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Chapter 8: Problem 27

Evaluate each integral using any algebraic method or trigonometric identity you think is appropriate, and then use a substitution to reduce it to a standard form. $$\int \frac{2 d x}{x \sqrt{1-4 \ln ^{2} x}}$$

Short Answer

Expert verified
The integral evaluates to \( \sin^{-1}(2 \ln x) + C \).

Step by step solution

01

Identify the type of integral

We are given the integral \( \int \frac{2 \, dx}{x \sqrt{1 - 4 \ln^2 x}} \). This integral involves a natural logarithm function, suggesting that a substitution using \( u = \ln x \) might simplify the problem.
02

Perform the substitution

Let \( u = \ln x \). Then, \( du = \frac{1}{x} \, dx \) or \( x \, du = dx \). Substitute into the integral, yielding \( \int \frac{2}{x \sqrt{1 - 4u^2}} \cdot x \, du = \int \frac{2 \, du}{\sqrt{1 - 4u^2}} \).
03

Simplify the integral

The integral becomes \( \int \frac{2 \, du}{\sqrt{1 - 4u^2}} \). Factor out the constant under the square root: \( \int \frac{2 \, du}{\sqrt{1 - (2u)^2}} \). This is a standard form that can be directly integrated.
04

Use a trigonometric substitution

Recognize \( \sqrt{1 - (2u)^2} \) as the form suitable for a trigonometric substitution. Set \( 2u = \sin \theta \), thus \( du = \frac{1}{2} \cos \theta \, d\theta \). Rewrite the integral in terms of \( \theta \): \( \int \frac{2 \cdot \frac{1}{2} \cos \theta \, d\theta}{\cos \theta} = \int d\theta \).
05

Integrate and back-substitute

Integrate to find \( \theta \): \( \int d\theta = \theta + C \). Since \( 2u = \sin \theta \) or \( u = \frac{1}{2} \sin \theta \), we substitute back to find \( \theta = \sin^{-1}(2u) \). Thus, the result is \( \sin^{-1}(2 \ln x) + C \).

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Key Concepts

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

Natural Logarithm
The natural logarithm, denoted as \( \ln x \), is a logarithm with base \( e \), where \( e \) is approximately equal to 2.71828. It is a fundamental function in mathematics because it can simplify expressions involving exponential growth or decay. The natural logarithm is the inverse function of the exponential function \( e^x \).
  • For any positive number \( x \), the natural logarithm is defined as the power to which \( e \) must be raised to obtain \( x \).
  • Mathematically, if \( y = \ln x \), then \( e^y = x \).
  • The derivative of the natural logarithm \( \ln x \) with respect to \( x \) is \( \frac{1}{x} \).
  • It's particularly useful in calculus, as seen in our exercise, where \( u = \ln x \) helped simplify the integral.
Understanding how to manipulate natural logarithms is critical in solving many calculus problems, especially those involving complex integrals.
Substitution Method
The substitution method is a technique used in integral calculus to simplify integrals. This involves substituting part of the integral with a new variable, making it easier to integrate.
  • Identify a portion of the integrand that can be substituted, often making the integral look like a standard form.
  • Replace this portion with a new variable \( u \), and rewrite the integral in terms of \( u \).
  • The goal is to transform a complicated integral into a simpler one, which can be integrated more easily.
In our problem, we used substitution by letting \( u = \ln x \), which simplified the integral significantly to \( \int \frac{2 \, du}{\sqrt{1 - 4u^2}} \). This transformation paved the way for further simplification using trigonometric substitution.
Trigonometric Substitution
Trigonometric substitution is a technique to simplify integrals involving square roots of quadratic expressions. By using trigonometric identities, this method transforms the problem into a trigonometric integral.
  • This approach works well when the integrand contains expressions like \( \sqrt{a^2 - x^2} \), \( \sqrt{x^2 + a^2} \), or \( \sqrt{x^2 - a^2} \).
  • Make a trigonometric substitution such as \( x = a \sin \theta \), \( x = a \tan \theta \), or \( x = a \sec \theta \), which can simplify the expression.
In our exercise, the substitution \( 2u = \sin \theta \) transformed \( \sqrt{1 - 4u^2} \) into \( \sqrt{1 - \sin^2 \theta} = \cos \theta \), leading to a much simpler integral \( \int d\theta \). This change made the integral more straightforward to solve.
Definite Integrals
Definite integrals calculate the accumulated quantity, such as area under a curve within given limits. However, our exercise focused on indefinite integration, which finds the antiderivative without specific limits.
  • The indefinite integral does not evaluate specific limits; instead, it represents a family of functions differing by a constant \( C \).
  • In contrast, definite integrals would require limits and give a specific value, such as area, volume, or another accumulated total.
While our problem didn't evaluate a definite integral, understanding the distinction helps in recognizing when to apply these important calculus tools. Integrating indefinitely, as we did, is crucial for finding general solutions and working with functions without set boundaries.

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Most popular questions from this chapter

Evaluate the integrals without using tables. $$\int_{0}^{2} \frac{d s}{\sqrt{4-s^{2}}}$$

Use integration, the Direct Comparison Test, or the limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{1}^{\infty} \frac{d x}{x^{3}+1}$$

Use integration, the Direct Comparison Test, or the limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{0}^{1} \frac{\ln x}{x^{2}} d x$$

Use integration, the Direct Comparison Test, or the limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{0}^{\ln 2} x^{-2} e^{-1 / x} d x$$

Use integration, the Direct Comparison Test, or the limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{0}^{2} \frac{d x}{1-x}$$
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