Evaluate the integrals in Exercises 23–32.
∫₀^π √(1 - cos²(θ)) dθ

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Recognize that the integrand is \( \sqrt{1 - \cos^2(\theta)} \). Using the Pythagorean identity, recall that \( \sin^2(\theta) + \cos^2(\theta) = 1 \), so \( 1 - \cos^2(\theta) = \sin^2(\theta) \).

Rewrite the integral using this identity: \( \int_0^{\pi} \sqrt{\sin^2(\theta)} \, d\theta \). Since the square root of a square is the absolute value, this becomes \( \int_0^{\pi} |\sin(\theta)| \, d\theta \).

Analyze the behavior of \( \sin(\theta) \) on the interval \( [0, \pi] \). Note that \( \sin(\theta) \) is non-negative on this interval, so \( |\sin(\theta)| = \sin(\theta) \) for \( \theta \in [0, \pi] \).

Simplify the integral to \( \int_0^{\pi} \sin(\theta) \, d\theta \). Now, find the antiderivative of \( \sin(\theta) \), which is \( -\cos(\theta) \).

Apply the Fundamental Theorem of Calculus by evaluating \( -\cos(\theta) \) from \( 0 \) to \( \pi \), i.e., compute \( [-\cos(\theta)]_0^{\pi} = -\cos(\pi) + \cos(0) \).

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주요 개념

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Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values within their domains. In this problem, recognizing that 1 - cos²(θ) equals sin²(θ) simplifies the integral significantly, allowing the square root to be expressed as |sin(θ)|.

Absolute Value in Integrals

When integrating expressions involving square roots of squared functions, the result is the absolute value of the original function. Since √(sin²(θ)) = |sin(θ)|, understanding how to handle absolute values over the interval [0, π] is essential, as sin(θ) changes sign within this range.

Definite Integration over Piecewise Functions

Definite integrals involving absolute values often require splitting the integral at points where the function inside the absolute value changes sign. For sin(θ) on [0, π], it is positive on [0, π/2] and positive on [π/2, π], so the integral can be evaluated by considering these intervals separately or by using symmetry.

Evaluate the integrals in Exercises 23–32.∫₀^π √(1 - cos²(θ)) dθ