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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 8 - Integration TechniquesProblem 8.R.38
Chapter 8, Problem 8.R.38

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.
38. ∫ (from π/4 to π/2) x csc²x dx

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Identify the integral to be solved: \(\int_{\pi/4}^{\pi/2} x \csc^{2}x \, dx\).
Recognize that this integral involves a product of a polynomial function \(x\) and a trigonometric function \(\csc^{2}x\), suggesting the use of integration by parts.
Recall the integration by parts formula: \(\int u \, dv = uv - \int v \, du\). Choose \(u = x\) (which simplifies upon differentiation) and \(dv = \csc^{2}x \, dx\) (which has a known antiderivative).
Compute \(du = dx\) and find \(v\) by integrating \(dv\): \(v = \int \csc^{2}x \, dx = -\cot x\).
Apply the integration by parts formula: \(\int x \csc^{2}x \, dx = -x \cot x + \int \cot x \, dx\). Then, evaluate the remaining integral \(\int \cot x \, dx\) and apply the definite integral limits \(\pi/4\) to \(\pi/2\) to find the final expression.

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Key Concepts

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

Integration by Parts

Integration by parts is a technique based on the product rule for differentiation. It transforms the integral of a product of functions into simpler integrals, using the formula ∫u dv = uv - ∫v du. This method is especially useful when integrating products like x and csc²x.
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Integral of csc²x

The integral of csc²x with respect to x is a standard integral that equals -cot x + C. Recognizing this allows simplification when integrating expressions involving csc²x, which is essential for solving the given integral.
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Definite Integrals and Evaluation at Limits

Definite integrals compute the net area under a curve between two bounds. After finding the antiderivative, you evaluate it at the upper and lower limits and subtract to find the exact value. Proper evaluation at π/4 and π/2 is crucial for the final answer.
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Related Practice
Textbook Question

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Textbook Question

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