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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 8 - Integration TechniquesProblema 8.4.9
Capitolo 8, Problema 8.4.9

7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.
9. ∫[5 to 5√3] √(100 - x²) dx

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Step 1: Recognize that the integral involves a square root of the form √(a² - x²), which suggests using a trigonometric substitution. Specifically, use the substitution x = a sin(θ), where a = 10 in this case, because √(100 - x²) matches the form √(a² - x²). This substitution simplifies the square root expression.
Step 2: Substitute x = 10 sin(θ) into the integral. Compute dx by differentiating x = 10 sin(θ), which gives dx = 10 cos(θ) dθ. Also, update the limits of integration: when x = 5, solve for θ using sin(θ) = x/10, which gives θ = arcsin(5/10) = π/6. Similarly, when x = 5√3, solve for θ using sin(θ) = x/10, which gives θ = arcsin(√3/2) = π/3.
Step 3: Rewrite the integral in terms of θ. Substitute √(100 - x²) = √(100 - (10 sin(θ))²) = √(100 - 100 sin²(θ)) = √(100(1 - sin²(θ))) = √(100 cos²(θ)) = 10 cos(θ). The integral becomes ∫[π/6 to π/3] 10 cos(θ) * 10 cos(θ) dθ = ∫[π/6 to π/3] 100 cos²(θ) dθ.
Step 4: Simplify the integral further using the trigonometric identity cos²(θ) = (1 + cos(2θ))/2. Substitute this identity into the integral: ∫[π/6 to π/3] 100 cos²(θ) dθ = ∫[π/6 to π/3] 100 * (1 + cos(2θ))/2 dθ = ∫[π/6 to π/3] 50 (1 + cos(2θ)) dθ.
Step 5: Split the integral into two parts: ∫[π/6 to π/3] 50 dθ + ∫[π/6 to π/3] 50 cos(2θ) dθ. Evaluate each part separately. For the first part, ∫[π/6 to π/3] 50 dθ, integrate directly to get 50θ evaluated from π/6 to π/3. For the second part, ∫[π/6 to π/3] 50 cos(2θ) dθ, use the substitution u = 2θ and adjust the limits accordingly, then integrate cos(u). Combine the results to complete the solution.

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Trigonometric Substitution

Trigonometric substitution is a technique used in calculus to simplify integrals involving square roots of quadratic expressions. By substituting a variable with a trigonometric function, such as x = a sin(θ) or x = a cos(θ), the integral can often be transformed into a more manageable form. This method is particularly useful for integrals that contain expressions like √(a² - x²), which can be simplified using the Pythagorean identity.
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Pythagorean Identity

The Pythagorean identity states that for any angle θ, sin²(θ) + cos²(θ) = 1. This fundamental relationship between sine and cosine is crucial when performing trigonometric substitutions, as it allows us to express one trigonometric function in terms of another. In the context of integrals, this identity helps to simplify expressions involving square roots, making it easier to evaluate the integral.
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Definite Integrals

Definite integrals represent the accumulation of quantities, such as area under a curve, over a specified interval [a, b]. When evaluating a definite integral, it is essential to compute the antiderivative of the integrand and then apply the Fundamental Theorem of Calculus, which states that the definite integral from a to b of a function f(x) is equal to F(b) - F(a), where F is the antiderivative of f. This process is integral to solving the given problem involving the limits of integration.
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