Skip to main content
Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
7장, 문제 7.6.71

Evaluate the integrals in Exercises 53–76.
71. ∫(from -π/2 to π/2) 2cosθ dθ/(1+(sinθ)²)

검증된 단계별 안내
1
Identify the integral to evaluate: \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2 \cos \theta}{1 + (\sin \theta)^2} \, d\theta\).
Consider the symmetry of the integrand. Note that \(\cos \theta\) is an even function and \((\sin \theta)^2\) is also even, so the entire integrand is an even function. This allows us to rewrite the integral as \(2 \int_0^{\frac{\pi}{2}} \frac{2 \cos \theta}{1 + (\sin \theta)^2} \, d\theta\).
Use the substitution \(u = \sin \theta\), which implies \(du = \cos \theta \, d\theta\). This substitution will simplify the integral because the denominator is in terms of \(\sin \theta\) and the numerator contains \(\cos \theta \, d\theta\).
Rewrite the integral in terms of \(u\): the limits change from \(\theta = 0\) to \(\theta = \frac{\pi}{2}\), which correspond to \(u = 0\) to \(u = 1\). The integral becomes \(2 \int_0^1 \frac{2}{1 + u^2} \, du\).
Recognize that \(\int \frac{1}{1 + u^2} \, du\) is the standard arctangent integral. Set up the integral accordingly and prepare to evaluate it using the antiderivative \(\arctan u\).

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
4m
도움이 되었나요?

주요 개념

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

Definite Integrals

A definite integral calculates the net area under a curve between two specified limits. It involves evaluating the integral function at the upper and lower bounds and subtracting these values. This concept is essential for solving integrals with given interval limits, such as from -π/2 to π/2.
추천 영상:
05:43
Definition of the Definite Integral

Trigonometric Functions and Identities

Understanding sine and cosine functions, their properties, and identities is crucial for simplifying integrals involving trigonometric expressions. For example, recognizing that (sin θ)² can be rewritten using identities helps in simplifying the denominator or integrand for easier integration.
추천 영상:
6:04
Introduction to Trigonometric Functions

Symmetry in Integrals

Symmetry properties of functions over symmetric intervals can simplify integration. Even functions satisfy f(-x) = f(x), and odd functions satisfy f(-x) = -f(x). Identifying whether the integrand is even or odd over [-π/2, π/2] can reduce the integral calculation or even determine if the integral is zero.
추천 영상:
06:18
Integration by Parts for Definite Integrals