Explanation Given information: ∫ x 1 − x d x Formula: sin 2 x cos 2 x = 4 sin 2 2 x sin 2 2 x = 1 − cos 4 x 2 sin 4 θ = 2 sin 2 θ cos 2 θ sin 2 θ = 2 sin θ cos θ cos 2 θ = 2 cos 2 θ − 1 ∫ x a d x = x a + 1 a + 1 + C ∫ cos a x d x = sin a x a + C Calculation: Express the integral as below. ∫ x 1 − x d x (1) Apply integral substitution by assuming the variable u as x . u = x x = u 2 Differentiate the equation with respect to u . d x d u = 2 u d x = 2 u d u Substitute 2 u d u for d x and u 2 for x in equation (1). ∫ x 1 − x d x = ∫ u 1 − u 2 ( 2 u ) d u ∫ x 1 − x d x = 2 ∫ u 2 1 − u 2 d u (2) Apply trigonometric substitution by assuming the variable u as sin θ . u = sin θ Differentiate the equation with respect to θ . d u d θ = cos θ d u = cos θ d θ Consider this term 1 − u 2 by substituting sin θ for u . 1 − u 2 = 1 − sin 2 θ = cos 2 θ = cos θ Substitute cos θ for 1 − u 2 , sin θ for u and cos θ d θ for d u in equation (2). ∫ x 1 − x d x = 2 ∫ sin 2 θ ( cos θ ) cos θ d θ = 2 ∫ sin 2 θ cos 2 θ d θ = 1 2 ∫ 4 sin 2 θ cos 2 θ d θ = 1 2 ∫ sin 2 2 θ d θ Substitute 1 − cos 4 θ 2 for sin 2 2 θ . ∫ x 1 − x d x = 1 2 ∫ [ 1 − cos 4 θ 2 ] d θ = 1 2 ∫ 1 2 d θ − 1 2 ∫ cos 4 θ 2 d θ = 1 4 θ − 1 4 [ sin 4 θ 4 ] + C Substitute 2 sin 2 θ cos 2 θ for sin 4 θ

University Calculus: Early Transcendentals, Books a la Carte Edition (3rd Edition)

3rd Edition

ISBN: 9780321999610

Author: Joel R. Hass, Maurice D. Weir, George B. Thomas Jr.

Publisher: PEARSON

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Chapter 8.3, Problem 47E

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