Problem 53 - Calculus of a Single Variable

Question

In Exercises 43–54, find the indefinite integral. $$ \int \frac{\operatorname{csch}(1 / x) \operatorname{coth}(1 / x)}{x^{2}} d x $$

Step-by-Step Solution

Answer

The indefinite integral of \( \frac{\operatorname{csch}(1 / x) \operatorname{coth}(1 / x)}{x^{2}} \) is \(-\log(\cosh(1/x)) + C \)

Step 1: Identify the function

First, we identify the overall function we need to integrate: \( \frac{\operatorname{csch}(1 / x) \operatorname{coth}(1 / x)}{x^{2}} \). Notice that this can be rewritten as \( \frac{1}{x^{2}} \cdot \operatorname{csch}(1 / x) \cdot \operatorname{coth}(1 / x) \).

Step 2: Recognizing the derivative of a function

This step is key to solve the integral. We recall that the derivative of \(-\log(\cosh(x))\) yields \(-\operatorname{csch}(x) \cdot \operatorname{coth}(x)\). Therefore, our integral bears a resemblance to the derivative of \(-\log(\cosh(x))\). However, our integrand involves \(1/x\) where \(x\) is in the argument of the csch and coth functions, so let's substitute \(1/x\) by \(u\), that is \(u=1/x\). The derivative of this substitution is \(-1/x^{2} dx = du\). Which is exactly the remaining part we wanted to get rid of in our integrand.

Step 3: Apply the substitution

Now, we rewrite the original integral using our substitution. So, original integral \( \int \frac{\operatorname{csch}(1 / x) \operatorname{coth}(1 / x)}{x^{2}} dx \) becomes \( - \int \operatorname{csch}(u) \operatorname{coth}(u) du \), which as we mentioned in Step 2, is exactly the derivative of \(-\log(\cosh(u))\).

Step 4: Indefinite Integral

Remembering that the indefinite integral is the antiderivative, and noticing that we have the derivative of \(-\log(\cosh(u))\), we can see that the integral of this function is merely \(-\log(\cosh(u))\).

Step 5: Back-substitute the variable

Replacing \(u\) with \(1/x\) gives us the final answer to the integral, \(-\log(\cosh(1/x))\).

Step 6: Include constant of integration

We must not forget to add the constant of integration C, considering we are finding an indefinite integral, that would give us the final solution \(-\log(\cosh(1/x)) + C \).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are mathematical functions that are analogs of trigonometric functions and can be essential in solving integrals. They have names similar to trigonometric functions but with a slight twist:

• Sinh: Represents the hyperbolic sine function, \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
• Cosh: Represents the hyperbolic cosine function, \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)
• Tanh: Represents the hyperbolic tangent function, which is \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \)
• Coth: Represents the hyperbolic cotangent function, which is \( \coth(x) = \frac{\cosh(x)}{\sinh(x)} \)
• Sech: Represents the hyperbolic secant, \( \operatorname{sech}(x) = \frac{1}{\cosh(x)} \)
• Csch: Represents the hyperbolic cosecant, \( \csch(x) = \frac{1}{\sinh(x)} \)

These functions are used especially in calculus, promoting easiness in differentiation and integration processes. They mimic the properties of trigonometric functions, thus helping with solving hyperbolic integrals.

Substitution Method

The substitution method simplifies the process of finding integrals by changing variables. This approach is especially helpful for complicated integrals, turning them into ones that are more straightforward to handle.
To employ substitution:

• Identify a function and its derivative within the integral.
• Choose a new variable, typically denoted as \( u \), to represent a part of the integrand. In our case, we chose \( u = \frac{1}{x} \).
• Differentiate your substitution equation to express \( dx \) in terms of \( du \), for example, \( du = -\frac{1}{x^2} \, dx \).
• Rewrite the entire integral in terms of this new variable \( u \), leading to a more manageable integral.
• Once integrated, revert back to the original variable by substituting back, to get to your final solution.

This method is an essential tool when dealing with integrals involving complex functions and derivative products.

Antiderivative

An antiderivative of a function is another function whose derivative is the original function. Finding the antiderivative is practically what indefinite integration accomplishes. With indefinite integration, we are looking for the family of functions, which differentiate back to the original integrand.

• In the given exercise, our tricky integral involved \( \operatorname{csch}(u) \cdot \operatorname{coth}(u) \), a derivative of a negative logarithmic function.
• Recognizing such derivatives helps in directly pinpointing the antiderivative. Here the antiderivative of \( -\operatorname{csch}(u) \cdot \operatorname{coth}(u) \) is \( -\log(\cosh(u)) \).

Finding the antiderivative can be seen as the reverse process of differentiation, essential in calculus for solving much broader mathematical problems.

Constant of Integration

When dealing with indefinite integrals, the constant of integration arises from the fact that differentiation removes the constant term, thus hiding information about the original function. Therefore, we must include it back during integration.

• Integration results in a general family of functions, all varying by a constant, denoted as \( C \).
• In our exercise, the final solution was \( -\log(\cosh(1/x)) + C \).
• This constant ensures our solution is complete, covering any potential original function that could have led to the initial derivative, making it crucial for representing all possible antiderivatives.

The constant of integration is an intrinsic part when providing a complete answer to an indefinite integral and is indeed pivotal in calculus for forming general solutions.