Explanation Given information: ∫ 0 1 / 2 1 + 1 − x 2 d x Formula: cos 2 θ = 1 − sin 2 θ a 2 − b 2 = ( a + b ) ( a − b ) ∫ u d v = u v − ∫ v d u (1) Calculation: Evaluate the integral given below. ∫ 0 1 / 2 1 + 1 − x 2 d x (2) Apply trigonometric substitution by assuming the variable x as sin θ . x = sin θ Differentiate the equation with respect to θ . d x d θ = cos θ d x = cos θ d θ Consider this term 1 − x 2 by assuming the variable x as sin θ . 1 − x 2 = 1 − sin 2 θ = cos 2 θ = cos θ For the variable x , the interval values are 0 and 0.5. For the variable θ , the interval values 0 and π 6 . Substitute cos θ for 1 − x 2 and cos θ d θ for d x and changing the limits in the equation (2). ∫ 0 1 / 2 1 + 1 − x 2 d x = ∫ 0 π / 6 1 + cos θ ( cos θ d θ ) Multiply and divide by 1 − cos θ term on the RHS. ∫ 0 1 / 2 1 + 1 − x 2 d x = ∫ 0 π / 6 1 + cos θ × 1 − cos θ 1 − cos θ ( cos θ d θ ) = ∫ 0 π / 6 1 − cos 2 θ 1 − cos θ cos θ d θ = ∫ 0 π / 6 sin 2 θ 1 − cos θ cos θ d θ ∫ 0 1 / 2 1 + 1 − x 2 d x = ∫ 0 π / 6 sin θ cos θ d θ 1 − cos θ (3) Apply trigonometric substitution by assuming u as cos θ . u = cos θ Differentiate the equation with respect to θ . d u d θ = − sin θ d u = − sin θ d θ Apply trigonometric substitution by assuming the term d v d θ as below. d v d θ = sin θ 1 − cos θ Integrate the above equation with respect to θ . ∫ d v d θ = ∫ sin θ 1 − cos θ v = 2 1 − cos θ Substitute cos θ for u , − sin θ d θ for d u , 2 1 − cos θ for v and sin θ 1 − cos θ d θ for d v in equation with the limits (1). ∫ 0 π / 6 sin θ cos θ d θ 1 − cos θ = [ cos θ ( 2 ( 1 − cos θ ) 1 / 2 ) ] 0 π / 6 − ∫ 0 π / 6 2 ( 1 − cos θ ) 1 / 2 ( − sin θ d θ ) ∫ 0 π / 6 sin θ cos θ d θ 1 − cos θ = [ 2 cos θ ( 1 − cos θ ) 1 / 2 ] 0 π / 6 + ∫ 0 π / 6 2 ( 1 − cos θ ) 1 / 2 ( sin θ d θ ) (4) Apply trigonometric substitution by assuming u as 1 − cos θ . u = 1 − cos θ Differentiate the equation with respect to θ

14th Edition

ISBN: 9780136208013

Author: Hass

Publisher: PEARSON

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Chapter 8, Problem 126PE

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