Problemset8s - MAT 137Y 2008-09 Winter Session Solutions to Problem Set 8 1(SHE 5.7 8 Let u = t 2 1 Then du = 2t dt so 3t(t 2 1)2 dt = 3 2 du 3 3 =

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MAT 137Y 2008-09 Winter Session, Solutions to Problem Set 8 1 (SHE 5.7) 8. Let u = t 2 + 1. Then du = 2 t dt , so Z 3 t ( t 2 - 1 ) 2 dt = 3 2 Z du u 2 = - 3 2 u + C = - 3 2 ( t 2 + 1 ) + C . 14. Let u = 1 - x 3 , so du = - 3 x 2 dx . Then Z x 2 ( 1 - x 3 ) 2 / 3 dx = - 1 3 Z du u 2 / 3 = - u 1 / 3 + C = - ( 1 - x 3 ) 1 / 3 + C . 18. Let u = x 2 + 3 x + 1. Then du = ( 2 x + 3 ) dx . Therefore, Z 4 x + 6 x 2 + 3 x + 1 dx = 2 Z du u = 4 u + C = 4 p x 2 + 3 x + 1 + C . 24. Let u = r 2 + 16, then du = 2 r dr . Hence, Z 3 0 r r 2 + 16 dr = 1 2 Z 25 16 du u = h u i 25 16 = 1 . 46. Let u = x 2 . Then du = 2 x dx , so Z x sec 2 x 2 dx = 1 2 Z sec 2 u du = 1 2 tan u + C = 1 2 tan x 2 + C . 52. I = Z ( 1 + tan 2 x ) sec 2 xdx = Z sec 2 xdx + Z tan 2 x sec 2 xdx . The second integral requires a sim- ple substitution u = tan x . Thus, I = tan x + Z u 2 du = tan x + 1 3 u 3 + C = tan x + 1 3 tan 3 x + C . 2
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This note was uploaded on 04/09/2010 for the course MAT 137 taught by Professor Uppal during the Spring '08 term at University of Toronto- Toronto.

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Problemset8s - MAT 137Y 2008-09 Winter Session Solutions to Problem Set 8 1(SHE 5.7 8 Let u = t 2 1 Then du = 2t dt so 3t(t 2 1)2 dt = 3 2 du 3 3 =

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