Question

Evaluate each integral. $$\int \frac{\cosh z}{\sinh ^{2} z} d z$$

Short answer

Question: Evaluate the following integral: \(\int \frac{\cosh z}{\sinh ^{2} z}\, dz\) Answer: \(-\frac{1}{2}(\coth^2 z) + C\)

Step-by-step solution

Recognize the derivative of a hyperbolic function

Notice that the integral resembles the derivative of \(\coth z\). We know that \(\frac{d(\coth z)}{dz} = -\frac{1}{\sinh^2 z}\).

Make a substitution

Let \(u = \coth z\), so \(du = -\frac{1}{\sinh^2 z}\, dz\). Therefore, the integral becomes: $$\int \cosh z \Big(-\frac{1}{\sinh^2 z}\Big)\, dz = -\int u\, du$$

Evaluate the integral

Now, we can evaluate the integral easily and get: $$-\int u\, du = -\frac{1}{2}u^2 + C = -\frac{1}{2}(\coth^2 z) + C$$

Write the final answer

The final result of the integral is: $$\int \frac{\cosh z}{\sinh ^{2} z}\, dz = -\frac{1}{2}(\coth^2 z) + C$$

Key concepts

Hyperbolic Functions

Hyperbolic functions, including the hyperbolic cosine (\(\text{cosh}\)) and the hyperbolic sine (\(\text{sinh}\)), are analogues of the trigonometric functions but for the hyperbola, much like trigonometric functions are for the circle.

These functions have properties similar to traditional trigonometric functions but are defined using the exponential function. For example, the hyperbolic cosine is defined as \(\text{cosh}(x) = \frac{e^x + e^{-x}}{2}\) and the hyperbolic sine as \(\text{sinh}(x) = \frac{e^x - e^{-x}}{2}\).

Understanding the derivatives and integrals of hyperbolic functions is crucial for evaluating hyperbolic integrals. For instance, the derivative of \(\text{cosh}(x)\) is \(\text{sinh}(x)\), and the derivative of \(\text{sinh}(x)\) is \(\text{cosh}(x)\). These functions frequently appear in calculus, particularly within integrals involving hyperbolic expressions, much like the textbook exercise at hand.

U-Substitution

U-substitution is a method often used to simplify integrals by making a substitution that converts the integral into a simpler form, usually involving a basic function whose integral is known.

In the method of u-substitution, we choose a part of the integrand (the function being integrated) to be our 'u'. We then derive what 'du' (the differential of 'u') would be. Substituting 'u' and 'du' back into the integral, we hope to get an integral in terms of 'u' that is easier to evaluate than the original.

This technique is very powerful and is akin to the reverse process of the chain rule from differentiation. A successful u-substitution often necessitates recognizing parts of the integrand that correspond to derivatives of functions familiar to us from basic differentiation rules.

Indefinite Integrals

Indefinite integrals, also known as antiderivatives, represent the general form of all the antiderivatives of a function. In practice, calculating an indefinite integral means finding a function that, when differentiated, gives the original function back.

Indefinite integrals are expressed with the integral sign followed by a function, without specific limits of integration. They always include a constant of integration, denoted as 'C', because the derivative of a constant is zero, and thus the antiderivative is not unique.

The indefinite integral of a hyperbolic function, for instance, yields another function (either hyperbolic or algebraic) plus C. It is important to become familiar with standard integrals and techniques such as u-substitution to tackle a broad range of indefinite integrals in calculus.