Problem 33 - Calculus Early Transcendentals

Question

Determine each indefinite integral. \(\int \frac{\sinh x}{1+\cosh x} d x\)

Step-by-Step Solution

Answer

Question: Determine the indefinite integral of the given function: $$\int \frac{\sinh x}{1+\cosh x} dx$$ Answer: The indefinite integral of the given function is $$\int \frac{\sinh x}{1+\cosh x} dx = \ln |1 + \cosh x| + C$$ where \(C\) is the constant of integration.

Step 1: Make an observation

Observe that the derivative of \(\cosh{x}\) is \(\sinh{x}\). This hints that we can use substitution to simplify the integral.

Step 2: Perform substitution

Let \(u = 1 + \cosh x\). Then, the derivative of \(u\) with respect to x is \(du/dx = \sinh x\). So, we can write \(dx = \frac{1}{\sinh x} du\). Now we can rewrite the integral in terms of \(u\): $$\int \frac{\sinh x}{1+\cosh x} d x = \int \frac{1}{u} du$$

Step 3: Integrate with respect to \(u\)

The integral \(\int \frac{1}{u} du\) is a basic integral with the solution being the natural logarithm function: $$\int \frac{1}{u} du = \ln |u| + C$$ where \(C\) is the constant of integration.

Step 4: Substitute back in terms of \(x\)

Now, let's substitute back the original variable \(x\). Recall that \(u = 1 + \cosh x\). So our final solution is: $$\int \frac{\sinh x}{1+\cosh x} d x = \ln |1 + \cosh x| + C$$

Key Concepts

Indefinite Integrals

Indefinite integrals represent a family of functions that are the antiderivatives of a given function. When we evaluate an indefinite integral, our goal is to find a function whose derivative matches the integrand. Indefinite integrals are always expressed with a constant of integration, symbolized by "C," since differentiating any constant results in zero. The notation for an indefinite integral is written as \[ \int f(x) \, dx \]where \( f(x) \) is the function to be integrated, and \( dx \) indicates the variable of integration. The result will then be \( F(x) + C \), where \( F(x) \) is the antiderivative.Understanding indefinite integrals is foundational, as they provide the building block for solving differential equations and evaluating areas under curves when moves to definite integrals.

Substitution Method

The substitution method is a powerful tool in integral calculus for simplifying complex integrals. It involves changing variables to make the integral easier to solve. The process is akin to applying the chain rule in reverse.First, identify a substitution that can simplify the integral, often setting \( u = g(x) \), where \( g'(x) \) appears or simplifies the integrand. Then, rewrite the differential, \( dx \), in terms of \( du \) by finding \( du = g'(x) \, dx \).Consider the original exercise:

• Observe the integral \( \int \frac{\sinh x}{1+\cosh x} d x \).
• Choose \( u = 1 + \cosh x \) and find that \( du = \sinh x \, dx \).
• Rewrite the integral as \( \int \frac{1}{u} \, du \).

This transformation simplifies the integral to a basic logarithmic form, making it straightforward to evaluate.

Hyperbolic Functions

Hyperbolic functions often resemble trigonometric functions, but they arise from hyperbolas rather than circles. They are widely used in calculus and many fields of mathematics and engineering for modeling exponential growth and dealing with differential equations.Common hyperbolic functions include:

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

These functions have derivatives and integrals similar in form to their trigonometric counterparts:

• The derivative of \( \sinh x \) is \( \cosh x \).
• The derivative of \( \cosh x \) is \( \sinh x \).

In the original exercise, recognizing these derivatives helped identify the appropriate substitution, making the integration process much more manageable.