Problem 50 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int \frac{\operatorname{csch}(\ln t) \operatorname{coth}(\ln t) d t}{t}$$

Step-by-Step Solution

Answer

The integral evaluates to \(-\operatorname{csch}(\ln t) + C\).

Step 1: Substitution

We will start by using a substitution method to simplify the integral. Let \( u = \ln t \). Then, \( du = \frac{1}{t} dt \) which means \( dt = t \cdot du \). Substituting this into the integral, we have:\[\int \operatorname{csch}(u) \operatorname{coth}(u) \cdot du\]

Step 2: Simplifying the Integral

Using the substitution from Step 1, we have simplified the integral into terms of \( u \). The integral becomes:\[\int \operatorname{csch}(u) \operatorname{coth}(u) \, du\]Recall that \( \operatorname{csch}(u) = \frac{1}{\sinh(u)} \) and \( \operatorname{coth}(u) = \frac{\cosh(u)}{\sinh(u)} \). Thus, the integrand simplifies to:\[\frac{\cosh(u)}{\sinh^2(u)}\]

Step 3: Direct Integration

We can now integrate directly with respect to \( u \). Notice that the derivative of \( \operatorname{csch}(u) \) is \( -\operatorname{csch}(u) \operatorname{coth}(u) \). Therefore, the integral of \( \operatorname{csch}(u) \operatorname{coth}(u) \, du \) is:\[-\operatorname{csch}(u) + C\]

Step 4: Reverse Substitution

Substitute back \( u = \ln t \) to express the answer in terms of \( t \). Thus, the expression becomes:\[-\operatorname{csch}(\ln t) + C\]

Step 5: Final Answer

The evaluated integral, given the substitution and integration steps, is:\[-\operatorname{csch}(\ln t) + C\]

Key Concepts

Substitution Method

The substitution method is a fundamental technique in calculus used to simplify solving integrals. It involves changing the variable of integration to make an integral easier to evaluate. In our given exercise, we used a substitution of the form:

• Letting \( u = \ln t \) transforms the variables and the functions within the integral.
• This means the differential changes as well, from \( dt \) to \( du \), following from \( du = \frac{1}{t} dt \), being equivalent to say \( dt = t \cdot du \).

This substitution transforms the original complex integral into a simpler one involving \( u \), allowing further simplification of the expression \[ \int \operatorname{csch}(u) \operatorname{coth}(u) \, du \].
By using the substitution method, we effectively translate a potentially unsolvable integral into a much more manageable form.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions, but for the hyperbola rather than the circle. They are vital in many areas of calculus, including evaluating integrals involving these types of functions. In our exercise, we encounter two specific hyperbolic functions:

• The hyperbolic cosecant, represented as \( \operatorname{csch}(u) \), is defined as \( \frac{1}{\sinh(u)} \).
• The hyperbolic cotangent, represented as \( \operatorname{coth}(u) \), is defined as \( \frac{\cosh(u)}{\sinh(u)} \).

These definitions are crucial for transforming and simplifying the integral. The simplification uses the derivative properties of these functions, specifically leveraging identities to integrate. The recognition that the derivative of \( \operatorname{csch}(u) \) is \( -\operatorname{csch}(u) \operatorname{coth}(u) \) helps in performing the integration directly.

Definite Integrals

Although the original exercise was an indefinite integral, understanding the concept of definite integrals can still be beneficial. Definite integrals provide the total accumulation or area under a function's curve between two points. This area is typically bounded by the limits of integration.
When dealing with definite integrals, certain techniques such as substitution are also applicable to simplify and evaluate the integral against its limits. In our problem, though dealing with an indefinite form, you could apply a similar approach sometimes necessary in definite evaluations.
It is important to learn how to translate these mathematical manipulations back to the original variable once integration is complete, as shown in the exercise, where we reverse-substitute \( u = \ln t \) to return to the original variable \( t \) after finding the indefinite integral solution.