Evaluate the integrals in Exercises 1–24 using integration by parts.
∫ arcsin(y) dy

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Identify the integral to solve: \(\int \arcsin(y) \, dy\).

Recall the integration by parts formula: \(\int u \, dv = uv - \int v \, du\).

Choose \(u = \arcsin(y)\) because its derivative simplifies, and \(dv = dy\) because it is easy to integrate.

Compute \(du\) and \(v\): - \(du = \frac{1}{\sqrt{1 - y^2}} \, dy\) (derivative of \(\arcsin(y)\)), - \(v = y\) (integral of \(dy\)).

Apply the integration by parts formula: \(\int \arcsin(y) \, dy = y \arcsin(y) - \int y \cdot \frac{1}{\sqrt{1 - y^2}} \, dy\). Next, focus on evaluating the remaining integral \(\int \frac{y}{\sqrt{1 - y^2}} \, dy\).

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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.

Derivative of Inverse Trigonometric Functions

Understanding the derivative of arcsin(y) is essential, as it helps in identifying du when applying integration by parts. The derivative of arcsin(y) with respect to y is 1/√(1 - y²), which is used to find du in the integration process.

Basic Integration Techniques

Familiarity with basic integrals, such as ∫ dy and integrals involving square roots, is important to solve the resulting integrals after applying integration by parts. This includes recognizing standard forms and using substitution if necessary.

Evaluate the integrals in Exercises 1–24 using integration by parts.∫ arcsin(y) dy