Skip to main content
Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.7.49

Evaluate the integrals in Exercises 41–60.
49. ∫(sech(√t)tanh(√t)dt)/√t

Guida verificata passo dopo passo
1
Start by examining the integral: \(\int \frac{\text{sech}(\sqrt{t}) \tanh(\sqrt{t})}{\sqrt{t}} \, dt\). Notice that the integrand involves \(\sqrt{t}\) inside the hyperbolic functions and also in the denominator.
Use the substitution method to simplify the integral. Let \(u = \sqrt{t}\), which means \(u = t^{1/2}\). Then, differentiate both sides with respect to \(t\) to find \(du\) in terms of \(dt\): \(du = \frac{1}{2\sqrt{t}} dt\) or equivalently, \(dt = 2u \, du\).
Rewrite the integral in terms of \(u\). Substitute \(\sqrt{t} = u\), \(dt = 2u \, du\), and replace the integrand accordingly: the denominator \(\sqrt{t}\) becomes \(u\), and the numerator remains \(\text{sech}(u) \tanh(u)\). So the integral becomes \(\int \frac{\text{sech}(u) \tanh(u)}{u} \cdot 2u \, du\).
Simplify the expression inside the integral. The \(u\) in the denominator and numerator cancel out, leaving \(\int 2 \text{sech}(u) \tanh(u) \, du\).
Recognize that the derivative of \(\text{sech}(u)\) is \(-\text{sech}(u) \tanh(u)\). Use this fact to rewrite the integral in terms of \(\text{sech}(u)\) and then integrate accordingly.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
1m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Hyperbolic Functions

Hyperbolic functions like sech(x) and tanh(x) are analogs of trigonometric functions but based on hyperbolas. Understanding their definitions, properties, and derivatives is essential for manipulating and integrating expressions involving these functions.
Video consigliato:
Percorso guidato
5:50
Asymptotes of Hyperbolas

Substitution Method in Integration

The substitution method simplifies integrals by changing variables to transform the integral into a more manageable form. Recognizing an inner function and its derivative within the integrand allows for effective substitution, especially when dealing with composite functions.
Video consigliato:
07:33
Euler's Method

Chain Rule and Its Role in Integration

The chain rule relates the derivative of a composite function to the derivatives of its inner and outer functions. In integration, recognizing the chain rule in reverse helps identify suitable substitutions and simplifies the integral of composite functions.
Video consigliato:
05:02
Intro to the Chain Rule