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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
Capitolo 8, Problema 8.3.40

9–61. Trigonometric integrals Evaluate the following integrals.
40. ∫[0 to π/6] tan⁵(2x) sec(2x) dx

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Step 1: Recognize that the integral involves a combination of trigonometric functions, specifically tan⁵(2x) and sec(2x). To simplify, recall that the derivative of tan(u) is sec²(u), which suggests a substitution method might be effective.
Step 2: Let u = tan(2x). Then, the derivative of u with respect to x is du/dx = 2 sec²(2x). Rearrange this to express dx in terms of du: dx = du / (2 sec²(2x)).
Step 3: Rewrite the integral using the substitution u = tan(2x). Since tan⁵(2x) = u⁵ and sec(2x) remains in the integral, substitute dx as well: ∫[0 to π/6] tan⁵(2x) sec(2x) dx = ∫[0 to π/6] u⁵ sec(2x) * (du / (2 sec²(2x))).
Step 4: Simplify the integral by canceling one sec(2x) term in the numerator with one in the denominator, leaving: ∫ u⁵ * (1 / 2 sec(2x)) du. Note that sec²(2x) is related to tan²(2x) via the identity sec²(θ) = 1 + tan²(θ), which can be used to express sec²(2x) in terms of u.
Step 5: Adjust the limits of integration to match the substitution. When x = 0, u = tan(2 * 0) = 0. When x = π/6, u = tan(2 * π/6) = tan(π/3). The integral now becomes ∫[0 to tan(π/3)] u⁵ / 2 du, which can be solved by applying the power rule for integration.

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