Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-54, perform the indicated integrations. \(\int \cosh 3 x d x\)

Step-by-Step Solution

Answer

The integral is \(\frac{1}{3} \sinh(3x) + C\).

Step 1: Identify the Integral Formula

The integral we need to solve is \( \int \cosh 3x \, dx \). The basic integral formula for the hyperbolic cosine function is \( \int \cosh(ax) \, dx = \frac{1}{a} \sinh(ax) + C \), where \(a\) is a constant and \(C\) is the integration constant.

Step 2: Determine the Value of 'a'

In the given integral, \(a\) corresponds to 3 because the function under integration is \(\cosh(3x)\). This means in applying the formula, \(a = 3\).

Step 3: Apply the Integral Formula

Using the formula from Step 1, we substitute \(a = 3\) into \(\frac{1}{a} \sinh(ax) + C\). This gives \(\frac{1}{3} \sinh(3x) + C\).

Step 4: Write the Final Result

Thus, the integral \(\int \cosh 3x \, dx \) is \(\frac{1}{3} \sinh(3x) + C\), where \(C\) is the constant of integration.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but are based on hyperbolas instead of circles. These functions are often used in calculus, especially in problems involving integration and differentiation. One of the essential hyperbolic functions is the hyperbolic cosine, denoted as \( \cosh(x) \). This function can be expressed in terms of exponential functions as:

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

This representation is useful because it reveals that hyperbolic functions have properties similar to trigonometric functions but are exponential in nature. Hyperbolic functions are prevalent in many areas, such as physics, where they describe hyperbolic motion and waves. They also appear in engineering problems, particularly in systems that model electrical currents and wave propagation.
In our exercise, we focus on the hyperbolic cosine, \( \cosh(3x) \). When integrating hyperbolic functions like \( \cosh(x) \), it is helpful to recall their definitions and properties. Knowing these formulas ensures that we approach each integration challenge confidently and accurately.

Indefinite Integrals

Indefinite integrals are integral calculus expressions without specified limits of integration. This makes them more general than definite integrals. They are represented in the form \( \int f(x) \, dx \) and result in a family of functions plus a constant, noted as \( C \). This constant appears because the process of integration involves reversing differentiation, which means multiple antiderivatives differ by a constant.

• For instance, the indefinite integral of \( f(x) = \cosh(3x) \) yields \( \frac{1}{3} \sinh(3x) + C \).

Understanding the concept of indefinite integrals helps solidify one's grasp of calculus fundamentals, as they are used to find the antiderivatives of functions. They are often a preliminary step in solving differential equations or evaluating definite integrals. In our example, calculating the indefinite integral of \( \cosh(3x) \) demonstrates using a specific formula within the broader framework of integrals.
By mastering indefinite integrals, students can systematically approach integration problems and develop solutions that apply broadly across different types of functions.

Integration Techniques

Integration techniques are various methods used to solve integrals. Each technique is designed to handle specific types of functions more effectively. Among the most common techniques are:

• Direct Integration: simplest form, using known integral formulas like \( \int \cosh(ax) \, dx = \frac{1}{a} \sinh(ax) + C \).
• Substitution: useful when the integral contains a composite function, making it easier by substituting variables.
• Integration by Parts: applied when the integrand is a product of two functions, utilizing the identity \( \int u \, dv = uv - \int v \, du \).

The specific problem in our original exercise uses the direct integration technique, appropriate for basic hyperbolic functions. The formula \( \int \cosh(ax) \, dx \) provides a straightforward approach, allowing us to substitute the value of \( a = 3 \) directly.
Learning different integration techniques equips students with a toolbox to tackle various integral problems efficiently. It also builds a stronger understanding of how integrals relate to the antiderivative nature of calculus. When applied correctly, these techniques simplify what might initially seem complex, enhancing problem-solving skills in mathematical analysis.