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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
Capitolo 8, Problema 8.PE.44

Evaluate the integrals in Exercises 37–44.
∫ eᵗ √[tan²(eᵗ) + 1] dt

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Recognize that the integral is of the form \(\int e^{t} \sqrt{\tan^{2}(e^{t}) + 1} \, dt\). Notice that inside the square root, we have \(\tan^{2}(e^{t}) + 1\), which can be simplified using a trigonometric identity.
Recall the Pythagorean identity: \(1 + \tan^{2}(x) = \sec^{2}(x)\). Applying this, rewrite the integrand as \(e^{t} \sqrt{\sec^{2}(e^{t})}\).
Since \(\sqrt{\sec^{2}(e^{t})} = |\sec(e^{t})|\), and assuming the domain where \(\sec(e^{t})\) is positive, the integrand simplifies to \(e^{t} \sec(e^{t})\).
Use substitution to simplify the integral: let \(u = e^{t}\). Then, $du = e^{t} dt$, which means $du = e^{t} dt$ or equivalently $e^{t} dt = du$. Substitute these into the integral to get \(\int \sec(u) \, du\).
Now, the integral reduces to \(\int \sec(u) \, du\), which is a standard integral. You can proceed by recalling the formula for \(\int \sec(u) \, du\) and then substitute back \(u = e^{t}\) to express the answer in terms of \(t\).

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