Problem 42 - College Algebra and Calculus: An Applied Approach

Question

Use any basic integration formula or formulas to find the indefinite integral. State which integration formula(s) you used to find the integral. $$ \int \frac{2\left(e^{x}-e^{-x}\right)}{\left(e^{x}+e^{-x}\right)^{2}} d x $$

Step-by-Step Solution

Answer

\(-\tanh(x) + C\)

Step 1: Recognize Hyperbolic Functions

First, we consider the hyperbolic cosine function \(\cosh(x)=\frac{e^{x}+e^{-x}}{2}\) and the hyperbolic sine function \(\sinh(x)=\frac{e^{x}-e^{-x}}{2}\). The integrand can be rewritten in terms of these hyperbolic functions. In fact, it looks like the derivative of \(-\tanh(x)\), since the derivative of \(-\tanh(x)\) is \(-\frac{1}{\cosh^{2}(x)}\) which is equal to \(\frac{1}{\cosh^{2}(x)}\cdot2\sinh(x)\cosh(x)\).

Step 2: Transform the Integral

We can rewrite the integrand in terms of hyperbolic functions, yielding the integral \(\int -2 \cdot \frac{\sinh(x)}{\cosh^2(x)}dx\). Notice this function corresponds to the derivative of the hyperbolic tangent function, \(-\tanh(x)\). Therefore, the integral matches to the concept of the antiderivative, that is, the integral of a function's derivative gives the original function (up to a constant).

Step 3: Integration

Since we derived that the integral to be solved is equal to the derivative of \(-\tanh(x)\), we know it must be equal to \(-\tanh(x)\) plus an integration constant C. Therefore, \(\int \frac{2\left(e^{x}-e^{-x}\right)}{\left(e^{x}+e^{-x}\right)^{2}} dx = -\tanh(x) + C\).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are mathematical functions that are analogs of the ordinary trigonometric, or circular, functions and are very useful in various fields, including calculus and complex analysis. They include six primary functions: hyperbolic sine (\( \sinh \)), hyperbolic cosine (\( \cosh \)), hyperbolic tangent (\( \tanh \)), and their reciprocals.

• The hyperbolic sine is defined as \( \sinh(x) = \frac{e^{x} - e^{-x}}{2} \).
• The hyperbolic cosine is defined as \( \cosh(x) = \frac{e^{x} + e^{-x}}{2} \).
• The hyperbolic tangent, then, is \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}} \).

These functions have properties and derivatives similar to their trigonometric counterparts, providing a powerful tool in solving differential equations and evaluating integrals.In this exercise, understanding hyperbolic functions allows us to simplify our integrand and identify it in terms of other functions, such as recognizing that the given integrand is the derivative of \( -\tanh(x) \). This insight paves the way for solving the integral efficiently.

Integration Formula

Integration, a core concept in calculus, involves finding functions called antiderivatives. In this exercise, applying a suitable integration formula is crucial for simplifying and solving the integral. One common integration formula used with hyperbolic functions is recognizing how closely they resemble derivative forms of other functions. Here, the process involves:

• Rewriting the integrand using hyperbolic sine and cosine:

\[\int \frac{2(e^{x} - e^{-x})}{(e^{x} + e^{-x})^{2}} \, dx \]can be described as:\[\int -2 \cdot \frac{\sinh(x)}{\cosh^2(x)} \, dx \]

• Identifying the integrand as the derivative of the hyperbolic tangent function (\(-\tanh(x)\)).

This recognition allows leveraging the rule that the antiderivative of a derivative function is the function itself, plus a constant (\( + C \)). Thus, understanding and applying the correct integration formula facilitates the computation process.

Antiderivative

The concept of an antiderivative is fundamental in integral calculus. An antiderivative of a function is essentially another function whose derivative gives the original function back. In the context of indefinite integration, this results in a family of functions differing by a constant. When dealing with the integral of the form given in this exercise, recognizing the structure is key:

• The integrand \(-2 \cdot \frac{\sinh(x)}{\cosh^2(x)}\) is the derivative of \(-\tanh(x)\),
• which indicates that \(-\tanh(x)\) is indeed an antiderivative of the integrand.

Thus, we write the solution as the original function plus a constant of integration:\[-\tanh(x) + C\]This represents how understanding and identifying derivatives can greatly aid in discovering antiderivatives, thereby solving indefinite integrals more effectively. By recognizing the relationship between derivatives and their corresponding antiderivatives, integration becomes an elegant mathematical operation.