Problem 77 - Thomas Calculus

Question

Evaluate each integral in Exercises \(71-82\) by using any technique you think is appropriate. $$ \int 3 \sinh \left(\frac{x}{2}+\ln 5\right) d x $$

Step-by-Step Solution

Answer

The evaluated integral is \(6 \cosh\left(\frac{x}{2} + \ln 5\right) + C\).

Step 1: Simplifying the Integral Expression

We start by recognizing that the integral involves a hyperbolic sine function. The integral can be expressed as follows using the linear properties of integration: \[\int 3 \sinh \left(\frac{x}{2}+\ln 5\right) dx = 3 \int \sinh \left(\frac{x}{2}+\ln 5\right) dx.\] We can factor out the constant, 3, from the integral.

Step 2: Using a Hyperbolic Identity

Recall the identity for the hyperbolic sine function: \(\sinh(a) = \frac{e^a - e^{-a}}{2}\). Let \(u = \frac{x}{2} + \ln(5)\), then \(\sinh(u) = \frac{e^u - e^{-u}}{2}\). This helps in visualizing the transformation needed.

Step 3: Substitution for Simplicity

Let \(v = \frac{x}{2} + \ln 5\). Then we have \(dv = \frac{1}{2}dx\), which implies \(dx = 2dv\). The integral becomes: \[3 \int \sinh(v) \cdot 2 \, dv,\] which simplifies to \(6 \int \sinh(v) \, dv\).

Step 4: Evaluating the Integral with Basic Identity

The antiderivative of \(\sinh(v)\) is \(\cosh(v)\), so \(\int \sinh(v) \, dv = \cosh(v) + C\). Hence, \(6 \int \sinh(v) \, dv = 6\cosh(v) + C\).

Step 5: Reverting to Original Variables

Substitute back for \(v\) in terms of \(x\): \(v = \frac{x}{2} + \ln 5\), so \(\cosh(v) = \cosh\left(\frac{x}{2} + \ln 5\right)\). Thus, the integral evaluates to: \[6 \cosh\left(\frac{x}{2} + \ln 5\right) + C.\]

Step 6: Final Answer

The evaluated integral is: \[6 \cosh\left(\frac{x}{2} + \ln 5\right) + C.\] This is the most simplified and general form of the answer, incorporating the constant of integration, \(C\).

Key Concepts

Hyperbolic Functions

Hyperbolic functions play a significant role in calculus, particularly with integration techniques. Just as we have trigonometric functions like sine and cosine, there are hyperbolic counterparts called hyperbolic sine (\(\sinh(x)\)) and hyperbolic cosine (\(\cosh(x)\)). These are defined using the exponential function and appear as:

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

The hyperbolic sine function, \(\sinh(x)\), is often encountered in integrals. It can be expressed via exponential functions, making it easier to integrate by recognizing the patterns that match with known function formulas. These functions
are essential in solving integrals involving exponential terms, similar to the one in our exercise.

Substitution Method

The substitution method is a powerful tool for simplifying integrals, particularly when dealing with composite functions. The core idea is to transform the variable of integration into a new one,
making the integral easier to solve. In the given problem, we substituted \(v = \frac{x}{2} + \ln 5\). We then determined \(dv = \frac{1}{2}dx\), leading us to \(dx = 2dv\). This transformation allowed us to rewrite the integral in terms of \(v\), removing the complexity of the compound expression

• New variable substitution clarifies the integral's structure
• Transforms a complex, nested integral into a simple form
• Facilitates easy application of known integral formulas

This simplification is a typical application in integrals involving functions where substitution matches the derivative of another term within the function. It's critical for reducing complexity
and reaching the solution efficiently.

Antiderivatives

Antiderivatives, or indefinite integrals, are the reverse operation of differentiation. They are critical in solving integration problems, as they provide a family of functions whose derivatives consist of
the integrand. Finding an antiderivative involves determining which function, when differentiated, returns the original integrand. For hyperbolic functions like \(\sinh(x)\), the antiderivative is \(\cosh(x)\).

• The basic principle: if \(f'(x) = g(x)\), then the antiderivative \(\int g(x)\, dx = f(x) + C\), where \(C\) is the constant of integration.
• Knowing standard derivatives of functions assists in identifying antiderivatives directly.
• The integration process simplifies when matched with basic function antiderivatives.

In this exercise, recognizing that \(\int \sinh(v) \, dv = \cosh(v) + C\) allowed for a smooth retrieval of the solution, demonstrating how essential antiderivatives are in integration tasks.