Problem 44 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(41-60\) $$\int 4 \cosh (3 x-\ln 2) d x$$

Step-by-Step Solution

Answer

\( \frac{4}{3} \sinh(3x - \ln 2) + C \)

Step 1: Understand Hyperbolic Cosine Function

The hyperbolic cosine function is given by \( \cosh(u) = \frac{e^u + e^{-u}}{2} \), where \( u = 3x - \ln 2 \) in this case. We will integrate the expression \( 4 \cosh(u) \).

Step 2: Setup the Integration

Write the integral as \( \int 4 \cosh(u) \frac{du}{dx} \ dx \). Here, \( u = 3x - \ln 2 \) and \( \frac{du}{dx} = 3 \). Thus, the differential \( dx \) can be expressed as \( \frac{du}{3} \).

Step 3: Substitute and Adjust the Integral

Substitute \( du = 3 dx \), giving \( dx = \frac{du}{3} \). The integral becomes \( \int 4 \cosh(u) \cdot \frac{du}{3} \).

Step 4: Simplify the Integral

Factor out the constant \( \frac{4}{3} \) to get \( \frac{4}{3} \int \cosh(u) \ du \). The integral of \( \cosh(u) \) is \( \sinh(u) + C \), where \( C \) is the integration constant.

Step 5: Solve the Integral

Integrate to find \( \frac{4}{3} \sinh(u) + C \).

Step 6: Substitute Back the Original Variable

Replace \( u \) with \( 3x - \ln 2 \) to get the final answer: \[ \frac{4}{3} \sinh(3x - \ln 2) + C \].

Key Concepts

Hyperbolic Functions

Hyperbolic functions are similar to trigonometric functions, but they are based on hyperbolas rather than circles. They are useful in various areas of mathematics, including calculus, due to their unique properties and relationships. In this exercise, we focus on the hyperbolic cosine function, denoted as \( \cosh(u) \).

• The hyperbolic cosine function is defined as \( \cosh(u) = \frac{e^u + e^{-u}}{2} \), where \( e \) is Euler's number, approximately equal to 2.71828.
• Unlike the regular cosine function, \( \cosh(u) \) is not periodic but is an even function, meaning \( \cosh(-u) = \cosh(u) \).
• This function is used in many calculus problems because of its smoothness and the simplicity of its derivative and integral relationships.

Understanding how \( \cosh(u) \) is derived and its properties will make it easier to work with these functions in integral calculus.

Integration Techniques

Integration is a fundamental concept in calculus that involves finding the area under a curve. Several techniques can simplify this process, one of which is utilized in the given exercise:

• Factoring Out Constants: Since constants such as \( 4 \) can be factored out of the integral, simplify the problem by handling the remaining expression separately.
• Integrating Hyperbolic Functions: The integral of hyperbolic cosine \( \cosh(u) \) is straightforward and leads to the hyperbolic sine function \( \sinh(u) \). This means that the integral \( \int \cosh(u) \, du = \sinh(u) + C \), where \( C \) is the constant of integration.

By understanding these techniques, solving integrals involving hyperbolic functions becomes more manageable, especially when combined with other methods, such as the substitution method.

Substitution Method

The substitution method is a powerful tool for solving integrals, especially when dealing with composite functions. It involves substituting a part of the integrand to simplify the integral.

• Identify the Inner Function: In this case, set \( u = 3x - \ln 2 \), a process that transforms the integral's appearance and simplifies the derivative calculation.
• Calculate the Derivative: We find \( \frac{du}{dx} = 3 \), allowing us to express \( dx \) in terms of \( du \), specifically as \( dx = \frac{du}{3} \).
• Substitute and Simplify: Replace the variables in the integral, resulting in a simpler form of \( \int 4 \cosh(u) \cdot \frac{du}{3} \).

Using substitution can greatly reduce the complexity of an integral, making the overall process easier and more intuitive. This method not only aids in solving problems more effectively but also deepens the understanding of the interdependence between functions.