Problem 112 - Calculus and its applications

Question

Evaluate. $$ \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} d x $$

Step-by-Step Solution

Answer

\( \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \, dx = \ln(e^{2x} + 1) + C \)

Step 1: Simplify the Integrand

The given integral is \( \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \, dx \). Observe that both the numerator and denominator can express hyperbolic functions.Recognizing hyperbolic identities, simplify: let \( u = e^x \), and thus \( e^{-x} = \frac{1}{u} \). The integrand becomes:\[\frac{u - \frac{1}{u}}{u + \frac{1}{u}} = \frac{u^2 - 1}{u^2 + 1}. \]

Step 2: Set up the Substitution

Let's substitute \( u = e^x \), which means \( du = e^x dx \) or \( dx = \frac{du}{u} \).Our integral becomes:\[\int \frac{u^2 - 1}{u^2 + 1} \cdot \frac{du}{u}. \]

Step 3: Simplify the New Integrand

Rewrite the expression:\[\int \frac{u(u^2 - 1)}{u(u^2 + 1)} \, du = \int \frac{u^2 - 1}{u^2 + 1} \, du.\]

Step 4: Split into Partial Fractions

We now decompose \( \frac{u^2 - 1}{u^2 + 1} \), although no need for a classic partial fraction since it's a simple formula of derivative format. Observe that the derivative of the denominator is twice the numerator:\[\text{The derivative of } (u^2 + 1) \text{ is } 2u,\]leading us directly to an integral form since it resembles \( \ln(u^2+1) \).

Step 5: Integrate the Function Directly

We can recognize this as a directly integrable form of natural logarithms. The integral of a derivative over its function is:\[\int \frac{u^2 - 1}{u^2 + 1} \, du = \ln(u^2 + 1) + C.\]

Step 6: Reverse the Substitution

Now, substitute back \( u = e^x \) into the integral result:\[\ln(e^{2x} + 1) + C.\]

Step 7: Final Answer

Thus, the evaluated integral is:\[\ln(e^{2x} + 1) + C.\]

Key Concepts

Definite Integrals

Definite integrals are an important part of calculus and involve finding the area under a curve between two points on the x-axis. They are different from indefinite integrals, which do not have specific limits. When solving definite integrals, we compute the value of the integral from a lower bound to an upper bound. This process essentially measures the total accumulation of a quantity, over a certain interval of the function.

In the exercise provided, even though it's set out as an indefinite integral for simplification purposes, definite integrals would be calculated by substituting in the limits at the end. This emphasizes the need to handle calculations carefully, especially with substitutions involved.

Key features of definite integrals include:

• Absolute values of integrals can determine positive or negative areas.
• The integration process becomes very clear when limits are given, as these inform the final result.
• The Fundamental Theorem of Calculus connects differentiation and integration by showing that integration can be reversed by differentiation.

When definite integrals are calculated, they provide practical results applicable to a variety of fields, such as physics, engineering, and economics.

Substitution Method

The substitution method is a common technique used in integration to simplify complex integrals. It involves changing variables to make the integral easier to solve. This method is much like reversing the chain rule used in differentiation.

In the step-by-step solution provided, the substitution method was used effectively. By letting \( u = e^x \), the function becomes more straightforward to integrate, turning the complex expression into a manageable form.

Here's how substitution works:

• Choose a substitution \( u = g(x) \) that simplifies the integrand.
• Calculate \( du \), which is derived from \( g'(x)dx \).
• Replace all occurrences of the original variable \( x \) in the integral with the new variable \( u \).
• Integrate with respect to \( u \).
• Finally, revert \( u \) back to \( x \) using the initial substitution to complete the problem.

This method is an invaluable tool in calculus as it transforms integrals that appear unsolvable at first glance into simpler algebraic forms. It’s especially powerful in situations involving hyperbolic or trigonometric functions.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but relate to hyperbolas rather than circles. They are very useful in various calculus problems due to their unique properties. The basic hyperbolic functions are:

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

In our original exercise, if we recognize that the integrand relates closely to hyperbolic identities, conversion becomes possible. Knowing these hyperbolic forms can significantly simplify complex expressions by reducing them to rational algebraic terms, which are easier to integrate.

Understanding hyperbolic functions helps in:

• Presenting solutions to differential equations elegantly.
• Modeling hyperbolic geometries and real-world structures.
• Resolving integration problems in calculus by simplifying convoluted expressions.

Because of these utilities, hyperbolic functions play a crucial role in both mathematics and applied sciences, offering a parallel to trigonometric analogs but expanding the scope and application of mathematical analysis.