{VERSION 6 0 "IBM INTEL NT" "6.0" }
{USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 
1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 
0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }
{CSTYLE "2D Input" 2 19 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE 
"2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 
256 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 
0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 0 0 0 0 0 
0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }
{CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 
263 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 
0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 0 1 0 0 0 0 
0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }
{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 
2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 
11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 
0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 12 1 {CSTYLE "" -1 
-1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 3 0 0 0 0 1 0 1 0 2 2 
0 1 }{PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 
2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }}
{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 256 7 "MAT 156" }}{PARA 256 "" 
0 "" {TEXT 263 6 " LAB 9" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 257 57 "Topi
c 1:  Different substitutions to the same integral.  " }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT 264 16 "Assignment 1    " }{TEXT -1 22 "Consider
 the integral " }{XPPEDIT 18 0 "int(sin*x*cos*x,x);" "6#-%$intG6$**%$s
inG\"\"\"%\"xGF(%$cosGF(F)F(F)" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" 
{TEXT -1 49 "a)  Solve the integral using the substitution    " }
{XPPEDIT 18 0 "u = sin*x;" "6#/%\"uG*&%$sinG\"\"\"%\"xGF'" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 46 "b)  Solve the inte
gral using the substitution " }{XPPEDIT 18 0 "u = cos*x;" "6#/%\"uG*&%
$cosG\"\"\"%\"xGF'" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 44 "c)  Explain why these answers are different." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 11 "" 1 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT 258 66 "Topic 2: The method of partial fr
actions and completing the square" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 81 "Quite often we have to integrate functions that ar
e quotients of two polynomials " }{XPPEDIT 18 0 "P(x)/Q(x);" "6#*&-%\"
PG6#%\"xG\"\"\"-%\"QG6#F'!\"\"" }{TEXT -1 122 ". These functions are c
alled rational functions and we can use the method of partial fraction
s to integrate them. Example:" }}{PARA 0 "> " 0 "" {XPPEDIT 19 1 "int(
(x^3-2*x^2+7)/(x^4-3*x^3+3*x^2-3*x+2),x);" "6#-%$intG6$*&,(*$%\"xG\"\"
$\"\"\"*&\"\"#F+*$F)F-F+!\"\"\"\"(F+F+,,*$F)\"\"%F+*&F*F+*$F)F*F+F/*&F
*F+*$F)F-F+F+*&F*F+F)F+F/F-F+F/F)" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 27 "We introduce the expression" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 41 "f:=(x^3-2*x^2+7)/(x^4-3*x^3+3*x^2-3*x+2);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG*&,(*$)%\"xG\"\"$\"\"\"F+*&\"\"#
F+)F)F-F+!\"\"\"\"(F+F+,,*$)F)\"\"%F+F+*&F*F+F(F+F/*&F*F+F.F+F+*&F*F+F
)F+F/F-F+F/" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "fpar:=conver
t(f, parfrac,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%fparG,(*&\"\"$
\"\"\",&%\"xGF(F(!\"\"F+F+*(\"\"&F+,&*&\"#8F(F*F(F(\"\"'F(F(,&*$)F*\"
\"#F(F(F(F(F+F(*(\"\"(F(F-F+,&F*F(F5F+F+F(" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 174 "fpar is equal to f, only it is written in a form that is
 easier to integrate. Now we can define the integral of fpar and ask M
aple to compute it. We recognize ourselves that " }{XPPEDIT 18 0 "x-2;
" "6#,&%\"xG\"\"\"\"\"#!\"\"" }{TEXT -1 30 " in the denominator will g
ive " }{XPPEDIT 18 0 "ln(x-2);" "6#-%#lnG6#,&%\"xG\"\"\"\"\"#!\"\"" }
{TEXT -1 10 "  and the " }{XPPEDIT 18 0 "x-1;" "6#,&%\"xG\"\"\"F%!\"\"
" }{TEXT -1 30 " in the denominator will give " }{XPPEDIT 18 0 "ln(x-1
);" "6#-%#lnG6#,&%\"xG\"\"\"F(!\"\"" }{TEXT -1 40 ". Also we can split
 the expression with " }{XPPEDIT 18 0 "x^2+1;" "6#,&*$%\"xG\"\"#\"\"\"
