Explanation Given information: ∫ 0 1 / 2 2 x sin − 1 ( x 2 ) d x Formula: ∫ u v d x = u ∫ v d x − ∫ d u d x ( ∫ v d x ) d x d f ( u ) d x = d f d u ⋅ d u d x d d u ( s i n − 1 ( u ) ) = 1 1 − u 2 ∫ a b f ( x ) d x = − ∫ b a f ( x ) d x , a < b d d x ( x a ) = a ⋅ x a − 1 ∫ x a d x = x a + 1 a + 1 Calculation: Evaluate the integral. 2 ∫ 0 1 / 2 x sin − 1 ( x 2 ) d x = 2 ( sin − 1 ( x 2 ) ∫ 0 1 / 2 x d x − ∫ 0 1 / 2 d ( sin − 1 ( x 2 ) ) d x ( ∫ x d x ) d x ) + C = 2 ( sin − 1 ( x 2 ) ( x 2 2 ) − ∫ 0 1 / 2 d ( sin − 1 ( x 2 ) ) d x ( x 2 2 ) d x ) + C = x 2 sin − 1 ( x 2 ) − 2 ∫ 0 1 / 2 d ( sin − 1 ( x 2 ) ) d x ( x 2 2 ) d x + C Apply the limits. 2 ∫ 0 1 / 2 x sin − 1 ( x 2 ) d x = [ x 2 sin − 1 ( x 2 ) ] 0 1 / 2 − ∫ 0 1 / 2 d ( sin − 1 ( x 2 ) ) d x x 2 d x + C = [ ( 1 2 ) 2 sin − 1 ( ( 1 2 ) 2 ) ] − ∫ 0 1 / 2 d ( sin − 1 ( x 2 ) ) d x x 2 d x + C = [ ( 1 2 ) sin − 1 ( 1 2 ) ] − ∫ 0 1 / 2 d ( sin − 1 ( x 2 ) ) d x x 2 d x + C = [ ( 1 2 ) ( π 6 ) ] − ∫ 0 1 / 2 d ( sin − 1 ( x 2 ) ) d x x 2 d x + C 2 ∫ 0 1 / 2 x sin − 1 ( x 2 ) d x = π 12 − ∫ 0 1 / 2 d ( sin − 1 ( x 2 ) ) d x x 2 d x + C (1) Consider d ( sin − 1 ( x 2 ) ) d x from the equation (1)

Thomas' Calculus: Early Transcendentals plus MyLab Math with Pearson eText -- Title-Specific Access Card Package (14th Edition)

14th Edition

ISBN: 9780134768496

Author: Joel R. Hass, Christopher E. Heil, Maurice D. Weir

Publisher: PEARSON

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