# HW3 - Math 191 HW 3 Solutions 5.5 4 Let u = 1 cos 1 2 ⇒...

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Unformatted text preview: Math 191 HW 3 Solutions 5.5: 4 Let u = 1- cos 1 2 ⇒ du = 1 2 sin 1 2 dt ⇒ 2 du = sin 1 2 dt . Then R (1- cos t 2 2 sin t 2 ) dt = R 2 u 2 du = 2 3 u 3 + C = 2 3 (1- cos t 2 ) 3 + C 5.5: 8 Let u = y 4 +4 y 2 +1 ⇒ du = (4 y 3 +8 y ) dy ⇒ 3 du = 12( y 3 +2 y ) dy . Then R 12( y 4 +4 y 2 +1) 2 ( y 3 +2 y ) dy = R 3 u 2 du = u 3 + C = ( y 4 + 4 y 2 + 1) 3 + C 5.5: 12 a. Let u = 5 x + 8 ⇒ du = 5 dx ⇒ 1 / 5 du = dx . Then R dx √ 5 x +8 = R 1 5 √ u du = 1 / 5 R u- 1 / 2 du = 1 / 5(2 u 1 / 2 ) + C = 2 / 5 u 1 / 2 + C- 2 / 5 √ 5 x + 8 + C b. Let u = √ 5 x + 8 ⇒ du = 1 / 2(5 x + 8)- 1 / 2 5 dx ⇒ 2 / 5 du = dx √ 5 x +8 . Then R dx √ 5 x +8 = R 2 / 5 du = 2 / 5 du + C = 2 / 5 √ 5 x + 8 + C 5.5: 16 Let u = 2- x ⇒ du =- dx ⇒ - du = dx . Then R 3 / (2- x ) 2 dx = R 3(- du ) /u 2 =- 3 R u- 2 du =- 3( u- 1 / (- 1)) + C = 3 / (2- x ) + C 5.5: 22 Let u = 1 + √ x ⇒ du = dx 2 √ x ⇒ 2 du = dx/ √ x . Then R (1+ √ x ) 3 √ x dx = R u 3 (2 du ) = 2( u 4 / 4) + C = 1 / 2(1 + √ x ) 4 + C 5.5: 36 Let u = 2 + sin t ⇒ du = cos tdt . Then R 6 cos t (2+sin t ) 3 dt = R 6 /u 3 du = 6 u (- 2) / (- 2) + C =- 3(2 + sin t )- 2 + C 5.5: 50 a. Let u = x- 1 ⇒ du = dx ; v = sin u ⇒ dv = cos udu ; w = 1 + v 2 ⇒ dw = 2 vdv ⇒ 1 / 2 dw = vdv...
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## This note was uploaded on 06/19/2008 for the course MATH 1910 taught by Professor Berman during the Spring '07 term at Cornell.

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HW3 - Math 191 HW 3 Solutions 5.5 4 Let u = 1 cos 1 2 ⇒...

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