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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 8 - Techniques of IntegrationProblem 8.3.28
Chapter 8, Problem 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))))

Verified step by step guidance
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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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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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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