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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
7장, 문제 7.2.49

Evaluate the integrals in Exercises 39–56.
49. ∫3sec²t/(6 + 3tan(t)) dt

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Identify the integral to solve: \(\int \frac{3 \sec^{2} t}{6 + 3 \tan t} \, dt\).
Notice that the denominator contains \(6 + 3 \tan t\), and the numerator contains \(3 \sec^{2} t\). Recall that the derivative of \(\tan t\) is \(\sec^{2} t\), which suggests a substitution.
Let \(u = 6 + 3 \tan t\). Then, compute \(du\): since \(\frac{d}{dt}(\tan t) = \sec^{2} t\), we have \(du = 3 \sec^{2} t \, dt\).
Rewrite the integral in terms of \(u\) and \(du\): the numerator \(3 \sec^{2} t \, dt\) becomes \(du\), and the denominator is \(u\), so the integral becomes \(\int \frac{1}{u} \, du\).
Integrate \(\int \frac{1}{u} \, du\) to get \(\ln|u| + C\), then substitute back \(u = 6 + 3 \tan t\) to express the answer in terms of \(t\).

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주요 개념

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