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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 7.6.75

Evaluate the integrals in Exercises 53–76.
75. ∫y dy/√(1-y^4)

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Identify the integral to solve: \(\int \frac{y \, dy}{\sqrt{1 - y^{4}}}\).
Consider a substitution to simplify the expression under the square root. Notice that the denominator involves \(1 - y^{4}\), which can be rewritten as \(1 - (y^{2})^{2}\), suggesting a substitution involving \(y^{2}\).
Let \(u = y^{2}\). Then, compute \(du\) in terms of \(dy\): since \(u = y^{2}\), we have \(du = 2y \, dy\), or equivalently, \(y \, dy = \frac{du}{2}\).
Rewrite the integral in terms of \(u\): substitute \(y \, dy\) with \(\frac{du}{2}\) and \(\sqrt{1 - y^{4}}\) with \(\sqrt{1 - u^{2}}\), so the integral becomes \(\int \frac{\frac{du}{2}}{\sqrt{1 - u^{2}}} = \frac{1}{2} \int \frac{du}{\sqrt{1 - u^{2}}}\).
Recognize that \(\int \frac{du}{\sqrt{1 - u^{2}}}\) is a standard integral whose antiderivative is \(\arcsin(u) + C\). After integrating, substitute back \(u = y^{2}\) to express the answer in terms of \(y\).

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