Integration by Substitutions

Transcription

1 NTEGRATON BY SUBSTTUTONS ntegration by Substitution the main idea One of the very powerful method for finding integrals analytically is the method of substitution There are many different substitutions that are very useful for evaluating integrals Some of the most important substitutions will be presented in the net chapters too The main aim of substitution using is to find a new integral equivalent to the given one that is easier to evaluate The idea is to change the independent variable in the integral f ( ) to a new variable t by means of a simple connecting formula ϕ () t f follows from this formula that: ( )' ϕ ( t ) ' t ϕ ' ( t ) dt and f ϕ( t) Consequently, f ( ) f ϕ( t ) ϕ' ( t) dt which is hopefully simple to evaluate t ψ Sometimes is better to use substitution Algorithm for evaluating the integral f by substitution Step Definite the best substitution formula depending on you problem Step Substitute t for in the integrand, calculate from the f t ' t dt substitution formula and find a new integral ϕ() ϕ () Step 3 Evaluate the new integral Step 4 Transform the answer F( t) to the variable Maple commands >:int(f,); >nt(f,); evaluating the integral of the function f and is a variable; >with(student):changevar(t^,); substituting the variable t for with a formula (in this eample t ) in the integral >:value(%);

2 evaluating the last result (integral) that is written in our Maple program 0 + Mathematical Solution The substitution is t > 0 t tdt Then tdt dt 0 arctgt + C arctg + C t + t t + Solution with Maple >0:int(/(*(+)*sqrt()),); 0 : arctg Detailed Solution with Maple ) definite the integral into the program STUDENT: >with(student): >0:nt(/(*(+)*sqrt()),); 0 : ( + ) ) substitute t for : >changevar(t^,0); t dt t + t 3) evaluate the new integral: >0:value(%); tarctan() t 0 t 4) go back to the variable : >0:subs(tsqrt(),0); arctan 0

3 4 Mathematical Solution The substitution is dt Then t t t t t t t dt dt 4 t t t arcsint C arcsin + + C Detailed Solution with Maple >restart: with(student): >:nt(/(*sqrt(^-4)),): >changevar(/t,); :value(%); >:subs(t/,); : arcsin 3 e e Mathematical Solution The substitution is { e t 0 } integral function from it t Then ln( t ) dt t 3ln( t ) e t ( t ) > because to make free the 3 dt dt t t t 3

4 ( t ) 3 t t dt t + + C 3 3 e e + e + C 3 Detailed Solution with Maple >:nt(ep(3*)/sqrt(-ep()),); >changevar(sqrt(-ep())t,); >:value(%); >:subs(tsqrt(-ep()),); ( / ) 4 ( 3 / : e e + e ) 3 sin cos 3 4 cos Mathematical Solution cos t The substitution is { } dt 3 arct gt + C t arctg cos Then dt cos sin + C Detailed Solution with Maple >3:nt(*sin()*cos()/(cos()^4-),); >changevar(cos()^t,3); >3:value(%); >3:subs(tcos()^,3); 3 : arctanh cos ( ) 4 sin

5 Mathematical Solution 4 t Then The substitution is { } t, 4t dt sint 4t dt 4 4 sintdt 4cos t C 3 t 4 4cos + C + Detailed Solution with Maple >4:nt(sin(^(/4))/()^(3/4),); >changevar(^(/4)t,4); 4:value(%); >4:subs(t^(/4),4); ( / 4) 4: 4cos ntegration of Function f a +b+ by Substitution There will be presented substitutions for evaluating the net kind of integrals: M + N J, a + b + c M + N J a + b + c t follows from b c b a + b + c + + a a 4a b b that the substitution is + t or t and dt a a +

6 Mathematical Solution t Then + t + and The substitution is { } ( t + ) t dt dt d t + t + t + ( ) t + ln t + + C ln + + C Detailed Solution with Maple >:nt((*-)/(^-*+),); >changevar(-t,); :value(%); >:subs(t-,); ( ) : ln + >simplify(); : ln( +) Mathematical Solution The substitution is { + t} Then t + 4 and dt t + 6 arctg + C arctg C + t + 4 Detailed Solution with Maple >6:nt(/(^+4*+8),); >changevar(+t,6); 6:value(%); >6:subs(t+,6); 6: arctan + 6

7 Mathematical Solution The substitution is { + t} Then + 4+ t + and 3t 7 t 7 dt 3 dt 7 dt t + t + t + 3 ( ) 7 t d t + + arctgt 3 ln t 7arctgt C ln 4 7arctg ( ) C Detailed Solution with Maple >7:nt((3*-)/(^+4*+),); >simplify(changevar(+t,7)); >7:value(%); >7:simplify(subs(t+,7)); 3 7: ln( + 4+ ) 7arctg( + ) Mathematical Solution 3 The substitution is t Then 3+ t 4 6 8t 8t 8 dt 8 dt t 6t 6 7

8 t dt dt ( t) ( t) ( 4 ) d t t t dt 4t ln ln 6t + C 4t + ln 4t ln 4t + ln 4t ln 4t + + C ln 4t 3ln 4t + + C ln 4 4 3ln 4 + C ln4 ln 3ln 3ln + C ln ln 3ln + C Detailed Solution with Maple >8:nt((7-8*)/(*^-3*+),); >8:simplify(changevar(-3/4t,8)); >8:value(%); >8:simplify(subs(t-3/4,8)); 8: ln 3ln ln 9 + Mathematical Solution The substitution is { t} Then + t + and ( t + ) t 9 dt dt d t + t + t + t + t + + C + +C Detailed Solution with Maple >9:nt((*-)/sqrt(^-*+),); >9:simplify(changevar(-t,9)); 8

9 >9:value(%); >9:simplify(subs(t-,9)); 9: Mathematical Solution The substitution is { t} Then ( 4 t ) and 3t 3 t 0 dt dt dt t t 4 t t 3 d( 4 t ) arcsin + C 4 t 3 t 34 ( t ) arcsin + C arcsin + C 3 Detailed Solution with Maple >0:nt((3*-)/sqrt(9+6*-3*^),); >0:simplify(changevar(-t,0)); >0:value(%); >0:simplify(subs(t-,0)); 0 : 3 arcsin Selftraining Problems Evaluate by substitutions the net integrals: 9

10 e, 4 e + 9 e Answer arctg + C 6 3, Answer ln + C 3 a a Answer 3 ± arccos + C a e + 4, e, substitution a ( t ) Answer ln e + C 4,, 4 + 3, + + ln arctg + + C, ln arctg + + C Answer ( ) 6 Answer + 7, Answer 7 ln C,

11 Answer 8 ( 3 3) ln arctg + + C Answer 3 9 ln + ln + C , Answer ln C + 3, + 3+ Answer C 4 +, + 3 Answer + + ln C 4 3, 4 3 Answer 3 ln C 4 Selfcontrol Test Evaluate the integrals: co s 4, 4

12 sin4, cos + 4 sin, 4 cos + 3 e, 3, , Selfcontrol Questions ) Eplain the idea of integration by substitutions ) Show an eample of integration by substitutions 3) Eplain the meaning of the Maple commands: with(student), changevar(t^,), simplify(changevar(t^,)), :value(%),:subs(tsqrt(),), :simplify(subs(tsqrt(),))