Textbook Question
Solve the initial value problems in Exercises 67–70 for x as a function of t.
(t + 1) (dx/dt) = x² + 1 (for t > -1), x(0) = 0
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 8.2.52
Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.
∫ x² tan⁻¹(x / 2) dx
Verified step by step guidance1
Identify the integral to solve: \(\int x^{2} \tan^{-1}\left(\frac{x}{2}\right) \, dx\).
Recognize that this integral involves a product of functions: a polynomial \(x^{2}\) and an inverse tangent function \(\tan^{-1}\left(\frac{x}{2}\right)\). This suggests using integration by parts.
Set up integration by parts using the formula: \(\int u \, dv = uv - \int v \, du\). Choose \(u = \tan^{-1}\left(\frac{x}{2}\right)\) because its derivative simplifies nicely, and \(dv = x^{2} \, dx\).
Compute \(du\) by differentiating \(u\): use the chain rule on \(\tan^{-1}\left(\frac{x}{2}\right)\), and compute \(v\) by integrating \(dv = x^{2} \, dx\).
Substitute \(u\), \(du\), \(v\), and \(dv\) into the integration by parts formula, then simplify and evaluate the resulting integral to complete the solution.
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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. Choosing u and dv wisely is crucial to simplify the integral effectively.
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Integration by Parts for Definite Integrals
Inverse Trigonometric Functions
Inverse trigonometric functions, like arctangent (tan⁻¹), are the inverses of trigonometric functions and often appear in integrals. Understanding their derivatives and properties helps in differentiating or integrating expressions involving these functions.
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06:35Derivatives of Other Inverse Trigonometric Functions
Polynomial Functions in Integration
Polynomial functions, such as x², are straightforward to integrate and often serve as parts of integrands. Recognizing how to handle polynomials within more complex integrals, especially when combined with other functions, is essential for applying integration techniques effectively.
Recommended video:
07:00Taylor Polynomials
Related Practice
Textbook Question
Volume of water in a swimming pool
A rectangular swimming pool is 30 ft wide and 50 ft long. The accompanying table shows the depth h(x) of the water at 5-ft intervals from one end of the pool to the other. Estimate the volume of water in the pool using the Trapezoidal Rule with n = 10 applied to the integral
V = ∫ from 0 to 50 of 30 · h(x) dx.
Textbook Question
The integrals in Exercises 1–44 are in no particular order. Evaluate each integral using any algebraic method, trigonometric identity, or substitution you think is appropriate.
∫ ((2ˣ - 1) / 3ˣ) dx
Textbook Question
Use the table of integrals at the back of the text to evaluate the integrals in Exercises 1–26.
∫ e^(-3t) sin(4t) dt
Textbook Question
The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.
∫₋∞^∞ (x dx) / (x² + 4)^(3/2)
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Textbook Question
The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.
∫₀⁴ dx / √(4 − x)
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