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# soln2 - MATH 138 Calculus 2 Solutions to Assignment 2 ln 3...

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MATH 138 Calculus 2, Solutions to Assignment 2 1: (a) Find ln 3 0 e x dx 1 + e x . Solution: Make the substitution u = e x so du = e x dx . Then ln 3 0 e x dx 1 + e x = 3 1 du 1 + u = ln(1 + u ) 3 1 = ln 4 - ln 2 = ln 2 . (b) Find π/ 3 0 sin 3 x cos 2 x dx . Solution: Make the substitution u = cos x so du = - sin x dx . Then π/ 3 0 sin 3 x cos 2 x dx = π/ 3 0 (1 - cos 2 x ) cos 2 x sin x dx = 1 / 2 1 - 1 - u 2 u 2 du = 1 / 2 1 - 1 u 2 + 1 du = 1 u + u 1 / 2 1 = ( 2 + 1 2 ) - (1 + 1) = 1 2 . 2: (a) Find 3 1 x x + 1 dx . Solution: Make the substitution u = x so u 2 = x and 2 u du = dx . Then 3 1 x dx x + 1 = 3 1 2 u 2 du u 2 + 1 = 3 1 2 - 2 u 2 + 1 du = 2 u - 2 tan - 1 u 3 1 = ( 2 3 - 2 π 3 ) - ( 2 - π 2 ) 2 3 - 2 - π 6 . (b) Find 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 e 1 (ln x ) 2 dx = x (ln x ) 2 - 2 ln x dx e 1 = x (ln x ) 2 - 2 x ln x + 2 x e 1 = ( e - 2 e + 2 e ) - 2 = e - 2 . 3: (a) Find 3 0 x 2 dx ( x + 1) 3 / 2 . Solution: Make the substitution u = x + 1 so x = u - 1 and dx = du . Then 3 0 x 2 dx ( x + 1) 3 / 2 = 4 1 ( u - 1) 2 u 3 / 2 du = 4 1 u 2 - 2 u + 1 u 3 / 2 du = 4 1 u 1 / 2 - 2 u - 1 / 2 + u - 3 / 2 du = 2 3 u 3 / 2 - 4 u 1 / 2 - 2 u - 1 / 2 4 1 = ( 16 3 - 8 - 1 ) - ( 2 3 - 4 - 2

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soln2 - MATH 138 Calculus 2 Solutions to Assignment 2 ln 3...

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