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

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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 \(\int \frac{x}{\sqrt{1 - x^2}} \, dx\).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.

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.

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.

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