Textbook Question
In Exercises 35–68, 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.
∫ from 0 to π/2 of (cot θ dθ)
Ch. 8 - Techniques of Integration
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 8, Problem 8.8.28
The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.
∫₀¹ (4r dr) / √(1 − r⁴)
Verified step by step guidance1
Recognize that the integral is of the form \(\int_0^1 \frac{4r}{\sqrt{1 - r^4}} \, dr\). The integrand involves a composite function with \(r^4\) inside the square root, suggesting a substitution related to \(r^2\) or \(r^4\).
Make the substitution \(u = r^2\). Then, compute the differential: \(du = 2r \, dr\), which implies \(r \, dr = \frac{du}{2}\). This will help simplify the integral.
Rewrite the integral in terms of \(u\): replace \(4r \, dr\) with \(4 \times \frac{du}{2} = 2 \, du\), and replace \(\sqrt{1 - r^4}\) with \(\sqrt{1 - u^2}\) since \(r^4 = (r^2)^2 = u^2\). The integral becomes \(\int_0^1 \frac{2}{\sqrt{1 - u^2}} \, du\).
Recognize that \(\int \frac{1}{\sqrt{1 - u^2}} \, du\) is the standard integral for \(\arcsin u\). Therefore, the integral simplifies to \(2 \int_0^1 \frac{1}{\sqrt{1 - u^2}} \, du = 2 [\arcsin u]_0^1\).
Evaluate the definite integral by substituting the limits into \(\arcsin u\): compute \(2 (\arcsin 1 - \arcsin 0)\). This will give the value of the original integral.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Improper Integrals and Convergence
An integral is improper if the interval is infinite or the integrand has an infinite discontinuity. Understanding convergence means determining whether the integral approaches a finite value. In this problem, the integrand involves a square root in the denominator, which may cause issues near the upper limit, so checking convergence is essential.
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Improper Integrals: Infinite Intervals
Substitution Method for Integration
The substitution method simplifies integrals by changing variables to transform the integral into a more manageable form. Here, substituting a function inside the square root (like u = r^4) can simplify the integrand and make the integral easier to evaluate without tables.
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Evaluating Definite Integrals
Evaluating definite integrals involves finding an antiderivative and applying the Fundamental Theorem of Calculus to compute the difference at the bounds. After substitution and integration, carefully substituting back and evaluating at the limits 0 and 1 yields the final value.
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Related Practice
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∫ dx / (x² + 2x)
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