9–40. Integration by parts Evaluate the following integrals using integration by parts.
23. ∫ x² sin(2x) dx

Guida verificata passo dopo passo

Identify the parts of the integral for integration by parts. Let \(u = x^{2}\) and \(dv = \sin(2x) \, dx\).

Compute the derivatives and integrals needed: find \(du = \frac{d}{dx}(x^{2}) \, dx = 2x \, dx\) and find \(v = \int \sin(2x) \, dx\).

Recall that \(\int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + C\), so here \(v = -\frac{1}{2} \cos(2x)\).

Apply the integration by parts formula: \(\int u \, dv = uv - \int v \, du\). Substitute the expressions for \(u\), \(v\), and \(du\).

Simplify the resulting integral \(\int v \, du\) and, if necessary, apply integration by parts again to evaluate it completely.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.

Durata del video:

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It transforms the integral of a product of functions into simpler integrals using the formula ∫u dv = uv - ∫v du. Choosing u and dv wisely is crucial to simplify the integral effectively.

Video consigliato:

Percorso guidato

06:18

Integration by Parts for Definite Integrals

Choosing u and dv

Selecting which part of the integrand to assign as u and which as dv affects the ease of solving the integral. Typically, u is chosen as a function that simplifies when differentiated (like polynomials), and dv is chosen as a function that is easy to integrate (like trigonometric functions).

Video consigliato:

Percorso guidato

07:51

Choosing a Convergence Test

Repeated Application of Integration by Parts

Some integrals, such as ∫x² sin(2x) dx, require applying integration by parts more than once. After the first application, the resulting integral may still be complex, necessitating a second iteration to fully evaluate the integral.

Video consigliato:

Percorso guidato