F'F'" }{TEXT -1 25 " in the denominator into " }{XPPEDIT 18 0 "6/5(x^2
+1);" "6#*&\"\"'\"\"\"-\"\"&6#,&*$%\"xG\"\"#F%F%F%!\"\"" }{TEXT -1 6 "
 plus " }{XPPEDIT 18 0 "13*x/5(x^2+1);" "6#*(\"#8\"\"\"%\"xGF%-\"\"&6#
,&*$F&\"\"#F%F%F%!\"\"" }{TEXT -1 34 ". The first term gives an arctan
 (" }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 26 "), while the second gi
ves " }{XPPEDIT 18 0 "ln(x^2+1);" "6#-%#lnG6#,&*$%\"xG\"\"#\"\"\"F*F*
" }{TEXT -1 37 "  with appropriate constant in front." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "fint:=int(fpar, x);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%%fintG,**&\"\"$\"\"\"-%#lnG6#,&%\"xGF(F(!\"\"F(F
.*&#\"#8\"#5F(-F*6#,&*$)F-\"\"#F(F(F(F(F(F(*&#\"\"'\"\"&F(-%'arctanG6#
F-F(F(*&#\"\"(F<F(-F*6#,&F-F(F8F.F(F(" }}}{EXCHG {PARA 11 "" 1 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 50 "Naturally Maple c
an do all these things by itself:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "gint:=int(f, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%%gintG,**&\"\"$\"\"\"-%#lnG6#,&%\"xGF(F(!\"\"F(F.*&#\"#8\"#5F(-F*6#,&
*$)F-\"\"#F(F(F(F(F(F(*&#\"\"'\"\"&F(-%'arctanG6#F-F(F(*&#\"\"(F<F(-F*
6#,&F-F(F8F.F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 262 21 
"Assignment Part 2:   " }{TEXT -1 54 "Find the partial fraction decomp
osition and integrate:" }}{PARA 0 "> " 0 "" {XPPEDIT 19 1 "int((x^8+2*
x-1)/((x-1)^3*(x^2+3)^2),x);" "6#-%$intG6$*&,(*$%\"xG\"\")\"\"\"*&\"\"
#F+F)F+F+F+!\"\"F+*&,&F)F+F+F.\"\"$,&*$F)F-F+F1F+F-F.F)" }{TEXT -1 3 "
 . " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,2*&\"\"#!\"\"%\"xGF%\"\"\"*&\"
\"$F(F'F(F(*&F(F(*&\"#;F(),&F'F(F(F&F%F(F&F&*&#\"#P\"#KF(-%#lnG6#F/F(F
(*(\"#[F&,&*&\"#!)F(F'F(F(\"\"'F&F(,&*$)F'F%F(F(F*F(F&F(*&#\"$n(\"$)GF
(*&F*#F(F%-%'arctanG6#,$*(F*F&F'F(F*FEF(F(F(F&*&F(F(*&F%F(F/F(F&F&*&#F
2\"#kF(-F56#F=F(F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 11 "" 1 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 226 "Sometimes this does not work
 because the roots of the denominator are not real numbers, or involve
 radicals. If the denominator does not have real roots, we can complet
e the square. The command is completesquare(expression, x)" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(student);" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#7@%\"DG%%DiffG%*DoubleintG%$IntG%&LimitG%(LineintG%(Pro
ductG%$SumG%*TripleintG%*changevarG%/completesquareG%)distanceG%'equat
eG%*integrandG%*interceptG%)intpartsG%(leftboxG%(leftsumG%)makeprocG%*
middleboxG%*middlesumG%)midpointG%(powsubsG%)rightboxG%)rightsumG%,sho
wtangentG%(simpsonG%&slopeG%(summandG%*trapezoidG" }}}{EXCHG {PARA 11 
"" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "f:=
1/(x^2+6*x+14);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG*&\"\"\"F&,(*
$)%\"xG\"\"#F&F&*&\"\"'F&F*F&F&\"#9F&!\"\"" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 29 "fpar:=convert(f, parfrac, x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%fparG*&\"\"\"F&,(*$)%\"xG\"\"#F&F&*&\"\"'F&F*F&F&\"#
9F&!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "newf:=completes
quare(f, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%newfG*&\"\"\"F&,&*$
),&%\"xGF&\"\"$F&\"\"#F&F&\"\"&F&!\"\"" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 31 "This suggests the substitution " }{XPPEDIT 18 0 "u = x+3;
" "6#/%\"uG,&%\"xG\"\"\"\"\"$F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" 
{TEXT 259 21 "Assignment Part 3:   " }{TEXT -1 50 "Perform this substi
tution and compute the integral" }{TEXT 265 1 "." }}}{EXCHG {PARA 11 "
" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 266 21 "Assignmen
t Part 4  : " }{TEXT -1 21 "Compute the integral " }{XPPEDIT 18 0 "int
((x^3+5*x^2-7*x+1)/(x^2+x+1),x);" "6#-%$intG6$*&,**$%\"xG\"\"$\"\"\"*&
\"\"&F+*$F)\"\"#F+F+*&\"\"(F+F)F+!\"\"F+F+F+,(*$F)F/F+F)F+F+F+F2F)" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 40 "using techniques we lear
ned in this lab." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
{PARA 11 "" 1 "" {TEXT -1 0 "" }}}}{MARK "27 1 0" 0 }{VIEWOPTS 1 1 0 
1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }