Problem 62 - Single Variable Calculus

Question

find the given integral. \(\int \frac{\sinh x}{1+\cosh x} d x\)

Step-by-Step Solution

Answer

The short answer is: \(\int \frac{\sinh x}{1+\cosh x} d x = \ln |1 + \cosh x| + C\).

Step 1: Substitute

Let's substitute \(u = 1 + \cosh x\). Now, we find the derivative of \(u\) with respect to \(x\): \( \frac{d u}{d x} = \frac{d (1 + \cosh x)}{d x} = \sinh x\).

Step 2: Rewrite the integral

Now let's rewrite the integral in terms of \(u\). Since \(\frac{d u}{d x} = \sinh x\), we can write \(d u = \sinh x d x\). Thus, the given integral becomes: \(\int \frac{\sinh x}{1+\cosh x} d x = \int \frac{1}{u} d u\).

Step 3: Integrate with respect to u

The integral \(\int \frac{1}{u} d u\) is a standard integral, and it is equal to \(\ln |u| + C\), where \(C\) is the integration constant.

Step 4: Substitute back

Now we need to substitute back \(u = 1 + \cosh x\) into our result: \(\ln |u| + C = \ln |1 + \cosh x| + C\).

Step 5: Final answer

Therefore, the integral of the given function is: \(\int \frac{\sinh x}{1+\cosh x} d x = \ln |1 + \cosh x| + C\).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for hyperbolas instead of circles. They are defined using exponential functions. The primary hyperbolic functions include:

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

Hyperbolic sine, \( \sinh x \), and hyperbolic cosine, \( \cosh x \), have properties similar to those of sine and cosine. For instance, the derivative of \( \sinh x \) is \( \cosh x \), and vice versa. These functions are useful in solving problems involving integrals, particularly when working with exponential growth and decay or certain geometrical problems, such as hyperbolic trigonometry.

Substitution Method

The substitution method is a powerful integration technique. It's especially useful when the integral seems more complicated than it needs to be. In simpler terms, substitution helps simplify the integral to something more manageable.
In the given problem, we used the substitution:

• Set \( u = 1 + \cosh x \) so that the integral becomes easier to solve.
• Find the derivative: \( \frac{du}{dx} = \sinh x \)
• This turns the integral into one with respect to \( u \), simplifying the process.

The goal is to change the variable from \( x \) to \( u \) to make the integration straightforward. Once integrated, we substitute back the original variable to find the solution for the initial problem.

Integration Constants

In any indefinite integral, you'll often encounter a constant of integration denoted by \( C \). This constant is crucial as it represents an entire family of solutions.
The reason for this is that when you differentiate a function, any constant disappears. Thus, integration only allows reconstruction of one particular antiderivative without specifying \( C \).
In the example, after integrating the expression \( \int \frac{1}{u} \, du \), we added the constant, resulting in \( \ln |u| + C \). This \( C \) ensures all possible vertical shifts of the antiderivative function are considered.

Definite and Indefinite Integrals

Integrals can be either definite or indefinite. Understanding the difference between them is vital:

• Indefinite Integrals: Represent a function plus a constant. They are written without bounds, indicating an entire family of functions. For example, \( \int f(x) \, dx = F(x) + C \).
• Definite Integrals: Represent the area under a curve between two points. They have limits of integration, like \( \int_{a}^{b} f(x) \, dx \), producing a numeric result.

In the original exercise, we calculated an indefinite integral. By integrating \( \frac{1}{u} \), we found \( \ln |1 + \cosh x| + C \), which provides an antiderivative without specific boundary limits, hence needing a constant \( C \).