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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.4.48

In Exercises 39–48, use an appropriate substitution and then a trigonometric substitution to evaluate the integrals.
∫ √(x - 2) / √(x - 1) dx

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Start by examining the integral \(\int \frac{\sqrt{x - 2}}{\sqrt{x - 1}} \, dx\). Notice the expressions under the square roots are similar, differing by 1. This suggests a substitution to simplify the radicals.
Make the substitution \(u = \sqrt{x - 1}\). Then, express \(x\) in terms of \(u\): \(x = u^2 + 1\). Also, find \(dx\) in terms of \(du\): differentiate both sides to get \(dx = 2u \, du\).
Rewrite the integral in terms of \(u\). The numerator becomes \(\sqrt{x - 2} = \sqrt{u^2 + 1 - 2} = \sqrt{u^2 - 1}\), and the denominator is \(\sqrt{x - 1} = u\). Substitute these and \(dx\) into the integral to get \(\int \frac{\sqrt{u^2 - 1}}{u} \cdot 2u \, du\).
Simplify the integral expression: the \(u\) in the denominator and numerator cancel, leaving \(\int 2 \sqrt{u^2 - 1} \, du\). Now, to evaluate this integral, use a trigonometric substitution suitable for \(\sqrt{u^2 - 1}\), such as \(u = \sec \theta\).
With \(u = \sec \theta\), express \(\sqrt{u^2 - 1}\) as \(\sqrt{\sec^2 \theta - 1} = \tan \theta\), and find \(du = \sec \theta \tan \theta \, d\theta\). Substitute these into the integral and simplify to an integral in terms of \(\theta\) that can be evaluated using standard trigonometric integrals.

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