Question

Evaluate the following integrals. \(\int \frac{\cosh z}{\sinh ^{2} z} d z\)

Short answer

Question: Evaluate the integral of the function \(\frac{\cosh z}{\sinh^2 z}\) with respect to \(z\). Answer: \(-\frac{1}{\sinh z} + C\)

Step-by-step solution

Recognize the derivative of \(\sinh z\) in the integrand

Notice that the derivative of \(\sinh z\) is \(\cosh z\). This observation hints that using substitution to change the variable of integration might be useful because the derivative will cancel with a term in the integrand.

Perform the substitution

Let's perform the substitution by defining a new variable \(u = \sinh z\). Then, the derivative of \(u\) with respect to \(z\) is given by: \(du = (\cosh z) dz\). Now we can rewrite the integral using the substitution: \(\int \frac{\cosh z}{\sinh^{2} z} dz = \int \frac{1}{u^2} du\).

Solve the integral

The integral now has the simple form of \(u^{-2}\) and is easy to solve. The antiderivative of this function is given by the power rule: \(\int \frac{1}{u^2} du = -\frac{1}{u} + C\), where \(C\) is the constant of integration.

Undo the substitution

Since we performed a substitution to find the antiderivative, we need to rewrite the result in terms of the original variable \(z\). Recall, we let \(u = \sinh z\). Therefore, the final result is: \(-\frac{1}{u} + C = -\frac{1}{\sinh z} + C\).

Key concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but they deal with hyperbolas instead of circles. They appear often in calculus, especially in problems involving integration and differentiation due to their intriguing properties. Two primary hyperbolic functions are the hyperbolic sine and cosine, \( \sinh z \) and \( \cosh z \), respectively.

These functions are defined as follows:

• The hyperbolic sine, \( \sinh z = \frac{e^z - e^{-z}}{2} \)
• The hyperbolic cosine, \( \cosh z = \frac{e^z + e^{-z}}{2} \)

Just as with standard trigonometric functions, hyperbolic functions have derivatives that connect to each other. The derivative of \( \sinh z \) is \( \cosh z \), and the derivative of \( \cosh z \) is \( \sinh z \).

These derivatives are extremely useful, particularly when solving integrals involving these functions. Since they behave like their trigonometric counterparts but are based on exponential functions, they make substitution and integration steps smoother.

Substitution Method

The substitution method is a powerful tool in calculus, designed to simplify integrals, making them easier to solve. It works by changing variables in an integral, transforming it into a simpler form.

Here's how it works:

• Identify a part of the integrand whose derivative also appears in the integral. In our problem, \( \sinh z \) and its derivative, \( \cosh z \), fit this description.
• Make a substitution. Define a new variable, say \( u \), such that \( u = \sinh z \). Consequently, \( du = \cosh z\,dz \).
• Rewrite the integral using this substitution. This often simplifies the integral, as in converting \( \int \frac{\cosh z}{\sinh^2 z} \, dz \) to \( \int \frac{1}{u^2} \, du \).
• Solve the simpler integral. For \( \int \frac{1}{u^2} \, du \), use the power rule to find the antiderivative.
• After integration, revert back to the original variable, replacing \( u \) with \( \sinh z \) if necessary.

By changing the variable of integration, substitutions make complex integrals manageable, turning challenging calculus problems into straightforward exercises.

Calculus

Calculus is a branch of mathematics focused on studying change, described through derivatives and integrals. It's fundamental in solving problems involving motion, growth, or area under curves.

There are two primary operations in calculus:

• Derivatives: Measure how functions change, representing the rate of change. For hyperbolic functions, you'll find that derivatives play a pivotal role when simplifying integrals.
• Integrals: Effectively reverse derivatives. They accumulate quantities such as areas, lengths, or volumes, which are essential when calculating antiderivatives.

Integration, such as in our exercise, often involves reversing processes established by differentiation. Calculus provides the techniques necessary to do this, with strategies like substitution smoothing the path.

By combining the rules of differentiation and integration, calculus enables us to manage equations involving complex functions—making it indispensable in many scientific fields.