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Homework7

# Homework7 - MA 115 Homework Solutions for Week 7 5.5#10 xex...

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MA 115 Homework Solutions for Week 7 5.5 #10 Let u = x 2 . Then du = 2 x, so integraltext xe x 2 dx = integraltext e u ( 1 2 du ) = 1 2 e u + C = 1 2 e x 2 + C 5.5 #20 Let u = 2 θ. Then du = 2 and = 1 2 du, so integraltext sec2 θ tan2 θdθ = integraltext sec u tan u ( 1 2 du ) = 1 2 sec u + C = 1 2 sec2 θ + C 5.5 #22 Let u = x 2 + 1 . Then du = 2 xdx and xdx = 1 2 du , so integraldisplay x x 2 + 1 dx = integraldisplay 1 2 du u = 1 2 ln | u | + C = 1 2 ln | x 2 + 1 | + C = 1 2 ln( x 2 + 1) + C [since x 2 + 1 > 0 for all x ] = ln( x 2 + 1) 1 / 2 + C = ln radicalbig x 2 + 1 + C 5.5 #24 Let u = tan - 1 x . Then du = dx 1+ x 2 , so integraltext tan - 1 x 1+ x 2 dx = integraltext udu = u 2 2 + C = (tan - 1 x ) 2 2 + C 5.5 #34 Let u = x 2 . Then du = 2 xdx , so integraltext x 1+ x 4 dx = integraltext 1 2 du 1+ u 2 = 1 2 tan - 1 u + C = 1 2 tan - 1 ( x 2 ) + C 5.5 #42 Let u = x 2 , so du = 2 xdx . When x = 0 , u = 0; when t = π, u = π . integraldisplay π 0 x cos( x 2 ) dx = integraldisplay π 0 cos u parenleftbigg 1 2 du parenrightbigg = 1 2 [sin u ] π 0 = 1 2 (sin π - sin0) = 0 Notice the change in the limits of integration when u = x 2 is substituted.

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Homework7 - MA 115 Homework Solutions for Week 7 5.5#10 xex...

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