Question

Integral formula Carry out the following steps to derive the formula \(\int \operatorname{csch} x \, d x=\ln \left|\tanh \frac{x}{2}\right|+C\) (Theorem 7.6) a. Change variables with the substitution \(u=\frac{x}{2}\) to show that $$\int \operatorname{csch} x \, d x=\int \frac{2 d u}{\sinh 2 u}$$ b. Use the identity for sinh \(2 u\) to show that \(\frac{2}{\sinh 2 u}=\frac{\operatorname{sech}^{2} u}{\tanh u}\) c. Change variables again to determine \(\int \frac{\operatorname{sech}^{2} u}{\tanh u} d u,\) and then express your answer in terms of \(x\).

Short answer

Answer: The integral formula for \(\int\operatorname{csch} x \, d x\) is \(\ln \left|\tanh \frac{x}{2}\right|+C\).

Step-by-step solution

Change variables with the substitution \(u=\frac{x}{2}\)

Start by applying the substitution \(u=\frac{x}{2}\), to show that \(\int\operatorname{csch} x \, d x=\int \frac{2 d u}{\sinh 2 u}\). We have \(x=2u\), \(\frac{dx}{du}=2\), so \(dx = 2 \, du\). The original integral, \(\int\operatorname{csch} x \, d x\), will be transformed into: $$\int\frac{1}{\sinh x} dx = \int\frac{1}{\sinh (2u)} (2 du) = \int \frac{2 du}{\sinh 2 u}$$ Now, we have obtained the new integral \(\int \frac{2 du}{\sinh 2 u}\).

Use the identity for \(\sinh 2 u\) to obtain a new expression

Use the identity for sinh \(2 u\), which is \(\sinh 2u = 2 \sinh u \cosh u\), to rewrite the integral as an expression with sech and tanh. We know that \(\frac{1}{\cosh^2 u} = \operatorname{sech}^2 u\) and \(\frac{\sinh u}{\cosh u} = \tanh u\). Therefore, the integral can be rewritten as: $$\int \frac{2 du}{\sinh 2 u} = \int \frac{2 du}{2 \sinh u \cosh u} = \int \frac{\operatorname{sech}^2 u}{\tanh u} du$$

Change variables again to determine the integral

Make another substitution to solve the integral: Let \(v = \tanh u\), so \(\frac{dv}{du} = \operatorname{sech}^2 u\). Thus, \(dv = \operatorname{sech}^2 u \, du\). The integral becomes: $$\int \frac{\operatorname{sech}^2 u}{\tanh u} du = \int \frac{1}{v} dv$$ Now, integrating with respect to \(v\), we obtain: $$\int \frac{1}{v} dv = \ln |v| + C = \ln |\tanh u| + C$$ Recall that \(u = \frac{x}{2}\), so we substitute back to get the result in terms of \(x\): $$\ln |\tanh u| + C = \ln \left|\tanh \frac{x}{2}\right|+C$$ Therefore, the integral formula has been derived: $$\int\operatorname{csch} x \, d x = \ln \left|\tanh \frac{x}{2}\right|+C$$

Key concepts

Hyperbolic Functions

When diving into the world of calculus, one cannot overlook hyperbolic functions, which are analogues of the ordinary trigonometric, or circular, functions. The two basic hyperbolic functions, sinh (hyperbolic sine) and cosh (hyperbolic cosine), are defined through the exponential function:

\[\[\begin{align*}\sinh x &= \frac{e^x - e^{-x}}{2} \cosh x &= \frac{e^x + e^{-x}}{2}\end{align*}\]\]
From these definitions, other hyperbolic functions like tanh (hyperbolic tangent), csch (hyperbolic cosecant), sech (hyperbolic secant), and coth (hyperbolic cotangent) can be derived. These functions have properties and identities similar to trigonometric functions, which make them useful in solving calculus problems, especially integrals and differential equations. For example, the identity \(\sinh 2u = 2 \sinh u \cosh u\) is crucial when integrating the hyperbolic cosecant function, \(\operatorname{csch} x\), as shown in the exercise.

Integration by Substitution

Integration by substitution, often called u-substitution, is a technique used to simplify integrals. The core idea is to find a substitution that makes the integral easier to evaluate. By replacing a part of the integral with a simpler expression, a new integral in terms of variable \(u\) can often be more straightforward to solve than the original one. Similar to the chain rule in differentiation, integration by substitution works by changing variables to provide a new perspective on the problem. In the given problem, substituting \(u = \frac{x}{2}\) transforms the original hyperbolic function into a new integral that then allows for further manipulation and eventual evaluation.

Trigonometric Identities

Trigonometric identities are key tools in calculus, especially for integration and solving equations. These identities involve trigonometric functions and relate to one another in various fashions. For hyperbolic functions, similar rules apply, and they aid in simplifying complicated expressions. As part of the integral of \(\operatorname{csch} x\), we employ the identity \(\sinh 2u = 2 \sinh u \cosh u\) to manipulate the integrand into a form involving a ratio of hyperbolic secant squared and hyperbolic tangent. This conceptual understanding of equivalent expressions is foundational for many problems encountered in calculus.

Indefinite Integrals

Indefinite integrals, or antiderivatives, are expressions representing the general form of the original function before differentiation, plus an arbitrary constant, \(C\). An indefinite integral represents all possible antiderivatives of a given function. Calculating these requires familiarity with a variety of techniques, such as basic integration rules, substitution, integration by parts, and trigonometric identities, to name a few. The indefinite integral of the hyperbolic cosecant function, \(\int \operatorname{csch} x \, d x\), results in a logarithmic form which showcases how such functions, despite seeming complex, can often be integrated to reveal elegant exponential and logarithmic relationships.