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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
8장, 문제 8.2.48

Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.
∫₀^π/2 x³ cos 2x dx

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Identify the integral to solve: \(\int_0^{\frac{\pi}{2}} x^3 \cos(2x) \, dx\).
Recognize that this integral involves a product of a polynomial (\(x^3\)) and a trigonometric function (\(\cos(2x)\)), which suggests using integration by parts.
Set up integration by parts by choosing \(u = x^3\) (which simplifies when differentiated) and \(dv = \cos(2x) \, dx\) (which can be integrated easily). Recall the formula: \(\int u \, dv = uv - \int v \, du\).
Compute \(du = 3x^2 \, dx\) and find \(v\) by integrating \(dv\): \(v = \int \cos(2x) \, dx = \frac{1}{2} \sin(2x)\).
Apply the integration by parts formula: \(\int_0^{\frac{\pi}{2}} x^3 \cos(2x) \, dx = \left. x^3 \cdot \frac{1}{2} \sin(2x) \right|_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} \frac{1}{2} \sin(2x) \cdot 3x^2 \, dx\). This reduces the original integral to a new integral involving \(x^2 \sin(2x)\), which may require repeating integration by parts.

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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 used to integrate products of functions. It is based on the product rule for differentiation and follows the formula ∫u dv = uv - ∫v du. Choosing u and dv wisely simplifies the integral, especially when one function becomes simpler upon differentiation.
추천 영상:
06:18
Integration by Parts for Definite Integrals

Definite Integrals

Definite integrals calculate the net area under a curve between two limits. They produce a numerical value and require evaluating the antiderivative at the upper and lower bounds. Understanding how to apply limits after integration is essential for solving definite integrals.
추천 영상:
05:43
Definition of the Definite Integral

Trigonometric Functions in Integration

Integrating functions involving trigonometric terms often requires using identities or special techniques. Recognizing how to handle integrals with cosine or sine, especially when multiplied by polynomials, helps in simplifying and solving the integral effectively.
추천 영상:
6:04
Introduction to Trigonometric Functions