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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 8 - Techniques of IntegrationProblem 8.8.40
Chapter 8, Problem 8.8.40

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)

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Identify the integral to be tested for convergence: \(\int_0^{\frac{\pi}{2}} \cot \theta \, d\theta\).
Recall that \(\cot \theta = \frac{\cos \theta}{\sin \theta}\), and note that the integrand has potential issues at the endpoints \(\theta = 0\) and \(\theta = \frac{\pi}{2}\) because \(\sin \theta\) approaches zero there.
Split the integral into two improper integrals to analyze the behavior near the problematic points: \(\int_0^{\frac{\pi}{2}} \cot \theta \, d\theta = \int_0^a \cot \theta \, d\theta + \int_a^{\frac{\pi}{2}} \cot \theta \, d\theta\) where \(a\) is some number between 0 and \(\frac{\pi}{2}\).
Use a comparison test by comparing \(\cot \theta\) near \(0\) to a simpler function that behaves like \(\frac{1}{\theta}\), since \(\sin \theta \approx \theta\) near zero, so \(\cot \theta \approx \frac{1}{\theta}\). Similarly, analyze the behavior near \(\frac{\pi}{2}\).
Determine whether the integral converges or diverges by evaluating the limits of the integral near the problematic points or by applying the Direct or Limit Comparison Test with the simpler functions identified.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Improper Integrals

Improper integrals involve integrals with infinite limits or integrands that become unbounded within the interval. To determine convergence, one must analyze the behavior of the function near points of discontinuity or infinite limits by taking limits of definite integrals.
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Improper Integrals: Infinite Intervals

Direct Comparison Test

The Direct Comparison Test determines convergence by comparing the given integral to another integral with a known behavior. If the integrand is less than or equal to a convergent function, the integral converges; if it is greater than or equal to a divergent function, it diverges.
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Direct Comparison Test

Limit Comparison Test

The Limit Comparison Test compares two functions by examining the limit of their ratio as the variable approaches a problematic point. If the limit is a positive finite number, both integrals either converge or diverge together, helping to determine the behavior of complex integrals.
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Limit Comparison Test
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