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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 8.3.30

Evaluate the integrals in Exercises 23–32.
∫_{π/2}^{3π/4} √(1 - sin(2x)) dx

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Start by examining the integrand: \(\sqrt{1 - \sin(2x)}\). Recognize that the expression inside the square root can be simplified using trigonometric identities.
Recall the double-angle identity for sine: \(\sin(2x) = 2\sin x \cos x\). However, a more useful identity here is the one involving cosine of a double angle: \(1 - \sin(2x) = 1 - 2\sin x \cos x\).
Use the identity \(1 - \sin(2x) = (\cos x - \sin x)^2\) because \((\cos x - \sin x)^2 = \cos^2 x - 2\sin x \cos x + \sin^2 x = 1 - 2\sin x \cos x = 1 - \sin(2x)\).
Rewrite the integral as \(\int_{\pi/2}^{3\pi/4} \sqrt{(\cos x - \sin x)^2} \, dx = \int_{\pi/2}^{3\pi/4} |\cos x - \sin x| \, dx\). Consider the absolute value carefully over the interval to determine the sign of \(\cos x - \sin x\).
Split the integral if necessary based on where \(\cos x - \sin x\) changes sign, then integrate \(\cos x - \sin x\) or its negative accordingly. Use the antiderivatives \(\int \cos x \, dx = \sin x\) and \(\int \sin x \, dx = -\cos x\) to evaluate the integral.

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