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The integrals in Exercises \(1-34\) converge. Evaluate the integrals without using tables. $$\int_{0}^{\infty} \frac{16 \tan ^{-1} x}{1+x^{2}} d x$$

Short Answer

Expert verified
The integral evaluates to \(2\pi^2\).

Step by step solution

01

Substitution Choice

To simplify the given integral \(\int_{0}^{\infty} \frac{16 \tan^{-1} x}{1+x^2} dx\), use a substitution. Let \( u = \tan^{-1} x \), which implies that \( du = \frac{1}{1+x^2} dx \). This substitution makes the integral in terms of \( u \).
02

Change the Limits of Integration

Since \( u = \tan^{-1} x \), as \( x \) ranges from 0 to \( \infty \), \( u \) ranges from \( \tan^{-1}(0) = 0 \) to \( \tan^{-1}(\infty) = \frac{\pi}{2} \).
03

Integral in Terms of New Variable

Substitute \( u \) and \( du \) into the integral, changing it to \[ \int_{0}^{\pi/2} 16u \, du. \] This is a straightforward integration problem.
04

Evaluating the Integral

Now, evaluate the integral \(16 \int_{0}^{\pi/2} u \, du\). The antiderivative of \(u\) is \(\frac{u^2}{2}\). Thus the integral becomes: \[ 16 \left[ \frac{u^2}{2} \right]_{0}^{\pi/2} = 16 \left[ \frac{(\pi/2)^2}{2} - \frac{0^2}{2} \right]. \]
05

Final Calculation

Calculate the expression obtained in Step 4: \[ 16 \left( \frac{\pi^2}{8} \right) = 2\pi^2. \] Thus, the integral evaluates to \( 2\pi^2 \).

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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 approach used to simplify complex integrals. It involves substituting a part of the integral with a new variable, which can make the integration process easier. In our original exercise, we employ substitution by letting \( u = \tan^{-1} x \). This choice is strategic because the derivative of \( \tan^{-1} x \), given by \( du = \frac{1}{1+x^2} dx \), conveniently matches a portion of the original integral's structure.

By rewriting the integral in terms of \( u \), we transform it from a potentially difficult problem into one that's more straightforward. The limits of integration must also change accordingly when the substitution is applied. This involves evaluating the new variable's limits, which shift from \( x = 0 \) and \( x = \infty \) to \( u = 0 \) and \( u = \frac{\pi}{2} \).
  • Choose a substitution that simplifies the integral.
  • Adjust the limits of integration to match the new variable.
  • Simplify and solve the integral in terms of the new variable.
This method is particularly useful in handling integrals involving inverse trigonometric functions, like arctangent in this exercise.
Trigonometric Integrals
Trigonometric integrals involve expressions with trigonometric functions, and they often require specialized techniques for simplification. The given integral includes \( \tan^{-1} x \), which is related to the inverse tangent.

These types of integrals are common in calculus, where they are frequently encountered in contexts involving periodic functions or when transforming between different forms. In our exercise, recognizing the integration of \( \tan^{-1} x \) combined with its derivative allows us to employ substitution effectively. Using a trigonometric identity or a known derivative can often transform the integral into a more workable form.
  • Identify the trigonometric function in the integral.
  • Consider using identities or properties of inverse trigonometric functions.
  • Apply substitution or other integration techniques as needed.
Mastery of trigonometric integrals helps in solving a wide range of problems where angles and periodicity are involved.
Integration Techniques
Integration techniques are methods used to solve integrals that are not easily evaluated by direct application of known formulas. In this example, substitution was the primary technique used, but there are others such as integration by parts, partial fractions, and trigonometric substitutions.

The key to choosing the right technique often lies in the form of the integrand. With practice, identifying the best method becomes intuitive. The goal is to simplify the integral into a basic form where the antiderivative can be easily determined.
  • Substitution: transforms the integral into an easier form.
  • Integration by parts: useful for products of functions.
  • Partial fractions: breaks down rational expressions.
  • Trigonometric substitutions: useful for integrals involving roots.
Familiarity with various integration techniques expands the toolkit for tackling complex integrals such as the one given.
Antiderivatives
Finding an antiderivative is at the heart of integration. It involves determining a function whose derivative is the given integrand. For the integral of a simple function like \( u \), where \( \ u = \tan^{-1} x \), the antiderivative is straightforward, \( \frac{u^2}{2} \).

Antiderivatives are crucial because they allow us to evaluate definite integrals using the Fundamental Theorem of Calculus. Once the antiderivative is found, it is used to calculate the definite integral by evaluating it at the given limits and then subtracting.
  • Identify the basic structure of the integral function.
  • Find a function whose derivative matches the integrand.
  • Use the antiderivative to compute definite integrals.
Understanding how to find and apply antiderivatives is essential for solving integrals efficiently and correctly.

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