Problem 43 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(41-60\) $$\int 6 \cosh \left(\frac{x}{2}-\ln 3\right) d x$$

Step-by-Step Solution

Answer

The integral evaluates to \( 2e^{\frac{x}{2}} - 18e^{-\frac{x}{2}} + C \).

Step 1: Identify the Integral

We need to evaluate the integral \( \int 6 \cosh \left( \frac{x}{2} - \ln 3 \right) \, dx \). The integrand involves a hyperbolic cosine function, \( \cosh \), and we will use integration techniques suitable for hyperbolic functions.

Step 2: Apply the Hyperbolic Identity

Recall that \( \cosh(u) = \frac{e^u + e^{-u}}{2} \). Here, the argument for \( \cosh \) is \( u = \frac{x}{2} - \ln 3 \). We can use this identity to express the integrand as a sum of exponentials.

Step 3: Rewrite the Integral

The integral becomes \( \int 6 \left( \frac{e^{\frac{x}{2} - \ln 3} + e^{-\left( \frac{x}{2} - \ln 3 \right)} }{2} \right) \, dx \). Simplify this to \( 3 \int \left( e^{\frac{x}{2} - \ln 3} + e^{-\left( \frac{x}{2} - \ln 3 \right)} \right) \, dx \).

Step 4: Simplify Exponential Terms

Recognize that \( e^{\frac{x}{2} - \ln 3} = \frac{e^{\frac{x}{2}}}{3} \) and \( e^{-\left( \frac{x}{2} - \ln 3 \right)} = 3e^{-\frac{x}{2}} \). Substitute these into the integral.

Step 5: Separate the Integral

Separate and simplify the integral: \( 3 \left( \int \frac{e^{\frac{x}{2}}}{3} \, dx + \int 3e^{-\frac{x}{2}} \, dx \right) \). This simplifies to \( \int e^{\frac{x}{2}} \, dx + 9 \int e^{-\frac{x}{2}} \, dx \).

Step 6: Integrate Each Term

For the first term, \( \int e^{\frac{x}{2}} \, dx \), use substitution: let \( u = \frac{x}{2} \), then \( du = \frac{1}{2}dx \) or \( dx = 2du \). The integral becomes \( 2 \int e^u \, du = 2e^u + C_1 \). Substitute back: \( 2e^{\frac{x}{2}} + C_1 \).

Step 7: Continue with the Second Term

For the second term, \( 9 \int e^{-\frac{x}{2}} \, dx \), use substitution: let \( v = -\frac{x}{2} \), then \( dv = -\frac{1}{2}dx \) or \( dx = -2dv \). The integral becomes \( -18 \int e^v \, dv = -18e^v + C_2 \). Substitute back: \( -18e^{-\frac{x}{2}} + C_2 \).

Step 8: Combine the Results

Combine the integrated terms: \( 2e^{\frac{x}{2}} - 18e^{-\frac{x}{2}} + C \), where \( C \) is the constant of integration, combining \( C_1 \) and \( C_2 \).

Step 9: Write the Final Answer

The evaluated integral is \( 2e^{\frac{x}{2}} - 18e^{-\frac{x}{2}} + C \).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola. They are crucial in calculus because they behave similarly to the well-known sine and cosine functions but use exponential components. Key hyperbolic functions include \( \sinh \) (hyperbolic sine) and \( \cosh \) (hyperbolic cosine). The function \( \cosh(x) \) is defined as:

• \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)

This identity is powerful because it simplifies expressions involving hyperbolic functions into exponential terms. When integrating hyperbolic functions like \( \int 6 \cosh \left( \frac{x}{2} - \ln 3 \right) \, dx \), transforming \( \cosh \) into exponential components allows the use of familiar techniques. Recognizing and applying hyperbolic identities effectively can simplify complex integrative processes.

Exponential Functions

Exponential functions are critical in calculus, defined as \( e^x \), where \( e \) is Euler's number, approximately 2.71828. These functions are potent tools because they model growth and decay processes in various fields.
When working with integrals, exponential functions enable simplification. By employing mathematical identities, such as rewriting hyperbolic functions into exponentials, integration becomes more manageable.In the given integral, terms are rewritten using exponential functions:

• \( e^{\frac{x}{2} - \ln 3} = \frac{e^{\frac{x}{2}}}{3} \)
• \( e^{-\left( \frac{x}{2} - \ln 3 \right)} = 3e^{-\frac{x}{2}} \)

These transformations simplify the integrand, facilitating the application of basic integration rules. Understanding these properties is essential for breaking down complex calculus problems into solvable parts.

Substitution Method

The substitution method, or "u-substitution," is a powerful technique for solving integrals, simplifying the process by changing variables. Essentially, substitution allows a complex expression to become simpler and solvable.
In this exercise, we apply substitution twice:

• For \( \int e^{\frac{x}{2}} \, dx \), set \( u = \frac{x}{2} \), yielding \( du = \frac{1}{2}dx \) or \( dx = 2du \).
• For \( 9 \int e^{-\frac{x}{2}} \, dx \), set \( v = -\frac{x}{2} \), yielding \( dv = -\frac{1}{2}dx \) or \( dx = -2dv \).

Substitutions are crucial because they transform difficult integrals into easier ones by changing the variable, leading to simpler forms. Knowing when and how to apply substitution effectively is an indispensable skill in calculus.

Definite and Indefinite Integrals

Integrals are a core component of calculus; they come in two types: definite and indefinite. An indefinite integral represents a family of functions and includes a constant of integration \( C \). This constant arises because integration is the reverse process of differentiation, which loses the original constant.
For example, integrating \( \int e^{\frac{x}{2}} \, dx \) results in \( 2e^{\frac{x}{2}} + C \). The constant \( C \) represents any possible vertical shift, ensuring all antiderivatives are accounted for.

• Indefinite Integral: \( \int f(x) \, dx = F(x) + C \) where \( F \) is the antiderivative of \( f \).
• Definite Integral: Represents the area under a curve from \( a \) to \( b \) without considering \( C \).

Understanding the differences and contexts for applying definite and indefinite integrals allows for the appropriate handling of various calculus problems, ranging from calculating areas to finding general solutions to differential equations.