/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 89 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 89

Evaluate the following integrals. $$\int \frac{\cosh z}{\sinh ^{2} z} d z$$

Short Answer

Expert verified
Question: Evaluate the integral of the function \(\frac{\cosh z}{\sinh^2 z}\) with respect to \(z\). Answer: \(\int \frac{\cosh z}{\sinh^2z} dz = \frac{1}{2} \ln|\cosh z-1| - \frac{1}{2} \ln|\cosh z+1| + C\)

Step by step solution

01

Recall the relationship between hyperbolic trigonometric functions

To rewrite \(\sinh^2z\), we need to use the identity \(\cosh^2z - \sinh^2z = 1\), which is the fundamental relationship between hyperbolic sine and hyperbolic cosine functions.
02

Rewrite \(\sinh^2z\) in terms of \(\cosh z\)

From the identity mentioned in step 1, we have \(\sinh^2z = \cosh^2z - 1\). Substitute this into our integral: $$ \int \frac{\cosh z}{\sinh^2z} dz = \int \frac{\cosh z}{\cosh^2z -1} dz $$
03

Make a substitution

Let \(u = \cosh z\), so \(\frac{du}{dz} = \sinh z\). Then the integral becomes: $$ \int \frac{u}{u^2 - 1} du $$
04

Solve the integral using partial fractions

We can rewrite the integrand using partial fractions as follows: $$ \frac{u}{u^2-1} = \frac{A}{u-1}+\frac{B}{u+1} $$ Multiplying both sides by \((u-1)(u+1)\), we get: $$ u = A(u+1)+B(u-1) $$ Setting \(u = 1\), we get \(A = \frac{1}{2}\). Setting \(u = -1\), we get \(B = -\frac{1}{2}\). Thus, our integrand becomes: $$ \frac{1}{2} \left(\frac{1}{u-1}-\frac{1}{u+1}\right) $$
05

Evaluate the integral

Now we can integrate easily: $$ \int \frac{u}{u^2 - 1} du = \frac{1}{2} \int \frac{1}{u-1} du - \frac{1}{2} \int \frac{1}{u+1} du $$ $$ = \frac{1}{2} \ln|u-1| - \frac{1}{2} \ln|u+1| + C $$
06

Substitute back the original variable

Finally, substitute back \(u = \cosh z\) into the previous equation to get the final answer: $$ \int \frac{\cosh z}{\sinh^2z} dz = \frac{1}{2} \ln|\cosh z-1| - \frac{1}{2} \ln|\cosh z+1| + C $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hyperbolic Functions
Hyperbolic functions are similar to trigonometric functions but are based on hyperbolas instead of circles. They are essential when working with hyperbolic planes and have applications in calculus, especially in integration. The two primary hyperbolic functions are hyperbolic sine, denoted as \( \sinh z \), and hyperbolic cosine, denoted as \( \cosh z \). They are defined by:
  • \( \sinh z = \frac{e^z - e^{-z}}{2} \)
  • \( \cosh z = \frac{e^z + e^{-z}}{2} \)
These functions feature prominently in the given integral expression, where the task is to integrate the quotient of \( \cosh z \) over \( \sinh^2 z \). An essential identity to remember, often used in such calculus problems, is \( \cosh^2 z - \sinh^2 z = 1 \), akin to Pythagorean identities used in trigonometry.
Substitution Method
The substitution method is a technique used in integration that simplifies a given function into a more manageable form. It involves replacing a part of the integrand with a single variable. In the given solution, the substitution \( u = \cosh z \) was made. This substitution transforms the differential \( \frac{du}{dz} = \sinh z \), leading to \( dz = \frac{du}{\sinh z} \). Once applied, the original integral with respect to \( z \) is expressed in terms of \( u \), offering a route to simplify complex expressions. This strategic change of variable often makes the integration process straightforward. Employing substitution in integrals is akin to rephrasing a problem in more familiar terms. It is an indispensable tool, especially when dealing with intricate functions.
Partial Fractions
Partial fractions are a technique where a complex fraction is rewritten as the sum of simpler fractions. This approach makes integration more straightforward, as each simpler fraction can often be handled individually. In the solution, the expression \( \frac{u}{u^2 - 1} \) is decomposed into partial fractions:
  • \( \frac{u}{u^2 - 1} = \frac{A}{u-1} + \frac{B}{u+1} \)
These simpler fractions allow for easier integration. The technique involves finding coefficients \( A \) and \( B \) by solving for values that balance the equation when multiplied back. This method is especially helpful when dealing with rational functions, allowing complex integrals to be broken down into sums or differences of simpler terms.
Fundamental Identities
Fundamental identities are core mathematical truths that simplify complex problems. In this integration problem, the identity \( \cosh^2 z - \sinh^2 z = 1 \) is crucial. This relationship is a hyperbolic analogue of the classic Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \).This identity is used to rewrite \( \sinh^2 z \) in terms of \( \cosh z \), leading to a simpler form of the integrand. Fundamental identities act as a powerful tool in calculus, creating a bridge between different aspects of a problem to make them easier to manage. Using these identities can help validate transformations and substitutions used during computations, ensuring accuracy and simplifying the path to solving integrals.

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Most popular questions from this chapter

A strong west wind blows across a circular running track. Abe and Bess start running at the south end of the track, and at the same time, Abe starts running clockwise and Bess starts running counterclockwise. Abe runs with a speed (in units of \(\mathrm{mi} / \mathrm{hr}\) ) given by \(u(\varphi)=3-2 \cos \varphi\) and \(\mathrm{Bess}\) runs with a speed given by \(v(\theta)=3+2 \cos \theta,\) where \(\varphi\) and \(\theta\) are the central angles of the runners. a. Graph the speed functions \(u\) and \(v,\) and explain why they describe the runners' speeds (in light of the wind). b. Compute the average value of \(u\) and \(v\) with respect to the central angle. c. Challenge: If the track has a radius of \(\frac{1}{10} \mathrm{mi}\), how long does it take each runner to complete one lap and who wins the race?

Use the following argument to show that \(\lim _{x \rightarrow \infty} \ln x=\infty\) and \(\lim _{x \rightarrow 0^{+}} \ln x=-\infty\). a. Make a sketch of the function \(f(x)=1 / x\) on the interval \([1,2] .\) Explain why the area of the region bounded by \(y=f(x)\) and the \(x\) -axis on [1,2] is \(\ln 2\) b. Construct a rectangle over the interval [1,2] with height \(\frac{1}{2}\) Explain why \(\ln 2>\frac{1}{2}\). c. Show that \(\ln 2^{n}>n / 2\) and \(\ln 2^{-n}<-n / 2\). d. Conclude that \(\lim _{x \rightarrow \infty} \ln x=\infty\) and \(\lim _{x \rightarrow 0^{+}} \ln x=-\infty\).

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