Skip to main content
Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
8장, 문제 8.3.10

Evaluate the integrals in Exercises 1–22.
∫₀^(π/6) 3cos⁵(3x) dx

검증된 단계별 안내
1
Recognize that the integral is \( \int_0^{\frac{\pi}{6}} 3 \cos^5(3x) \, dx \). The constant 3 can be factored out: \( 3 \int_0^{\frac{\pi}{6}} \cos^5(3x) \, dx \).
Use the substitution \( u = 3x \), which implies \( du = 3 \, dx \) or \( dx = \frac{du}{3} \). Also, change the limits of integration: when \( x=0 \), \( u=0 \); when \( x=\frac{\pi}{6} \), \( u=\frac{\pi}{2} \).
Rewrite the integral in terms of \( u \): \( 3 \int_0^{\frac{\pi}{6}} \cos^5(3x) \, dx = 3 \int_0^{\frac{\pi}{2}} \cos^5(u) \cdot \frac{du}{3} = \int_0^{\frac{\pi}{2}} \cos^5(u) \, du \).
To integrate \( \cos^5(u) \), express it as \( \cos^4(u) \cos(u) \) and use the identity \( \cos^2(u) = 1 - \sin^2(u) \) to rewrite \( \cos^4(u) = (\cos^2(u))^2 = (1 - \sin^2(u))^2 \).
Use the substitution \( t = \sin(u) \), so \( dt = \cos(u) \, du \). The integral becomes \( \int_0^{\frac{\pi}{2}} \cos^5(u) \, du = \int_0^1 (1 - t^2)^2 \, dt \). Expand the integrand and integrate term-by-term.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
6m
도움이 되었나요?

주요 개념

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

Integration of Trigonometric Functions

This involves finding the integral of functions involving sine, cosine, or other trig functions. Techniques often include using identities to simplify powers or products of trig functions before integrating.
추천 영상:
6:04
Introduction to Trigonometric Functions

Trigonometric Power-Reduction Formulas

Power-reduction formulas express powers of sine or cosine in terms of functions with lower powers or multiple angles, making integration manageable. For example, cos^5(θ) can be rewritten using these formulas to simplify the integral.
추천 영상:
6:04
Introduction to Trigonometric Functions

Definite Integration and Limits

Definite integration calculates the exact area under a curve between two bounds. After finding the antiderivative, you evaluate it at the upper and lower limits (here 0 and π/6) and subtract to get the final value.
추천 영상:
05:43
Definition of the Definite Integral