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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
8장, 문제 8.R.1d

Explain why or why not. Determine whether the following statements are true and give an explanation or counterexample.
d. ∫2 sin x cos x dx = −(1/2) cos 2x + C.

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1
Step 1: Recall the trigonometric identity for the product of sine and cosine: 2 sin(x) cos(x) = sin(2x). This identity simplifies the integrand.
Step 2: Substitute the identity into the integral: ∫2 sin(x) cos(x) dx = ∫sin(2x) dx. This reduces the problem to finding the integral of sin(2x).
Step 3: Recall the formula for the integral of sin(kx): ∫sin(kx) dx = -(1/k) cos(kx) + C. Here, k = 2.
Step 4: Apply the formula to compute the integral: ∫sin(2x) dx = -(1/2) cos(2x) + C. This matches the given expression.
Step 5: Conclude that the statement is true because the integral calculation aligns with the given result: ∫2 sin(x) cos(x) dx = -(1/2) cos(2x) + C.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
2m
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주요 개념

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

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Integration of trigonometric functions often involves using identities to simplify the integrand. In this case, the identity sin(2x) = 2 sin(x) cos(x) can be applied to rewrite the integral, making it easier to evaluate. Understanding these identities is crucial for solving integrals involving products of sine and cosine.
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The Fundamental Theorem of Calculus links differentiation and integration, stating that if F is an antiderivative of f on an interval, then the integral of f from a to b is F(b) - F(a). This theorem is essential for confirming the correctness of an integral's evaluation and understanding the relationship between a function and its integral.
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Constant of Integration

When performing indefinite integration, it is important to include a constant of integration (C) because the process of differentiation eliminates constant terms. This constant represents the family of antiderivatives and is crucial for expressing the general solution to an integral, ensuring that all possible functions that could yield the same derivative are accounted for.
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