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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 8.2.48

Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.
∫₀^π/2 x³ cos 2x dx

Guida verificata passo dopo passo
1
Identify the integral to solve: \(\int_0^{\frac{\pi}{2}} x^3 \cos(2x) \, dx\).
Recognize that this integral involves a product of a polynomial (\(x^3\)) and a trigonometric function (\(\cos(2x)\)), which suggests using integration by parts.
Set up integration by parts by choosing \(u = x^3\) (which simplifies when differentiated) and \(dv = \cos(2x) \, dx\) (which can be integrated easily). Recall the formula: \(\int u \, dv = uv - \int v \, du\).
Compute \(du = 3x^2 \, dx\) and find \(v\) by integrating \(dv\): \(v = \int \cos(2x) \, dx = \frac{1}{2} \sin(2x)\).
Apply the integration by parts formula: \(\int_0^{\frac{\pi}{2}} x^3 \cos(2x) \, dx = \left. x^3 \cdot \frac{1}{2} \sin(2x) \right|_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} \frac{1}{2} \sin(2x) \cdot 3x^2 \, dx\). This reduces the original integral to a new integral involving \(x^2 \sin(2x)\), which may require repeating integration by parts.

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