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Ch. 5 - Integrals
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 5, Problema 5.PE.45

Evaluate the integrals in Exercises 37–46.


∫(sin 2θ - cos 2θ)/(sin 2θ + cos 2θ)³dθ

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Start by examining the integral \( \int \frac{\sin 2\theta - \cos 2\theta}{(\sin 2\theta + \cos 2\theta)^3} \, d\theta \). Notice that the numerator and denominator involve \( \sin 2\theta + \cos 2\theta \) and its derivative might be related to the numerator.
Set \( u = \sin 2\theta + \cos 2\theta \). Then, compute \( \frac{du}{d\theta} \) to find the differential \( du \). Since \( \frac{d}{d\theta}(\sin 2\theta) = 2\cos 2\theta \) and \( \frac{d}{d\theta}(\cos 2\theta) = -2\sin 2\theta \), we get \( \frac{du}{d\theta} = 2\cos 2\theta - 2\sin 2\theta = 2(\cos 2\theta - \sin 2\theta) \).
Rewrite the numerator \( \sin 2\theta - \cos 2\theta \) in terms of \( \cos 2\theta - \sin 2\theta \) to relate it to \( du/d\theta \). Notice that \( \sin 2\theta - \cos 2\theta = - (\cos 2\theta - \sin 2\theta) \).
Express the integral in terms of \( u \) and \( du \) by substituting the numerator and denominator accordingly. The integral becomes \( \int \frac{- (\cos 2\theta - \sin 2\theta)}{u^3} \, d\theta \). Using the expression for \( du \), solve for \( d\theta \) and substitute into the integral.
Simplify the integral to a function of \( u \) and \( du \), which should be easier to integrate. After integration, substitute back \( u = \sin 2\theta + \cos 2\theta \) to express the answer in terms of \( \theta \).

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Trigonometric identities are equations involving trigonometric functions that hold true for all values within their domains. They help simplify expressions, such as rewriting sin 2θ and cos 2θ in terms of other functions or combining terms. Recognizing and applying these identities is essential to simplify the integral's numerator and denominator.
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