Textbook Question
Evaluate the integrals in Exercises 51–56 by making a substitution (possibly trigonometric) and then applying a reduction formula.
∫ (from 0 to 1/√3) dt / (t² + 1)^(7/2)
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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 8.2.12
Evaluate the integrals in Exercises 1–24 using integration by parts.
∫ arcsin(y) dy
Verified step by step guidance1
Identify the integral to solve: \(\int \arcsin(y) \, dy\).
Recall the integration by parts formula: \(\int u \, dv = uv - \int v \, du\).
Choose \(u = \arcsin(y)\) because its derivative simplifies, and \(dv = dy\) because it is easy to integrate.
Compute \(du\) and \(v\):
- \(du = \frac{1}{\sqrt{1 - y^2}} \, dy\) (derivative of \(\arcsin(y)\)),
- \(v = y\) (integral of \(dy\)).
Apply the integration by parts formula:
\(\int \arcsin(y) \, dy = y \arcsin(y) - \int y \cdot \frac{1}{\sqrt{1 - y^2}} \, dy\).
Next, focus on evaluating the remaining integral \(\int \frac{y}{\sqrt{1 - y^2}} \, dy\).
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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 derived from the product rule of differentiation. It transforms the integral of a product of functions into simpler integrals using the formula ∫u dv = uv - ∫v du. Choosing u and dv wisely is crucial to simplify the integral effectively.
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06:18
Integration by Parts for Definite Integrals
Derivative of Inverse Trigonometric Functions
Understanding the derivative of arcsin(y) is essential, as it helps in identifying du when applying integration by parts. The derivative of arcsin(y) with respect to y is 1/√(1 - y²), which is used to find du in the integration process.
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06:35Derivatives of Other Inverse Trigonometric Functions
Basic Integration Techniques
Familiarity with basic integrals, such as ∫ dy and integrals involving square roots, is important to solve the resulting integrals after applying integration by parts. This includes recognizing standard forms and using substitution if necessary.
Recommended video:
06:07Basic Rules for Definite Integrals
Related Practice
Textbook Question
[Technology Exercise] When solving Exercises 33-40, you may need to use a calculator or a computer.
Find, to two decimal places, the areas of the surfaces generated by revolving the curves in Exercises 35 and 36 about the x-axis.
y = x²/4, 0 ≤ x ≤ 2
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Textbook Question
Use the table of integrals at the back of the text to evaluate the integrals in Exercises 1–26.
∫ x (7x + 5)^(3/2) dx
Textbook Question
Evaluate the integrals in Exercises 51–56 by making a substitution (possibly trigonometric) and then applying a reduction formula.
∫ csc³(√θ) / √θ dθ
Textbook Question
[Technology Exercise] When solving Exercises 33-40, you may need to use a calculator or a computer.
Find, to two decimal places, the areas of the surfaces generated by revolving the curves in Exercises 35 and 36 about the x-axis.
y = sin x, 0 ≤ x ≤ π
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Textbook Question
Use any method to evaluate the integrals in Exercises 15–38. Most will require trigonometric substitutions, but some can be evaluated by other methods.
∫ √(9 - w²) dw / w²