Skip to main content
Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.2.20

9–40. Integration by parts Evaluate the following integrals using integration by parts.
20. ∫ sin⁻¹(x) dx

Guida verificata passo dopo passo
1
Identify the integral to solve: \(\int \sin^{-1}(x) \, dx\).
Recall the integration by parts formula: \(\int u \, dv = uv - \int v \, du\).
Choose \(u\) and \(dv\) wisely. Let \(u = \sin^{-1}(x)\) because its derivative simplifies, and let $dv = dx$ because it is easy to integrate.
Compute \(du\) and \(v\): - \(du = \frac{1}{\sqrt{1 - x^2}} \, dx\) (derivative of \(\sin^{-1}(x)\)), - \(v = x\) (integral of \(dx\)).
Apply the integration by parts formula: \(\int \sin^{-1}(x) \, dx = x \sin^{-1}(x) - \int x \cdot \frac{1}{\sqrt{1 - x^2}} \, dx\). Next, focus on evaluating the remaining integral \(\int \frac{x}{\sqrt{1 - x^2}} \, dx\).

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:
8m

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

Inverse Trigonometric Functions

Inverse trigonometric functions, like sin⁻¹(x), are the inverses of trigonometric functions and have specific derivatives. For example, the derivative of sin⁻¹(x) is 1/√(1 - x²). Understanding these derivatives helps in setting up the parts for integration.
Video consigliato:
06:35
Derivatives of Other Inverse Trigonometric Functions

Differentiation and Integration of Algebraic and Transcendental Functions

Integrating functions involving inverse trigonometric terms often requires combining algebraic manipulation with knowledge of derivatives and integrals of transcendental functions. Recognizing how to differentiate and integrate these functions is essential for applying integration by parts successfully.
Video consigliato:
Percorso guidato
04:22
Integrals Resulting in Basic Trig Functions Example 1