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Unformatted text preview: Solutions to quiz 2A 1. Evaluate the integral. (a) www.math.ufl.edu/~mathguy cos4 ()d 1 (Solution) We will use the fact that cos2 () = 2 (1 + cos(2)). Thus, cos4 ()d = = = = = = = 1 4 1 4 (cos2 ())2 d 1 (1 + cos(2)) 2
2 d 1 + 2 cos(2) + cos2 (2)d 1d + 2 cos(2)d + cos2 (2)d 1 1 + sin(2) + 4 2 (1 + cos(4))d 1 1 1 + sin(2) + + sin(4) + C 4 2 8 3 1 1 + sin(2) + sin(4) + C1 8 4 32 (b) ex sin(x)dx (Solution) Here we will use integration by parts. u = sin(x) du = cos(x)dx ex sin(x)dx = udv = u v  dv = ex dx v = ex vdu = sin(x)ex  ex cos(x)dx. dv1 = ex dx v1 = ex v1 du1 = cos(x)ex + ex sin(x)dx ex cos(x)dx Now, we will use integration by parts on u1 = cos(x) du1 =  sin(x)dx ex cos(x)dx = u1 dv1 = u1 v1  Thus, putting everything together, we have: ex sin(x)dx = sin(x)ex  = sin(x)ex  ex cos(x)dx cos(x)ex + ex sin(x)dx ex sin(x)dx = sin(x)ex  cos(x)ex  So, 2 ex sin(x)dx = sin(x)ex  cos(x)ex + c ex sin(x)dx = 1 1 sin(x)ex  cos(x)ex + c1 2 2 ...
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This note was uploaded on 07/11/2011 for the course MAC 2312 taught by Professor Bonner during the Spring '08 term at University of Florida.
 Spring '08
 Bonner
 Math

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