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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
Tutti i libri di testoHass 15th EditionCh. 7 - Transcendental FunctionsProblema 7.6.71
Capitolo 7, Problema 7.6.71

Evaluate the integrals in Exercises 53–76.
71. ∫(from -π/2 to π/2) 2cosθ dθ/(1+(sinθ)²)

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Identify the integral to evaluate: \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2 \cos \theta}{1 + (\sin \theta)^2} \, d\theta\).
Consider the symmetry of the integrand. Note that \(\cos \theta\) is an even function and \((\sin \theta)^2\) is also even, so the entire integrand is an even function. This allows us to rewrite the integral as \(2 \int_0^{\frac{\pi}{2}} \frac{2 \cos \theta}{1 + (\sin \theta)^2} \, d\theta\).
Use the substitution \(u = \sin \theta\), which implies \(du = \cos \theta \, d\theta\). This substitution will simplify the integral because the denominator is in terms of \(\sin \theta\) and the numerator contains \(\cos \theta \, d\theta\).
Rewrite the integral in terms of \(u\): the limits change from \(\theta = 0\) to \(\theta = \frac{\pi}{2}\), which correspond to \(u = 0\) to \(u = 1\). The integral becomes \(2 \int_0^1 \frac{2}{1 + u^2} \, du\).
Recognize that \(\int \frac{1}{1 + u^2} \, du\) is the standard arctangent integral. Set up the integral accordingly and prepare to evaluate it using the antiderivative \(\arctan u\).

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Definite Integrals

A definite integral calculates the net area under a curve between two specified limits. It involves evaluating the integral function at the upper and lower bounds and subtracting these values. This concept is essential for solving integrals with given interval limits, such as from -π/2 to π/2.
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Trigonometric Functions and Identities

Understanding sine and cosine functions, their properties, and identities is crucial for simplifying integrals involving trigonometric expressions. For example, recognizing that (sin θ)² can be rewritten using identities helps in simplifying the denominator or integrand for easier integration.
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Symmetry in Integrals

Symmetry properties of functions over symmetric intervals can simplify integration. Even functions satisfy f(-x) = f(x), and odd functions satisfy f(-x) = -f(x). Identifying whether the integrand is even or odd over [-π/2, π/2] can reduce the integral calculation or even determine if the integral is zero.
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