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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 8.4.65

60–69. Completing the square Evaluate the following integrals.
65. ∫[1/2 to (√2 + 3)/(2√2)] dx / (8x² - 8x + 11)

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Start by examining the quadratic expression in the denominator: \(8x^{2} - 8x + 11\). To simplify the integral, we want to complete the square for this quadratic.
Factor out the coefficient of \(x^{2}\) from the first two terms: \(8(x^{2} - x) + 11\).
Complete the square inside the parentheses: take half of the coefficient of \(x\), which is \(-1\), divide by 2 to get \(-\frac{1}{2}\), then square it to get \(\left(-\frac{1}{2}\right)^{2} = \frac{1}{4}\). Add and subtract this inside the parentheses:
\(8\left(x^{2} - x + \frac{1}{4} - \frac{1}{4}\right) + 11 = 8\left(\left(x - \frac{1}{2}\right)^{2} - \frac{1}{4}\right) + 11\).
Distribute the 8 and simplify the constant terms: \(8\left(x - \frac{1}{2}\right)^{2} - 8 \times \frac{1}{4} + 11 = 8\left(x - \frac{1}{2}\right)^{2} - 2 + 11 = 8\left(x - \frac{1}{2}\right)^{2} + 9\).
Rewrite the integral using this completed square form: \(\int_{\frac{1}{2}}^{\frac{\sqrt{2} + 3}{2\sqrt{2}}} \frac{dx}{8\left(x - \frac{1}{2}\right)^{2} + 9}\). Next, factor out the 9 to express the denominator in a form suitable for an arctangent substitution.

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Completing the Square

Completing the square is a technique used to rewrite quadratic expressions in the form ax² + bx + c as a perfect square plus or minus a constant. This simplifies integration by transforming the denominator into a form that matches standard integral formulas, especially those involving inverse trigonometric functions.
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