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Chapter 8: Problem 66

The integration formulas for \(1 / \sqrt{u^{2}-1}\) in Theorem 8.8 .6 are valid for \(u>1 .\) Show that the following formula is valid for \(u<-1:\) \(\int \frac{d u}{\sqrt{u^{2}-1}}=-\cosh ^{-1}(-u)+C=\ln |u+\sqrt{u^{2}-1}|+C\)

Short Answer

Expert verified
Substitute \( v = -u \), \( u < -1 \): \( \int \frac{du}{\sqrt{u^2-1}} = -\cosh^{-1}(-u) + C = \ln|u + \sqrt{u^2-1}| + C \).

Step by step solution

01

Analyze the Given Problem

We are given the integration of the function \( \frac{1}{\sqrt{u^2 - 1}} \) for \( u > 1 \) and need to prove it valid for \( u < -1 \). The formula given is \( \int \frac{du}{\sqrt{u^2 - 1}} = -\cosh^{-1}(-u) + C = \ln |u + \sqrt{u^2 - 1}| + C \). We'll use a substitution method to derive this formula.
02

Consider Symmetry and Substitution

Notice the symmetry of the function \( \sqrt{u^2 - 1} \). By considering \( v = -u \), where \( v > 1 \) when \( u < -1 \). Thus, \(-u = v\), \( du = -dv \). This provides a way to relate the integration of \(-u\) with \(v\).
03

Substitute and Integrate for v > 1

Substituting \( v = -u \), our integral becomes \( \int \frac{-dv}{\sqrt{v^2 - 1}} \). The negative sign changes the integration bounds (not necessary here, just note direction), resulting in \( -\cosh^{-1}(v) + C \).
04

Substitute Back to Original Variable

Revert substitution with \( v = -u \) in \(-\cosh^{-1}(v) + C\) to get \(-\cosh^{-1}(-u) + C\). This matches the given result to prove.
05

Show Logarithmic Expression

The solution also consists of a logarithmic expression. Since \( \cosh^{-1}(x) = \ln|x + \sqrt{x^2 - 1}| \), apply this to \( -u \), simplifying: \(-\cosh^{-1}(-u) = \ln|-u + \sqrt{(-u)^2 - 1}| = \ln|u + \sqrt{u^2 - 1}|\). This verifies the formula.

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Key Concepts

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

Substitution in Integration
Substitution is a clever technique in integration. It often turns a complex integral into a simpler one. When integrating, particularly when dealing with functions that don't fit basic formulas, substitution is a go-to method. It involves changing the variable into another form that makes the integration more straightforward.
For instance, if dealing with an integral like \( \int \frac{1}{\sqrt{u^2 - 1}} \, du \), converting the variable with a suitable substitution can simplify the process.
  • Choose a new variable to represent a part of the original expression.
  • Calculate the differential of the new variable.
  • Replace the original variable and differential in the integral.
In this exercise, the substitution \( v = -u \) showed symmetry and simplified the expression under the square root, making integration feasible.
Inverse Hyperbolic Functions
Inverse hyperbolic functions, like \( \cosh^{-1}(x) \), are analogues of inverse trigonometric functions. These functions are useful when dealing with integrals of expressions like \( \sqrt{u^2-1} \). In the context of this problem, \( \cosh^{-1}(-u) \) is significant for describing the integral's result.
These functions serve as solutions to hyperbolic equations, much like inverse trigonometric functions for circle-based equations.
  • They can be expressed in terms of natural logarithms. For example, \( \cosh^{-1}(x) = \ln |x + \sqrt{x^2 - 1}| \).
  • This forms a direct connection between hyperbolic functions and logarithmic expressions, which is key in solving the integral given \( u < -1 \).
Definite Integrals
Definite integrals involve working not just with functions but also limits. They give the signed area under a curve. Though the given exercise is about indefinite integrals, a grasp of definite integrals provides insight into the nature of integration as a whole.
For expressions involving hyperbolic functions or those under square roots, evaluating a definite integral often involves checking symmetry or substituting to simplify the expression.
Here is a simple process:
  • Determine the limits of integration based on the context.
  • Perform substitution if necessary.
  • Evaluate antiderivatives correctly.
  • Compute the difference to find the area or value over the interval.
Logarithmic Integration
Logarithmic integration is used when an integral results in a function of the form involving logarithms. It鈥檚 particularly common when integrating rational functions or those involving inverse hyperbolic functions.
In the context of this exercise, logarithmic expressions like \( \ln |u + \sqrt{u^2 - 1}| \) emerge from evaluating functions like \( \cosh^{-1}(-u) \).
  • Logarithmic integration often surfaces in solutions involving substitutive variables, especially when hyperbolic identities are used.
  • Recognizing that \( \cosh^{-1}(x) \) is equivalent to a log expression aids in verifying integral results.
Understanding the interconnectedness of logarithms and hyperbolic functions enriches comprehension of the underlying patterns in complex integrals.

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