Chapter 4: Problem 40
Find the integral. \(\int \frac{\sinh x}{1+\sinh ^{2} x} d x\)
Short Answer
Expert verified
\(\int \frac{\sinh x}{1+\sinh ^{2} x} d x = \arctan(\sinh(x)) + C\
Step by step solution
01
Substitution
Let \(u = \sinh(x)\), so \(du = \cosh(x) dx\). The integral then simplifies to \(\int \frac{u}{1+u^2} du\)
02
Integral Calculation
The new form of the integral is a standard form integral. It can be solved by recognizing that it is the derivative of arctangent: The integral of \(\frac{u}{1+u^2}\) is equal to \(\arctan(u)\), so our integral becomes \(\arctan(u)\)
03
Back Substitution
Replace \(u\) with the original expression that we substituted: \(\arctan(u)\) becomes \(\arctan(\sinh(x))\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful technique used in calculus to simplify the process of finding integrals, particularly when the function involved is complex. By carefully choosing a new variable to replace an existing expression, the integral can often be transformed into a more familiar or simpler form.
In our example, the initial integral is \( \int \frac{\sinh x}{1+\sinh^{2} x} dx \). Choosing \( u = \sinh(x) \) as the substitution not only simplifies the integral to \( \int \frac{u}{1+u^2} du \), but it also allows us to see a resemblance to the derivative of the arctangent function, which is a great clue on how to proceed with the integration.
Key aspects of the substitution method include determining the new variable \( u \), differential substitution \( du \), and the eventual back substitution to return to the original variable. Keeping track of these substitutions is vital to properly solve the integral.
In our example, the initial integral is \( \int \frac{\sinh x}{1+\sinh^{2} x} dx \). Choosing \( u = \sinh(x) \) as the substitution not only simplifies the integral to \( \int \frac{u}{1+u^2} du \), but it also allows us to see a resemblance to the derivative of the arctangent function, which is a great clue on how to proceed with the integration.
Key aspects of the substitution method include determining the new variable \( u \), differential substitution \( du \), and the eventual back substitution to return to the original variable. Keeping track of these substitutions is vital to properly solve the integral.
Integral Calculation
Integral calculation is the process of finding the antiderivative or integral of a function. This process is fundamental in calculus and has wide-ranging applications in physics, engineering, and beyond.
Our exercise involves calculating the integral of a hyperbolic function, which often requires recognizing patterns that match basic integral forms. The ability to recognize when an integral matches the derivative of a known function, like the arctangent or inverse trigonometric functions, is a valuable skill. In the given problem, we identify that \( \frac{u}{1+u^2} \) matches the derivative of the arctangent function, \( \arctan(u) \), hence our integral simplifies to \( \arctan(u) + C \), where \( C \) is the constant of integration.
It's essential to be familiar with a range of integral formulas and techniques, such as direct integration, integration by parts, and trigonometric substitutions, to efficiently tackle diverse integral problems.
Our exercise involves calculating the integral of a hyperbolic function, which often requires recognizing patterns that match basic integral forms. The ability to recognize when an integral matches the derivative of a known function, like the arctangent or inverse trigonometric functions, is a valuable skill. In the given problem, we identify that \( \frac{u}{1+u^2} \) matches the derivative of the arctangent function, \( \arctan(u) \), hence our integral simplifies to \( \arctan(u) + C \), where \( C \) is the constant of integration.
It's essential to be familiar with a range of integral formulas and techniques, such as direct integration, integration by parts, and trigonometric substitutions, to efficiently tackle diverse integral problems.
Inverse Trigonometric Functions
Inverse trigonometric functions, also known as arcfunctions, allow us to find angles when we know the values of trigonometric functions. In relation to integration, they frequently appear as antiderivatives, particularly when the integrand is in a form that suggests a trigonometric substitution.
For instance, the integral of \( \frac{1}{1+x^2} \) is \( \arctan(x) + C \), as was seen in our step-by-step solution where we integrated \( \frac{u}{1+u^2} \) to obtain \( \arctan(u) \). In a broader sense, recognizing when to apply inverse trigonometric functions can simplify the integration process and lead to straightforward solutions.
The main inverse trigonometric functions include arcsine \( (\arcsin) \), arccosine \( (\arccos) \), and arctangent \( (\arctan) \), each with their respective domains and ranges that must be considered when solving integrals.
For instance, the integral of \( \frac{1}{1+x^2} \) is \( \arctan(x) + C \), as was seen in our step-by-step solution where we integrated \( \frac{u}{1+u^2} \) to obtain \( \arctan(u) \). In a broader sense, recognizing when to apply inverse trigonometric functions can simplify the integration process and lead to straightforward solutions.
The main inverse trigonometric functions include arcsine \( (\arcsin) \), arccosine \( (\arccos) \), and arctangent \( (\arctan) \), each with their respective domains and ranges that must be considered when solving integrals.
Hyperbolic Trigonometry
Hyperbolic trigonometry involves hyperbolic functions, such as \( \sinh \), \( \cosh \), and \( \tanh \), which are analogues of the familiar trigonometric functions but relate to a hyperbola rather than a circle.
These functions have properties and relationships akin to those of trigonometric functions, such as the hyperbolic identity \( \cosh^2(x) - \sinh^2(x) = 1 \), analogous to the Pythagorean identity in trigonometry. In our exercise, the function \( \sinh(x) \) was pivotal, and understanding its behavior was key to solving the integral.
Just like their trigonometric counterparts, hyperbolic functions have inverse functions that are frequently used in calculus. For example, \( \text{arsinh} \) or \( \text{arcosh} \) may be used when integrating expressions involving hyperbolic functions. Mastering these concepts is essential for students to tackle integrals that involve hyperbolic terms.
These functions have properties and relationships akin to those of trigonometric functions, such as the hyperbolic identity \( \cosh^2(x) - \sinh^2(x) = 1 \), analogous to the Pythagorean identity in trigonometry. In our exercise, the function \( \sinh(x) \) was pivotal, and understanding its behavior was key to solving the integral.
Just like their trigonometric counterparts, hyperbolic functions have inverse functions that are frequently used in calculus. For example, \( \text{arsinh} \) or \( \text{arcosh} \) may be used when integrating expressions involving hyperbolic functions. Mastering these concepts is essential for students to tackle integrals that involve hyperbolic terms.
