Problem 43 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

The integral is \( 12 \sinh\left(\frac{x}{2} - \ln 3\right) + C \).

Step 1: Identify the Integral Expression

We're given the integral \( \int 6 \cosh \left(\frac{x}{2} - \ln 3\right) dx \). Our goal is to find the antiderivative of the function.

Step 2: Simplify the Integral

To simplify, let us use a substitution. Set \( u = \frac{x}{2} - \ln 3 \), which leads to \( du = \frac{1}{2} dx \). This implies \( dx = 2 du \). Substitute into the integral.

Step 3: Apply Substitution

Substituting, we have the integral \( 6 \int \cosh(u) \cdot 2 \, du \), which simplifies to \( 12 \int \cosh(u) \, du \).

Step 4: Integrate \( \cosh(u) \)

The integral of \( \cosh(u) \) is simply \( \sinh(u) \). Thus, \( 12 \int \cosh(u) \, du = 12 \sinh(u) + C \), where \( C \) is the constant of integration.

Step 5: Substitute Back for \( x \)

Recall the substitution \( u = \frac{x}{2} - \ln 3 \). Thus, \( \sinh(u) = \sinh(\frac{x}{2} - \ln 3) \). So, the integral becomes \( 12 \sinh\left(\frac{x}{2} - \ln 3\right) + C \).

Key Concepts

Substitution Method

The substitution method is quite useful when solving integrals, especially when the integrand is complex or cumbersome. Here's a friendly breakdown of how the substitution works.

The main idea of the substitution method is to transform the integrand into something simpler. This is done by changing the variable of integration. By choosing a suitable substitution, we can often make an integral easier to handle.

• Pick a new variable, commonly denoted as \( u \), to replace a complicated part of the integrand.
• Calculate \( du \), which represents the differential of the new variable. This step involves taking the derivative of the \( u \) expression with respect to the original variable.
• Substitute both \( u \) and \( du \) into the integral. This changes the integral from a function of \( x \) to a function of \( u \). This transformation can simplify the expression significantly.

Finally, after integrating with respect to \( u \), always remember to transform back to the original variable (\( x \) in this case), so that the solution is in terms of the variable given in the problem statement.

Antiderivatives

Finding antiderivatives is essentially what integration is all about. An antiderivative is a function whose derivative is the given function. In other words, if \( F'(x) = f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \).

When you see an integral, you're tasked with finding an antiderivative. In this context, the integral sign \( \int \) is a request to find an antiderivative.

• An indefinite integral, like in this exercise, yields a general antiderivative plus a constant \( C \). This constant represents any constant value that could be part of the original function.
• The reason the constant \( C \) is included is that the derivative of a constant is zero. Thus, multiple functions with different constants could share the same derivative.

The key is selecting suitable antiderivatives that match the function inside the integral. For example, the integral of \( \cosh(u) \) became \( \sinh(u) \), noted by how these hyperbolic functions are antiderivatives of each other.

Hyperbolic Functions

Hyperbolic functions, like their trigonometric cousins, are essential in calculus. They come into play in various integration problems due to their unique properties.

The core hyperbolic functions are \( \sinh(x) \) and \( \cosh(x) \). They are defined as:

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

What makes hyperbolic functions interesting is their relationship with exponential functions. They are not only structurally similar to sine and cosine, but they also connect to exponential growth.

In the context of the integral \( \int 6 \cosh(\frac{x}{2} - \ln 3) \, dx \), the hyperbolic cosine function \( \cosh(u) \) was integral to simplifying the problem. Recognizing that the antiderivative of \( \cosh(u) \) is \( \sinh(u) \) helps streamline these calculations.

Understanding hyperbolic functions can greatly benefit calculus students, as they provide a powerful tool for solving complex integrals much like trigonometric functions do in geometry contexts.