Homework7

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

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Unformatted text preview: 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 . Then du = 2 d and d = 1 2 du, so integraltext sec 2 tan 2 d = integraltext sec u tan u ( 1 2 du ) = 1 2 sec u + C = 1 2 sec 2 + 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 = , u = ....
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This note was uploaded on 04/07/2008 for the course MA 115 taught by Professor Mahalanobis during the Fall '08 term at Stevens.

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

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