Problem 40 - Calculus Early Transcendentals

Question

Evaluate each definite integral. $$\int_{\ln 2}^{\ln 3} \operatorname{csch} y d y$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral of the hyperbolic cosecant function from the natural logarithm of 2 to the natural logarithm of 3. Answer: The definite integral is equal to $\operatorname{arccoth}(2) - \operatorname{arccoth}(\sqrt{3})$.

Step 1: Rewrite the hyperbolic cosecant function

Rewrite the hyperbolic cosecant function using the definition, csch(y) = 1/sinh(y): $$\int_{\ln 2}^{\ln 3} \frac{1}{\sinh y} d y$$

Step 2: Use substitution method

We will use substitution method to evaluate the integral. Let's set: $$u = \sinh y$$ Then, find the derivative with respect to y: $$\frac{d u}{d y} = \cosh y$$ Now, solve for dy: $$d u = \cosh y d y$$ $$d y = \frac{d u}{\cosh y}$$ Replace the bounds of the integral: - When y = ln(2): u = sinh(ln(2)) - When y = ln(3): u = sinh(ln(3)) Now, substitute y and dy into the integral: $$\int_{\sinh(\ln 2)}^{\sinh(\ln 3)} \frac{1}{u} \frac{d u}{\cosh (\operatorname{arcsinh} u)}$$

Step 3: Simplify the integral

Using the identity \(\cosh (\operatorname{arcsinh} u) = \sqrt{1+u^2}\), we simplify the integral: $$\int_{\sinh(\ln 2)}^{\sinh(\ln 3)} \frac{1}{u \sqrt{1 + u^2}} d u$$

Step 4: Evaluate the integral

This integral is a standard form. The antiderivative of \(\frac{1}{u\sqrt{1+u^2}}\) is \(\operatorname{arccoth}(\sqrt{1+u^2})\) (sometimes also denoted as \(\operatorname{arctanh}(\frac{1}{\sqrt{1+u^2}})\) ). Therefore, we have: $$\operatorname{arccoth}(\sqrt{1+(\sinh(\ln 3))^2}) - \operatorname{arccoth}(\sqrt{1+(\sinh(\ln 2))^2}) $$

Step 5: Simplify the final expression

Now, simplify the final expression: $$\operatorname{arccoth}(\sqrt{1+3}) - \operatorname{arccoth}(\sqrt{1+2})$$ $$\operatorname{arccoth}(2) - \operatorname{arccoth}(\sqrt{3})$$ So, the definite integral of the given function is: $$\int_{\ln 2}^{\ln 3} \operatorname{csch} y d y = \operatorname{arccoth}(2) - \operatorname{arccoth}(\sqrt{3})$$

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogues of the ordinary trigonometric, or circular, functions. The basic hyperbolic functions are hyperbolic sine (\text{sinh}) and hyperbolic cosine (\text{cosh}), from which others are derived. For instance, hyperbolic cosecant (\text{csch}) is the reciprocal of \text{sinh}, just like cosecant is the reciprocal of sine in trigonometry.

These functions might seem exotic at first glance, but they occur naturally in various areas of mathematics, including the solutions of certain differential equations and the descriptions of hyperbolic geometry. Another intriguing aspect of hyperbolic functions is their relation to the unit hyperbola, similar to how trigonometric functions are related to the unit circle.

Understanding hyperbolic functions can be facilitated by learning their definitions, such as \(\text{sinh}(x) = \frac{e^{x} - e^{-x}}{2}\) and \(\text{cosh}(x) = \frac{e^{x} + e^{-x}}{2}\), and their properties are analogous to those of trigonometric functions. For example, the hyperbolic identity \(\text{cosh}^2(x) - \text{sinh}^2(x) = 1\) mirrors the Pythagorean identity \(\text{sin}^2(x) + \text{cos}^2(x) = 1\).

Substitution Method in Integration

The substitution method, also known as u-substitution, is a technique used to solve integrals by simplifying them into a form that is easier to integrate. It is the reverse process of the chain rule in differentiation and is particularly handy when dealing with composite functions.

To use this method, we generally choose a part of the integral as 'u', which simplifies the integral when substituted back. The goal is to rewrite the entire integral in terms of 'u', making the integral more manageable. In some cases, substitution can turn a seemingly complex integral into a basic form whose antiderivative is known.

For instance, if you have an integral that involves \(\text{sinh}(y)\), you might let \(u = \text{sinh}(y)\), then find \(du/dy\) to express \(dy\) in terms of \(du\) and substitute these back into the integral. The boundaries of the integral must also be changed to correspond to the new variable 'u'. The beauty of u-substitution lies in its ability to transform integrals into familiar forms, allowing the use of known antiderivatives to find the solution.

Antiderivatives

Antiderivatives, also known as indefinite integrals, play a central role in calculus, representing the operation inverse to differentiation. When we find the antiderivative of a function, we are essentially determining a function whose derivative is the given function. This process is crucial for solving differential equations and computing areas under curves, among other applications.

In the context of definite integrals, once we evaluate an indefinite integral and find an antiderivative, we can use the Fundamental Theorem of Calculus to compute the area under the curve over a certain interval. To find the definite integral, we evaluate the antiderivative at the upper and lower limits of integration and subtract the latter from the former. This procedure allows us to determine the accumulation of a quantity, such as distance, area, volume, or other physical and geometrical quantities.

Therefore, understanding antiderivatives is not only about finding a function but also about connecting various concepts in calculus to solve practical problems. Knowing standard antiderivatives, like \(\text{sin}(x)\), \(\text{cos}(x)\), or \(e^x\), and applying techniques like the substitution method can vastly simplify the process of integration.