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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
Capitolo 7, Problema 8.6.22

Use the table of integrals at the back of the text to evaluate the integrals in Exercises 1–26.
∫ sin(2x) cos(3x) dx

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1
Recognize that the integral involves the product of sine and cosine functions with different arguments: \(\int \sin(2x) \cos(3x) \, dx\).
Use the product-to-sum trigonometric identity to rewrite the product of sine and cosine as a sum of sines: \(\sin(A) \cos(B) = \frac{1}{2} [\sin(A+B) + \sin(A-B)]\).
Apply the identity with \(A = 2x\) and \(B = 3x\) to get: \(\sin(2x) \cos(3x) = \frac{1}{2} [\sin(5x) + \sin(-x)]\).
Simplify \(\sin(-x)\) using the odd property of sine: \(\sin(-x) = -\sin(x)\), so the expression becomes \(\frac{1}{2} [\sin(5x) - \sin(x)]\).
Rewrite the integral as \(\int \sin(2x) \cos(3x) \, dx = \frac{1}{2} \int [\sin(5x) - \sin(x)] \, dx\), then integrate each sine term separately using the integral formula \(\int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C\).

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