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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.R.57

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.
57. ∫ (from 0 to √3/2) 4/(9 + 4x²) dx

Guida verificata passo dopo passo
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Step 1: Recognize the integral's form. The integrand resembles the formula for the derivative of the arctangent function: \( \frac{1}{a^2 + x^2} \). Specifically, \( \int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C \). Here, the denominator is \( 9 + 4x^2 \), which can be rewritten as \( (3)^2 + (2x)^2 \).
Step 2: Factor out constants to simplify the integral. Rewrite the integrand as \( \frac{4}{9 + 4x^2} = \frac{4}{(3)^2 + (2x)^2} \). Factor out \( 4 \) from the numerator and denominator: \( \int \frac{4}{(3)^2 + (2x)^2} dx = 4 \int \frac{1}{(3)^2 + (2x)^2} dx \).
Step 3: Perform a substitution to simplify further. Let \( u = 2x \), so \( du = 2 dx \). This substitution transforms the integral into \( 4 \int \frac{1}{(3)^2 + u^2} \cdot \frac{du}{2} \), which simplifies to \( 2 \int \frac{1}{(3)^2 + u^2} du \).
Step 4: Apply the arctangent formula. Using \( \int \frac{1}{a^2 + u^2} du = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C \), where \( a = 3 \), the integral becomes \( 2 \cdot \frac{1}{3} \arctan\left(\frac{u}{3}\right) + C \), or \( \frac{2}{3} \arctan\left(\frac{u}{3}\right) + C \).
Step 5: Substitute back \( u = 2x \) and evaluate the definite integral. Replace \( u \) with \( 2x \) to get \( \frac{2}{3} \arctan\left(\frac{2x}{3}\right) \). Evaluate this expression at the bounds \( x = 0 \) and \( x = \frac{\sqrt{3}}{2} \) to find the final result.

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Definite Integrals

A definite integral represents the signed area under a curve between two specified limits. It is denoted as ∫ from a to b f(x) dx, where 'a' and 'b' are the lower and upper limits, respectively. The result of a definite integral is a numerical value that quantifies the accumulation of the function's values over the interval [a, b].
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Integration techniques are methods used to evaluate integrals that may not be solvable by basic antiderivatives. Common techniques include substitution, integration by parts, and trigonometric substitution. Understanding these methods is crucial for simplifying complex integrals into forms that can be easily evaluated.
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Trigonometric substitution is a technique used to simplify integrals involving square roots or quadratic expressions by substituting a variable with a trigonometric function. For example, in integrals involving the form √(a² - x²), one might use x = a sin(θ). This method transforms the integral into a trigonometric form that is often easier to evaluate.
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