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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.PE.43

Evaluate the integrals in Exercises 31–78.
43. ∫tan(ln v)/v dv

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Recognize that the integral is of the form \(\int \frac{\tan(\ln v)}{v} \, dv\). Notice the presence of \(\ln v\) inside the tangent function and the \(\frac{1}{v}\) factor outside, which suggests a substitution involving \(\ln v\).
Let \(u = \ln v\). Then, compute the differential \(du\): since \(u = \ln v\), we have \(du = \frac{1}{v} dv\). This means \(\frac{1}{v} dv = du\).
Rewrite the integral in terms of \(u\): substituting \(u\) and \(du\) gives \(\int \tan(u) \, du\).
Recall the integral formula for \(\int \tan(u) \, du\). The integral of \(\tan(u)\) is \(-\ln|\cos(u)| + C\). Use this to express the integral in terms of \(u\).
Finally, substitute back \(u = \ln v\) to write the answer in terms of the original variable \(v\). The result will be \(-\ln|\cos(\ln v)| + C\).

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Integration by Substitution

Integration by substitution is a method used to simplify integrals by changing variables. It involves identifying a part of the integrand as a new variable, which transforms the integral into a simpler form. This technique is especially useful when the integral contains a composite function, such as tan(ln v).
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Logarithmic functions, like ln(v), have specific properties that help in integration, such as their derivatives and the chain rule application. Understanding how ln(v) behaves and how its derivative 1/v appears in the integral is crucial for choosing the correct substitution.
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