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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
Tutti i libri di testoHass 15th EditionCh. 8 - Techniques of IntegrationProblema 8.3.28
Capitolo 8, Problema 8.3.28

Evaluate the integrals in Exercises 23–32.
∫₀^(π/6) √(1 + sin(x)) dx
(Hint: Multiply by √((1 - sin(x)) / (1 - sin(x))))

Guida verificata passo dopo passo
1
Start with the integral: \(\int_0^{\frac{\pi}{6}} \sqrt{1 + \sin(x)} \, dx\).
Use the hint to multiply the integrand by \(\sqrt{\frac{1 - \sin(x)}{1 - \sin(x)}}\) to simplify the expression under the square root:
\[\sqrt{1 + \sin(x)} = \sqrt{\frac{(1 + \sin(x))(1 - \sin(x))}{1 - \sin(x)}} = \frac{\sqrt{1 - \sin^2(x)}}{\sqrt{1 - \sin(x)}}.\]
Recall the Pythagorean identity \(\sin^2(x) + \cos^2(x) = 1\), so \(\sqrt{1 - \sin^2(x)} = |\cos(x)|\). Since \(x\) is in \([0, \frac{\pi}{6}]\), \(\cos(x)\) is positive, so \(\sqrt{1 - \sin^2(x)} = \cos(x)\).
Rewrite the integral as \(\int_0^{\frac{\pi}{6}} \frac{\cos(x)}{\sqrt{1 - \sin(x)}} \, dx\). Then, use the substitution \(u = 1 - \sin(x)\), find \(du\), change the limits accordingly, and express the integral in terms of \(u\) to proceed with integration.

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Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values within their domains. In this problem, identities like 1 - sin²(x) = cos²(x) help simplify expressions under the square root, making the integral more manageable.
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Verifying Trig Equations as Identities

Techniques of Integration - Rationalizing the Integrand

Rationalizing the integrand involves multiplying by a conjugate expression to simplify the integrand, especially when it contains square roots. Here, multiplying by √((1 - sin(x)) / (1 - sin(x))) transforms the integrand into a form easier to integrate.
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Completing the Square to Rewrite the Integrand

Definite Integrals and Limits of Integration

Definite integrals compute the net area under a curve between specified limits. Understanding how to evaluate the integral from 0 to π/6 requires applying integration techniques and then substituting the upper and lower bounds to find the exact value.
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Definition of the Definite Integral
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