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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 8.4.42

In Exercises 39–48, use an appropriate substitution and then a trigonometric substitution to evaluate the integrals.
∫ dy / (y√(1 + (ln y)²)) from 1 to e

Guida verificata passo dopo passo
1
Identify the integral: \( \int_{1}^{e} \frac{dy}{y \sqrt{1 + (\ln y)^2}} \). Notice the presence of \( \ln y \) inside the square root, which suggests a substitution involving \( \ln y \).
Make the substitution \( u = \ln y \). Then, compute \( du = \frac{1}{y} dy \), which implies \( dy = y du \). Substitute into the integral to rewrite it in terms of \( u \).
After substitution, the integral becomes \( \int_{u=0}^{u=1} \frac{y du}{y \sqrt{1 + u^2}} = \int_{0}^{1} \frac{du}{\sqrt{1 + u^2}} \). The limits change because when \( y=1 \), \( u=\ln 1=0 \), and when \( y=e \), \( u=\ln e=1 \).
Recognize that the integral \( \int \frac{du}{\sqrt{1 + u^2}} \) is a standard form that can be evaluated using a trigonometric substitution. Use the substitution \( u = \tan \theta \), which implies \( du = \sec^2 \theta d\theta \) and \( \sqrt{1 + u^2} = \sec \theta \).
Rewrite the integral in terms of \( \theta \), simplify, and then integrate with respect to \( \theta \). Finally, convert back to the variable \( u \) and then to \( y \) to express the answer in the original variable.

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