Evaluate the integrals in Exercises 1–24 using integration by parts.
∫ e^(-y) cos(y) dy

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Identify the integral to solve: \(\int e^{-y} \cos(y) \, dy\).

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

Choose \(u\) and \(dv\) wisely. For this integral, let \(u = \cos(y)\) and \(dv = e^{-y} dy\).

Compute \(du\) and \(v\): differentiate \(u\) to get \(du = -\sin(y) dy\), and integrate \(dv\) to get \(v = \int e^{-y} dy = -e^{-y}\).

Apply the integration by parts formula: substitute \(u\), \(v\), \(du\) into \(\int u \, dv = uv - \int v \, du\) and simplify the resulting integral.

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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 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.

Integration of Exponential Functions

Exponential functions like e^(-y) have straightforward antiderivatives, often involving a constant factor adjustment. Recognizing how to integrate e^(-y) is essential, as it frequently appears in integrals combined with trigonometric functions.

Integration of Trigonometric Functions

Integrals involving trigonometric functions such as cos(y) often require knowledge of their antiderivatives, like sin(y) for cos(y). When combined with other functions, these integrals may need repeated application of integration techniques.

Evaluate the integrals in Exercises 1–24 using integration by parts.∫ e^(-y) cos(y) dy