Problem 39 - Calculus Early Transcendentals

Question

Evaluate each definite integral. $$\int_{0}^{\ln 2} \tanh x d x$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral of the hyperbolic tangent function from 0 to ln(2): $$\int_{0}^{\ln 2} \tanh x d x.$$ Answer: The definite integral is equal to: $$\int_{0}^{\ln 2} \tanh x d x = \ln \frac{5}{4}.$$

Step 1: Find the antiderivative of \(\tanh x\)

Recall that the hyperbolic tangent function, \(\tanh x\), can be written as the quotient of the hyperbolic sine and cosine functions, i.e., $$\tanh x = \frac{\sinh x}{\cosh x}.$$ Now, let's find the antiderivative of \(\tanh x\). To do this, we can use a known antiderivative formula: $$\int \tanh x d x = \ln(\cosh x) + C.$$

Step 2: Apply the Fundamental Theorem of Calculus

Now that we have the antiderivative of \(\tanh x\), we can use the Fundamental Theorem of Calculus to evaluate the definite integral. Recall that the theorem states that if an antiderivative of a function \(f(x)\) is \(F(x)\) and we want to evaluate \(\int_{a}^{b} f(x) d x\), then it is equal to \(F(b) - F(a)\). Applying the Fundamental Theorem of Calculus for our problem, we get: $$\int_{0}^{\ln 2} \tanh x d x = \ln (\cosh(\ln 2)) - \ln (\cosh(0)).$$

Step 3: Evaluate the definite integral

We evaluate the definite integral by finding the values of \(\cosh(\ln 2)\) and \(\cosh(0)\), and then taking their natural logarithms: $$\cosh(\ln 2) = \frac{e^{\ln 2}+e^{-\ln 2}}{2} = \frac{2 + \frac{1}{2}}{2} = \frac{5}{4}.$$ Since \(\cosh(0) = 1\), we have \(\ln(\cosh(0)) = \ln(1) = 0\). Therefore, the definite integral is: $$\int_{0}^{\ln 2} \tanh x d x = \ln (\frac{5}{4}) - 0 = \ln \frac{5}{4}.$$

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of the trigonometric functions but for hyperbolas, rather than circles. They include functions like hyperbolic sine (\( \sinh x \)), hyperbolic cosine (\( \cosh x \)), and hyperbolic tangent (\( \tanh x \)).
These functions have applications in many areas, such as engineering, physics, and mathematics. They provide solutions to various problems, particularly those involving hyperbolic shapes.
The hyperbolic tangent function, \( \tanh x \), is defined as:

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

It ranges from -1 to 1 and, similar to the regular tangent function, it has a slew of interesting properties. For example, it provides the slope of the line on the hyperbolic tangent curve.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus establishes a crucial connection between differentiation and integration, two of the main operations in calculus.
It is divided into two parts:

• The first part allows you to compute the derivative of an antiderivative, effectively tying integration with differentiation.
• The second part provides a way to evaluate definite integrals via antiderivatives.

Using this theorem, to find the definite integral of a function over an interval, we find an antiderivative, calculate its value at the boundary points, and take the difference.
In our example involving the integral of \( \tanh x \), we found the antiderivative, \( \ln(\cosh x) \), and used it to find the definite integral between the limits 0 and \( \ln 2 \). This is the power of the Fundamental Theorem of Calculus: it transforms definite integral problems into simpler arithmetic.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are the reverse process of differentiation. Finding a function's antiderivative means finding a function whose derivative is the original function. For instance, if the derivative of \( F(x) \) is \( f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \).
The process of finding antiderivatives is fundamental to solving integrals, especially when we deal with definite integrals using the Fundamental Theorem of Calculus. This theorem helps to calculate the total accumulation of a quantity, such as area under the curve.
When calculating the definite integral of \( \tanh x \), we initially needed its antiderivative, which was determined to be \( \ln(\cosh x) \).
To handle integrations of hyperbolic functions efficiently, one often memorizes common antiderivative forms, making it easier to tackle complex calculus problems.