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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 8.2.52

Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.
∫ x² tan⁻¹(x / 2) dx

Guida verificata passo dopo passo
1
Identify the integral to solve: \(\int x^{2} \tan^{-1}\left(\frac{x}{2}\right) \, dx\).
Recognize that this integral involves a product of functions: a polynomial \(x^{2}\) and an inverse tangent function \(\tan^{-1}\left(\frac{x}{2}\right)\). This suggests using integration by parts.
Set up integration by parts using the formula: \(\int u \, dv = uv - \int v \, du\). Choose \(u = \tan^{-1}\left(\frac{x}{2}\right)\) because its derivative simplifies nicely, and \(dv = x^{2} \, dx\).
Compute \(du\) by differentiating \(u\): use the chain rule on \(\tan^{-1}\left(\frac{x}{2}\right)\), and compute \(v\) by integrating \(dv = x^{2} \, dx\).
Substitute \(u\), \(du\), \(v\), and \(dv\) into the integration by parts formula, then simplify and evaluate the resulting integral to complete the solution.

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Integration by Parts

Integration by parts is a technique based on the product rule for 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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Inverse Trigonometric Functions

Inverse trigonometric functions, like arctangent (tan⁻¹), are the inverses of trigonometric functions and often appear in integrals. Understanding their derivatives and properties helps in differentiating or integrating expressions involving these functions.
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Polynomial Functions in Integration

Polynomial functions, such as x², are straightforward to integrate and often serve as parts of integrands. Recognizing how to handle polynomials within more complex integrals, especially when combined with other functions, is essential for applying integration techniques effectively.
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