Question

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

Short answer

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

Step-by-step solution

Recognize the Hyperbolic Identity

The given integral can be expressed in terms of the hyperbolic tangent function. Recall that the hyperbolic tangent is defined by \( \tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \). Thus, we can rewrite the integral as \( \int \tanh(x) \, dx \).

Determine the Integral of \( \tanh(x) \)

The integral of \( \tanh(x) \) is a standard result. It is known to be \( \ln(\cosh(x)) \), as the derivative of \( \ln(\cosh(x)) \) is \( \tanh(x) \).

Write the Integral Solution

From Step 2, apply the standard result to find the integral of \( \tanh(x) \). Therefore, the integral \( \int \tanh(x) \, dx = \ln(\cosh(x)) + C \), where \( C \) is the constant of integration.

Key concepts

Hyperbolic Functions

Hyperbolic functions are analogs of the trigonometric functions but for hyperbolas, rather than circles. They are defined using exponential functions, much like how the traditional trigonometric functions are defined using the unit circle.
Some commonly used hyperbolic functions include:

• The hyperbolic sine: \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
• The hyperbolic cosine: \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)
• The hyperbolic tangent: \( \tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \)

Just like the trigonometric functions, hyperbolic functions have specific properties and identities that make them useful in calculus.
For instance, \( \tanh(x) \) can be particularly useful, as shown in the original exercise, where the given integral is rewritten in terms of \( \tanh(x) \). Recognizing such representations simplifies the integration process considerably.

Substitution Method

The substitution method is a powerful integration technique that transforms a complicated integral into a simpler one. It involves changing variables to make the integral more manageable.
This method often begins by identifying a substitution that corresponds to a part of the integrand. In the exercise, noticing that the integrand resembles the hyperbolic tangent function \( \tanh(x) \) is the key.

• First, identify a potential substitution: Recognize that the expression \( \frac{e^x - e^{-x}}{e^x + e^{-x}} \) matches \( \tanh(x) \).
• Next, rewrite the integral: Utilize this substitution to simplify the integral to \( \int \tanh(x) \, dx \).
• Finally, solve the simplified integral: Knowing the integral of \( \tanh(x) \) is \( \ln(\cosh(x)) \) directly provides the solution.

Recognizing appropriate substitutions, such as those involving hyperbolic functions, is fundamental to tackling integrals that may initially appear complex.

Logarithmic Composition

Logarithmic compositions are often the result of integration where the antiderivative involves logarithms. In integral calculus, understanding how to derive these compositions is crucial.
In the context of our example, after simplifying the integral to \( \int \tanh(x) \, dx \), the solution involves recognizing that the antiderivative is \( \ln(\cosh(x)) \).

• The function \( \cosh(x) \) naturally appears due to its relationship with \( \tanh(x) \) since the derivative of \( \ln(\cosh(x)) \) is \( \tanh(x) \).
• The appearance of the natural logarithm here indicates a composition since the integration transforms \( \tanh(x) \) into \( \ln(\cosh(x)) \).
• Understanding this composition helps in handling integrals that involve exponential expressions, particularly those resembling hyperbolic functions.

The use of logarithmic compositions in integration highlights the interconnectedness of exponential, hyperbolic, and logarithmic functions.