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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 8, Problem 8.2.23

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

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1
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.

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Key Concepts

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

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.
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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).
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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.
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Repeated Integration by Parts