Problem 354 - Calculus Volume 1

Question

In the following exercises, use appropriate substitutions to express the trigonometric integrals in terms of compositions with logarithms. $$\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} d x$$

Step-by-Step Solution

Answer

The integral is \( \ln|\cosh(x)| + C \).

Step 1: Recognize a Hyperbolic Function

The given expression \[ \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \, dx \]can be expressed using hyperbolic functions. Notice that\[ \sinh(x) = \frac{e^{x} - e^{-x}}{2} \quad \text{and} \quad \cosh(x) = \frac{e^{x} + e^{-x}}{2}. \]Thus, \( \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} = \frac{2\sinh(x)}{2\cosh(x)} = \tanh(x). \)

Step 2: Rewrite the Integral

Using the substitution from Step 1, rewrite the integral as:\[ \int \tanh(x) \, dx. \]

Step 3: Integrate Using Hyperbolic Identity

Recall the integral of the hyperbolic tangent function:\[ \int \tanh(x) \, dx = \ln|\cosh(x)| + C, \]where \( C \) is the constant of integration. Therefore:\[ \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \, dx = \ln|\cosh(x)| + C. \]

Step 4: Result

The final result of the integral is:\[ \ln|\cosh(x)| + C. \]

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of the trigonometric functions but for the hyperbola instead of the circle. The two basic hyperbolic functions are hyperbolic sine and hyperbolic cosine, denoted as \( \sinh(x) \) and \( \cosh(x) \) respectively. These functions come up frequently when dealing with expressions involving exponentials of the form \( e^{x} \) and \( e^{-x} \).
- The hyperbolic sine is given by: \[ \sinh(x) = \frac{e^{x} - e^{-x}}{2} \] - The hyperbolic cosine is given by: \[ \cosh(x) = \frac{e^{x} + e^{-x}}{2} \] When you see expressions containing \( e^{x} \) and \( e^{-x} \), think about these hyperbolic functions. They can simplify the problem significantly.
In this context, we noticed that the original integral \( \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \, dx \) can be rewritten using \( \sinh(x) \) and \( \cosh(x) \), showing a direct usage of hyperbolic functions to simplify integrals.

Tangent Function Integration

Integrating hyperbolic functions follows similar rules as integrating trigonometric functions. In our step-by-step solution, once we identified the expression as a hyperbolic tangent, or \( \tanh(x) \), we simplified our task significantly.
The hyperbolic tangent function, \( \tanh(x) \), is defined as:\[ \tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \]The integral of the hyperbolic tangent function is a well-known result:\[ \int \tanh(x) \, dx = \ln|\cosh(x)| + C \] where \( C \) is the constant of integration.
Why does this work? For integrals involving \( \tanh(x) \), remembering this particular result allows you to proceed without having to derive from scratch every time. It can be derived by recalling that the derivative of \( \cosh(x) \) is \( \sinh(x) \), and that \( \tanh(x) = \sinh(x)/\cosh(x) \), which simplifies our integration using basic calculus identity. This is handy when solving these types of integrals.

Logarithmic Expression in Integrals

Integrals involving functions like \( \tanh(x) \) often lead to logarithmic expressions. This is partly because of the fundamental relationships between hyperbolic functions and their derivatives. In our problem, the use of the integral \( \int \tanh(x) \, dx = \ln|\cosh(x)| + C \) highlights this transformation.
Logarithmic expressions usually come from integrals that involve functions whose derivative forms a fraction, like with \( \tanh(x) \). The expression \( \ln|\cosh(x)| \) is the result of integrating \( \tanh(x) \), and represents a standard form you often see when hyperbolic functions are integrated.
This is because of the relationship between logarithmic functions and natural exponentials. Our solution is essentially leveraging this close relationship, transforming an otherwise complete expression into a simpler form involving a logarithm. Understanding these patterns will make handling integrals involving hyperbolic or trigonometric functions more intuitive.