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.
∫ dx / (x² √(x² + 1))
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 8.2.48
Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.
∫₀^π/2 x³ cos 2x dx
Verified step by step guidance1
Identify the integral to solve: \(\int_0^{\frac{\pi}{2}} x^3 \cos(2x) \, dx\).
Recognize that this integral involves a product of a polynomial (\(x^3\)) and a trigonometric function (\(\cos(2x)\)), which suggests using integration by parts.
Set up integration by parts by choosing \(u = x^3\) (which simplifies when differentiated) and \(dv = \cos(2x) \, dx\) (which can be integrated easily). Recall the formula: \(\int u \, dv = uv - \int v \, du\).
Compute \(du = 3x^2 \, dx\) and find \(v\) by integrating \(dv\): \(v = \int \cos(2x) \, dx = \frac{1}{2} \sin(2x)\).
Apply the integration by parts formula: \(\int_0^{\frac{\pi}{2}} x^3 \cos(2x) \, dx = \left. x^3 \cdot \frac{1}{2} \sin(2x) \right|_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} \frac{1}{2} \sin(2x) \cdot 3x^2 \, dx\). This reduces the original integral to a new integral involving \(x^2 \sin(2x)\), which may require repeating integration by parts.
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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 used to integrate products of functions. It is based on the product rule for differentiation and follows the formula ∫u dv = uv - ∫v du. Choosing u and dv wisely simplifies the integral, especially when one function becomes simpler upon differentiation.
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Integration by Parts for Definite Integrals
Definite Integrals
Definite integrals calculate the net area under a curve between two limits. They produce a numerical value and require evaluating the antiderivative at the upper and lower bounds. Understanding how to apply limits after integration is essential for solving definite integrals.
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Trigonometric Functions in Integration
Integrating functions involving trigonometric terms often requires using identities or special techniques. Recognizing how to handle integrals with cosine or sine, especially when multiplied by polynomials, helps in simplifying and solving the integral effectively.
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Related Practice
Textbook Question
In Exercises 39–48, use an appropriate substitution and then a trigonometric substitution to evaluate the integrals.
∫ √(x - 2) / √(x - 1) dx
Textbook Question
Find the value of the constant c so that the given function is a probability density function for a random variable X over the specified interval.
f(x) = (1/x) over [c, c + 1]
Textbook Question
Evaluate the integrals in Exercises 1–22.
∫ cos³(4x) dx
Textbook Question
Evaluate the integrals in Exercises 33–52.
∫ cot⁶(2x) dx
Textbook Question
In Exercises 67–73, use integration by parts to establish the reduction formula.
∫ x^n sin(x) dx = -x^n cos(x) + n ∫ x^(n-1) cos(x) dx