soln2 - MATH 138 Calculus 2, Solutions to Assignment 2 1:...

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Unformatted text preview: MATH 138 Calculus 2, Solutions to Assignment 2 1: (a) Find Z ln 3 e x dx 1 + e x . Solution: Make the substitution u = e x so du = e x dx . Then Z ln 3 e x dx 1 + e x = Z 3 1 du 1 + u = h ln(1 + u ) i 3 1 = ln 4- ln 2 = ln 2 . (b) Find Z / 3 sin 3 x cos 2 x dx . Solution: Make the substitution u = cos x so du =- sin x dx . Then Z / 3 sin 3 x cos 2 x dx = Z / 3 (1- cos 2 x ) cos 2 x sin x dx = Z 1 / 2 1- 1- u 2 u 2 du = Z 1 / 2 1- 1 u 2 + 1 du = h 1 u + u i 1 / 2 1 = ( 2 + 1 2 )- (1 + 1) = 1 2 . 2: (a) Find Z 3 1 x x + 1 dx . Solution: Make the substitution u = x so u 2 = x and 2 u du = dx . Then Z 3 1 x dx x + 1 = Z 3 1 2 u 2 du u 2 + 1 = Z 3 1 2- 2 u 2 + 1 du = h 2 u- 2 tan- 1 u i 3 1 = ( 2 3- 2 3 )- ( 2- 2 ) 2 3- 2- 6 . (b) Find Z e 1 (ln x ) 2 dx . Solution: Integrate by parts using u = (ln x ) 2 , du = 2 ln x x dx , v = x and dv = dx to get Z e 1 (ln x ) 2 dx = x (ln x ) 2- Z 2 ln x dx e 1 = h x (ln x ) 2- 2 x ln x + 2 x i e 1 = ( e- 2...
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This note was uploaded on 09/18/2008 for the course MATH MAth 138 taught by Professor Marshman during the Spring '08 term at Waterloo.

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soln2 - MATH 138 Calculus 2, Solutions to Assignment 2 1:...

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