Question

Integral proof Prove the formula \(\int\) coth \(x \, d x=\ln |\sinh x|+C\) of Theorem 7.6.

Short answer

Question: Prove that the integral of the function coth(x) is equal to ln|sinh(x)| + C. Answer: To prove this, we first rewrite coth(x) as a ratio of hyperbolic functions: coth(x) = cosh(x) / sinh(x). Then, we apply the substitution method, where u = sinh(x) and du/dx = cosh(x). After substituting and solving the integral, we find that the integral of coth(x) equals ln|sinh(x)| + C.

Step-by-step solution

Rewrite the function coth(x) as a ratio of hyperbolic functions

Recall that \(\text{coth}(x) = \frac{\text{cosh}(x)}{\text{sinh}(x)}\). Our integral becomes: $$ \int \text{coth}(x) \, dx = \int \frac{\text{cosh}(x)}{\text{sinh}(x)} dx$$

Apply the substitution method

Let \(u=\text{sinh}(x)\), then \(\frac{du}{dx}=\text{cosh}(x)\). Rearrange for \(dx\): $$dx = \frac{du}{\text{cosh}(x)}$$ Now substitute in our integral: $$\int \frac{\text{cosh}(x)}{\text{sinh}(x)} dx = \int \frac{1}{u} du$$

Solve the integral

The integral of \(\frac{1}{u}\) is straightforward: $$\int \frac{1}{u} du = \ln|u| + C$$

Substitute back the original function

Recall that \(u = \text{sinh}(x)\), so we have: $$\ln |u| + C = \ln |\text{sinh}(x)| + C$$ Hence, we have proved the formula: $$\int \text{coth}(x) \, dx = \ln |\text{sinh}(x)| + C$$

Key concepts

Hyperbolic Functions

Hyperbolic functions are mathematical functions that are analogs to trigonometric functions but for hyperbolas rather than circles. These functions are important in various fields such as calculus, physics, and engineering.
Here are the main hyperbolic functions:

• Hyperbolic sine, \(\sinh(x)\), is defined by \(\sinh(x) = \frac{e^x - e^{-x}}{2}\).
• Hyperbolic cosine, \(\cosh(x)\), is \(\cosh(x) = \frac{e^x + e^{-x}}{2}\).
• The hyperbolic tangent, \(\tanh(x)\), combines them as \(\tanh(x) = \frac{\sinh(x)}{\cosh(x)}\).
• Similarly, the hyperbolic cotangent, \(\coth(x)\), which is used in the original exercise, is given by \(\coth(x) = \frac{\cosh(x)}{\sinh(x)}\).

Just like their trigonometric cousins, hyperbolic functions have unique identities and properties that make them useful in integration and solving differential equations.
Understanding these functions and their relations gives students a strong foundation for tackling complex integrals and proofs, like the one in the exercise.

Substitution Method

The substitution method is a powerful technique in calculus used to simplify an integral by changing variables. It's akin to reverse chain rule for derivatives. The method replaces a complex expression with a single variable, making the integral easier to evaluate.
### The Substitution Process

• Identify the part of the integral to substitute. In our problem, this was \(u = \sinh(x)\).
• Differentiate this substitution with respect to the original variable, \(x\), resulting in \(\frac{du}{dx} = \cosh(x)\).
• Express \(dx\) in terms of \(du\), yielding \(dx = \frac{du}{\cosh(x)}\).
• Rewrite the integral using the new variables. Here, \(\int\coth(x) \, dx\) transformed into \(\int \frac{1}{u} \, du\).

The substitution method simplifies the problem, especially when direct integration is challenging. In this case, it converted a hyperbolic function based integral into a natural logarithm, which is straightforward.

Definite and Indefinite Integrals

Integrals in calculus can either be definite or indefinite, each serving distinct purposes.
### Indefinite Integrals

• An indefinite integral, like in our exercise, is represented as \(\int f(x) \, dx = F(x) + C\), where \(C\) is an arbitrary constant.
• It represents a family of functions and is equivalent to finding the antiderivative or original function before differentiation.

### Definite Integrals

• Definite integrals compute a numerical value and are represented as \(\int_a^b f(x) \, dx\).
• This process involves finding the area under the curve between two points \(a\) and \(b\) on a graph.

The solution of \(\int \coth(x) \, dx = \ln |\sinh(x)| + C\) demonstrates an indefinite integral. Understanding the distinction between definite and indefinite integrals helps students appreciate the comprehensive use of integration in mathematical analysis.