Problem 50 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(41-50\) . $$ \int \frac{\operatorname{csch}(\ln t) \operatorname{coth}(\ln t) d t}{t} $$

Step-by-Step Solution

Answer

The integral evaluates to \(-\frac{1}{\sinh(\ln t)} + C\).

Step 1: Understand the Integrals

In this exercise, we need to evaluate the integral of the function given by \( \int \frac{\operatorname{csch}(\ln t) \operatorname{coth}(\ln t) \, dt}{t} \). This involves hyperbolic functions expressed in terms of the natural logarithm of \( t \).

Step 2: Apply the Change of Variables

To simplify, let's use a substitution. Let \( u = \ln t \) so that \( du = \frac{1}{t} \, dt \). This substitution transforms the integral into \( \int \operatorname{csch}(u) \operatorname{coth}(u) \ du \).

Step 3: Simplify the Hyperbolic Functions Integral

We know that \( \operatorname{csch}(u) = \frac{1}{\sinh(u)} \) and \( \operatorname{coth}(u) = \frac{\cosh(u)}{\sinh(u)} \). Therefore, \( \operatorname{csch}(u) \operatorname{coth}(u) = \frac{\cosh(u)}{\sinh^2(u)} \).

Step 4: Integrate Using Known Derivatives

Recall that the derivative of \( -\frac{1}{\sinh(u)} \) is \( \operatorname{csch}(u) \operatorname{coth}(u) \). So, the integral \( \int \operatorname{csch}(u) \operatorname{coth}(u) \, du \) simplifies to \( -\frac{1}{\sinh(u)} + C \), where \( C \) is an integration constant.

Step 5: Reverse the Substitution

Substitute \( u = \ln t \) back into the result. Since \( \sinh(u) = \sinh(\ln t) \), the integral becomes \( -\frac{1}{\sinh(\ln t)} + C \).

Step 6: Final Simplification (if applicable)

For a clean final answer, express the result using hyperbolic functions defined with \( t \).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are mathematical functions that are analogs of trigonometric functions. While trigonometric functions are based on a circle, hyperbolic functions are based on a hyperbola. The hyperbolic sine function, \( \sinh(x) \), is defined as \( \sinh(x) = \frac{e^x - e^{-x}}{2} \). Likewise, the hyperbolic cosine function, \( \cosh(x) \), is defined as \( \cosh(x) = \frac{e^x + e^{-x}}{2} \). Derived from these two, hyperbolic functions like the hyperbolic cotangent, \( \operatorname{coth}(x) \), and the hyperbolic cosecant, \( \operatorname{csch}(x) \), are defined as follows:

• \( \operatorname{coth}(x) = \frac{\cosh(x)}{\sinh(x)} \)
• \( \operatorname{csch}(x) = \frac{1}{\sinh(x)} \)

In various integration problems, these functions help describe the behavior of curves and areas where standard trigonometric functions might not apply.Utilizing hyperbolic identities, we can simplify and evaluate integrals expressed in terms of hyperbolic functions effectively.

Change of Variables

The change of variables technique is a helpful method in calculus to simplify integrals. It involves substituting variables in an integral with more convenient ones. This technique is particularly useful when dealing with integrals involving complex expressions or functions. In our exercise, we encounter the integral \( \int \frac{\operatorname{csch}(\ln t) \operatorname{coth}(\ln t) \, dt}{t} \). By setting \( u = \ln t \), we change the variable from \( t \) to \( u \). This substitution transforms the differential \( dt \) to \( du = \frac{1}{t} \, dt \). Thus, the original integral becomes \( \int \operatorname{csch}(u) \operatorname{coth}(u) \, du \).Using this change, integrals that initially seem complex can be rewritten in simpler terms, making them easier to solve.

Substitution Method

The substitution method is a fundamental integration technique used to evaluate integrals by replacing part of the integrand with a new variable. This simplification can make it easier to find a solution.In our exercise, we used the substitution \( u = \ln t \). This resulted in the differential \( du = \frac{1}{t} \, dt \), which perfectly matched the terms in the original integrand. Transforming the given integral \( \int \frac{\operatorname{csch}(\ln t) \operatorname{coth}(\ln t) \, dt}{t} \) into \( \int \operatorname{csch}(u) \operatorname{coth}(u) \, du \) allows us to solve it using known identities. This technique can convert complicated integrands into forms that are simpler to integrate, often involving standard integrals with well-known solutions.

Integration Techniques

Integration techniques are methods utilized to solve integrals, which are crucial for determining areas under curves, among other applications. There are several techniques, ranging from basic antiderivatives to advanced integral transformations.One common technique, as applied in our example, is using known derivatives of functions. For instance, to integrate \( \operatorname{csch}(u) \operatorname{coth}(u) \), we utilize the fact that the derivative of \(-\frac{1}{\sinh(u)}\) is indeed \( \operatorname{csch}(u) \operatorname{coth}(u) \). This straightforward approach is often effective when dealing with hyperbolic functions.Moreover, reversing substitutions often marks the final step. Substituting back \( u = \ln t \), our integral transforms into the solution \(-\frac{1}{\sinh(\ln t)} + C\). By applying these techniques, complex integrals can often be reduced to straightforward expressions, making them manageable even for challenging problems.