan t \). Then, \( du/dt = \sec^2 t \), which implies \( du = \sec^2 t \, dt \). Substitute \( u \) and \( du \) into the integral: \[ \int \frac{\sec^2 t}{2 + \tan t} \, dt = \int \frac{1}{u} \, du. \] This substitution simplifies the integral to the natural logarithmic form.

Step 3: Evaluate the Integral

Now evaluate the integral \( \int \frac{1}{u} \, du \). This is a standard integral that results in the natural logarithm: \[ \int \frac{1}{u} \, du = \ln |u| + C, \] where \( C \) is the constant of integration.

Step 4: Back-Substitute to Original Variable

Substitute back \( u = 2 + \tan t \) to express the integral in terms of the original variable: \[ \ln |u| + C = \ln |2 + \tan t| + C. \] This is the solution to the original integral problem with the integration constant.

Key Concepts

Trigonometric Integration

Trigonometric integration involves integrating functions that include trigonometric functions such as sine, cosine, tangent, and their reciprocals. These functions often appear in calculus problems, and knowing how to deal with them is essential. In this exercise, we notice the presence of the secant (\( \sec t \)) and tangent (\( \tan t \)) functions.To tackle trigonometric integrals, one often uses substitution to simplify the expression, as we did by substituting \( u = 2 + \tan t \). This kind of integration by substitution is especially useful for integrals involving compositions of trigonometric and algebraic functions. Simplifying the integrand by factoring out common terms also reduces complexity, making the integral more approachable.

Definite and Indefinite Integrals

Integrals can be either definite or indefinite. In our exercise, we dealt with an indefinite integral, which is essentially the antiderivative of a function. This process does not have upper and lower limits of integration, and it comes with a constant of integration, \( C \), representing a family of functions.

• Indefinite Integrals: They provide the general form of antiderivatives and require adding \( C \), as seen when we found the solution \( \ln |2 + \tan t| + C \)
• Definite Integrals: In contrast, result in a specific numerical value, as they have limits applied to the antiderivative evaluation.

Understanding the difference between these types of integrals helps students apply the correct method in different contexts.

Natural Logarithmic Functions

The natural logarithm is a function that plays a crucial role in calculus, particularly in integration. It arises naturally when integrating the reciprocal of a linear function, like \( \int \frac{1}{u} \, du \), which yields \( \ln |u| + C \).In our exercise, after simplifying the integrand with substitution methods, we converted it into a form where applying the natural logarithm was straightforward. Natural logarithmic functions are essential for solving problems involving growth models, radioactive decay, and in scenarios where exponential functions are present.Recognizing when a trigonometric or algebraic function can translate into a natural logarithm through integration can significantly simplify solving complex calculus problems.