Problem 34 - Calculus Early Transcendentals

Question

Determine each indefinite integral. $$\int \operatorname{coth}^{2} x \operatorname{csch}^{2} x d x$$

Step-by-Step Solution

Answer

The indefinite integral of \(\operatorname{coth}^2{x}\operatorname{csch}^2{x}\,dx\) is \(-\frac{1}{3}\operatorname{coth}^3 x - \frac{1}{5}\operatorname{coth}^5 x + C\), where \(C\) is the constant of integration.

Step 1: Simplify using hyperbolic identity

Recall the hyperbolic identity: \(\operatorname{csch}^2 x = 1 + \operatorname{coth}^2 x\). Using this identity, we can rewrite the given expression as $$\int \operatorname{coth}^2 x (\operatorname{csch}^2 x) \, d x = \int \operatorname{coth}^2 x (1 + \operatorname{coth}^2 x) \, d x.$$

Step 2: Distribute and simplify

Distribute \(\operatorname{coth}^2 x\) in the resulting expression: $$\int \operatorname{coth}^2 x (1 + \operatorname{coth}^2 x) \, d x = \int (\operatorname{coth}^2 x + \operatorname{coth}^4 x) \, d x.$$

Step 3: Use substitution

Let's use the substitution \(u = \operatorname{coth} x\). Then, we have \(-\operatorname{csch}^2 x \, d x = d u\) (as the derivative of \(\operatorname{coth} x\) is \(-\operatorname{csch}^2 x\)). Substitute \(u\) back into the simplified expression: $$ \int (\operatorname{coth}^2x + \operatorname{coth}^4x) \, dx = \int (u^2 + u^4) (-d u). $$

Step 4: Integrate and substitute back

Now, we can integrate the expression with respect to \(u\): $$ \int (u^2 + u^4) (-d u) = -\frac{1}{3} u^3 - \frac{1}{5} u^5 + C. $$ Lastly, substitute \(\operatorname{coth} x\) back for \(u\), and we get the indefinite integral: $$ -\frac{1}{3} (\operatorname{coth} x)^3 - \frac{1}{5} (\operatorname{coth} x)^5 + C = -\frac{1}{3} \operatorname{coth}^3 x - \frac{1}{5} \operatorname{coth}^5 x + C. $$ Therefore, the indefinite integral is: $$ \int \coth^2 x \,\operatorname{csch}^2 x \, d x = -\frac{1}{3} \operatorname{coth}^3 x - \frac{1}{5} \operatorname{coth}^5 x + C. $$

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of the well-known trigonometric functions but for the hyperbola, rather than the circle. Some of the commonly used hyperbolic functions are the hyperbolic sine ( \( \sinh(x) \) ) and hyperbolic cosine ( \( \cosh(x) \) ). These functions have properties similar to sine and cosine functions but are defined using exponential functions.

• Hyperbolic Cotangent: The hyperbolic cotangent ( \( \coth(x) \) ) is analogous to the cotangent function. It is defined as the ratio of the hyperbolic cosine and hyperbolic sine: \( \coth(x) = \frac{\cosh(x)}{\sinh(x)} \).
• Hyperbolic Cosecant: Similarly, the hyperbolic cosecant ( \( \csch(x) \) ) is the reciprocal of hyperbolic sine: \( \csch(x) = \frac{1}{\sinh(x)} \).

These functions have unique identities and properties, such as \( \operatorname{csch}^2 x = 1 + \operatorname{coth}^2 x \), which are crucial in calculus for simplifying complex integral expressions.
Understanding these identities is essential, as they allow for the transformation of integrals into more manageable forms.

Substitution Method

The substitution method is a powerful technique in integral calculus used to simplify the process of finding an antiderivative. It is analogous to the chain rule used in differentiation and helps in handling integrals, especially when dealing with complex functions.

Here's a brief overview of how it works:

• Select a substitution: Pick a function contained within the integral to represent as a new variable ( \( u \) ). For example, in the given problem, \( u = \operatorname{coth} x \).
• Differentiate to find \( du \): This involves differentiating the substitution equation to express differential \( dx \) in terms of differential \( du \). In our case, \(-\operatorname{csch}^2 x \, dx = du \).
• Replace and integrate: Substitute \( u \) and \( du \) back into the integral expression, which often simplifies the integral significantly. This allows us to solve the integral in terms of \( u \).
• Substitute back the original variable: After integrating, substitute the original function back in for \( u \) to express the antiderivative in terms of the initial variable.

This method simplifies complex integrals and is especially useful when the integral involves compositions of functions, allowing for easier computation and clear solutions.

Integral Calculus

Integral calculus is one of the two main branches of calculus, with the other being differential calculus. While differential calculus focuses on the concept of derivatives and rates of change, integral calculus is concerned with the concept of integration.

• Definite vs Indefinite Integrals: Indefinite integrals represent a family of functions and include a constant of integration, \( C \). In contrast, definite integrals compute the area under a curve over a specified interval.
• The Process of Integration: Integration reverses differentiation. It involves finding an antiderivative of a given function. In practical terms, this means determining a function whose derivative matches the integrand.
• Applications: Integral calculus has many applications, such as calculating areas under curves, volumes of solids of revolution, and solving differential equations.

In the exercise provided, we focused on finding an indefinite integral involving hyperbolic functions, which required simplification through identities and substitution. This showcases how integral calculus helps in deriving precise solutions for complex functions.