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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 8.4.44

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

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Identify the integral: \(\int \frac{\sqrt{1 - (\ln x)^2}}{x \ln x} \, dx\).
Use the substitution \(u = \ln x\). Then, compute \(du = \frac{1}{x} dx\), which implies \(dx = x \, du\).
Rewrite the integral in terms of \(u\): substitute \(\ln x\) with \(u\) and \(dx\) with \(x \, du\). The integral becomes \(\int \frac{\sqrt{1 - u^2}}{x u} \cdot x \, du = \int \frac{\sqrt{1 - u^2}}{u} \, du\).
Now, to handle the integral \(\int \frac{\sqrt{1 - u^2}}{u} \, du\), use a trigonometric substitution. Since the integrand contains \(\sqrt{1 - u^2}\), let \(u = \sin \theta\), which implies \(du = \cos \theta \, d\theta\).
Rewrite the integral in terms of \(\theta\): substitute \(u = \sin \theta\) and \(du = \cos \theta \, d\theta\). The integral becomes \(\int \frac{\sqrt{1 - \sin^2 \theta}}{\sin \theta} \cdot \cos \theta \, d\theta\). Simplify the square root using the Pythagorean identity and proceed to integrate with respect to \(\theta\).

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