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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 8.3.32

Evaluate the integrals in Exercises 23–32.
∫₋π^π (1 - cos²(t))^(3/2) dt

Guida verificata passo dopo passo
1
Recognize that the integrand is \( (1 - \cos^{2}(t))^{3/2} \). Recall the Pythagorean identity \( \sin^{2}(t) + \cos^{2}(t) = 1 \), which allows us to rewrite the integrand in terms of \( \sin(t) \).
Rewrite the integrand as \( (\sin^{2}(t))^{3/2} = |\sin(t)|^{3} \) because \( (\sin^{2}(t))^{3/2} = (\sin^{2}(t))^{1 \cdot 3/2} = |\sin(t)|^{3} \).
Since the integral is from \( -\pi \) to \( \pi \), and \( |\sin(t)|^{3} \) is an even function (because \( |\sin(-t)| = |\sin(t)| \)), use the property of even functions to simplify the integral: \( \int_{-\pi}^{\pi} |\sin(t)|^{3} dt = 2 \int_{0}^{\pi} |\sin(t)|^{3} dt \).
On the interval \( [0, \pi] \), \( \sin(t) \) is non-negative, so \( |\sin(t)|^{3} = \sin^{3}(t) \). Thus, the integral becomes \( 2 \int_{0}^{\pi} \sin^{3}(t) dt \).
To evaluate \( \int \sin^{3}(t) dt \), use the reduction formula or rewrite \( \sin^{3}(t) \) as \( \sin(t) \cdot \sin^{2}(t) = \sin(t)(1 - \cos^{2}(t)) \), then use substitution \( u = \cos(t) \) to integrate.

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Trigonometric identities are equations involving trigonometric functions that hold true for all values within their domains. In this problem, recognizing that 1 - cos²(t) equals sin²(t) simplifies the integrand, making the integral easier to evaluate.
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