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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
Tutti i libri di testoHass 15th EditionCh. 8 - Techniques of IntegrationProblema 8.2.78
Capitolo 8, Problema 8.2.78

Use the formula ∫ f⁻¹(x) dx = x f⁻¹(x) - ∫ f(y) dy, y = f⁻¹(x)
To evaluate the integrals in Exercises 77-80. Express your answers in terms of x.
∫ arctan x dx

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1
Identify the function and its inverse: Here, the function inside the integral is \( f^{-1}(x) = \arctan x \). This means \( f(y) = \tan y \) because \( y = \arctan x \) implies \( x = \tan y \).
Recall the formula given: \( \int f^{-1}(x) \, dx = x f^{-1}(x) - \int f(y) \, dy \), where \( y = f^{-1}(x) \). Substitute \( f^{-1}(x) = \arctan x \) and \( y = \arctan x \) into the formula.
Write the integral in terms of \( y \): \( \int \arctan x \, dx = x \arctan x - \int \tan y \, dy \). Now focus on evaluating \( \int \tan y \, dy \).
Recall the integral of \( \tan y \): \( \int \tan y \, dy = -\ln |\cos y| + C \). Use this to express the integral in terms of \( y \).
Finally, substitute back \( y = \arctan x \) into the expression to write the answer entirely in terms of \( x \). This completes the evaluation of the integral \( \int \arctan x \, dx \).

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Inverse Functions and Their Properties

An inverse function reverses the effect of the original function, such that f(f⁻¹(x)) = x. Understanding how to identify and work with inverse functions is essential, especially when applying integration formulas involving inverses, like the one given for ∫ f⁻¹(x) dx.
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Integration by Parts

Integration by parts is a technique based on the product rule for differentiation, used to integrate products of functions. It is often applied when integrating inverse functions, such as arctan x, by expressing the integral in a form that simplifies evaluation.
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Integral Formula for Inverse Functions

The formula ∫ f⁻¹(x) dx = x f⁻¹(x) - ∫ f(y) dy, where y = f⁻¹(x), provides a method to integrate inverse functions by relating the integral of the inverse to an integral involving the original function. Applying this formula requires understanding substitution and the relationship between f and f⁻¹.
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