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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 7.7.71a

Evaluate the integrals in Exercises 67–74 in terms of
a. inverse hyperbolic functions.
71. ∫(from 1/5 to 3/13)dx/(x√(1-16x²))

Guida verificata passo dopo passo
1
Identify the integral to be evaluated: \(\int_{\frac{1}{5}}^{\frac{3}{13}} \frac{dx}{x \sqrt{1 - 16x^2}}\).
Recognize that the integrand has the form \(\frac{1}{x \sqrt{1 - a^2 x^2}}\) with \(a = 4\), which suggests using a substitution or a known formula involving inverse hyperbolic functions.
Recall the inverse hyperbolic function identity: \(\operatorname{arcosh}(z) = \ln(z + \sqrt{z^2 - 1})\) and the derivative of \(\operatorname{arcosh}(x)\) is \(\frac{1}{\sqrt{x^2 - 1}}\), but here the integrand involves \(\sqrt{1 - 16x^2}\), so consider the substitution \(x = \frac{1}{4} \sin \theta\) or relate it to \(\operatorname{arcosh}\) or \(\operatorname{arcsinh}\) forms.
Use the substitution \(x = \frac{1}{4} \sin \theta\) to rewrite the integral in terms of \(\theta\), then simplify the integral to a form involving \(\csc \theta\) or \(\cot \theta\), which can be integrated to inverse hyperbolic functions.
After integrating with respect to \(\theta\), back-substitute to express the answer in terms of \(x\), and then evaluate the definite integral by plugging in the limits \(x = \frac{1}{5}\) and \(x = \frac{3}{13}\).

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