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

Verified step by step guidance

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 \( \ln y \).
2. 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 \( u \).
3. 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 \( y=e \), \( u=\ln e=1 \).
4. 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 \( \sqrt{1 + u^2} = \sec \theta \).
5. 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.

Key Concepts

Substitution Method in Integration

The substitution method simplifies integrals by changing variables to transform the integral into a more manageable form. It involves choosing a substitution u = g(y) such that the integral's expression becomes easier to integrate. This technique is especially useful when the integral contains composite functions.

Trigonometric Substitution

Trigonometric substitution replaces algebraic expressions involving square roots with trigonometric functions to simplify integration. It is commonly used when the integrand contains expressions like √(a² + x²), √(a² - x²), or √(x² - a²). This substitution leverages trigonometric identities to rewrite the integral in terms of trigonometric functions.

Integration of Functions Involving Logarithms

Integrals involving logarithmic functions often require careful substitution to handle the ln(y) term. Recognizing how to express the integral in terms of ln(y) and its derivative helps in simplifying the integral. Understanding the derivative of ln(y), which is 1/y, is crucial for effective substitution.