ratsional funksiyani (kasrni) integrallash. ba`zi bir trigonometrik ifodalarni integrallash

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1576157184.doc n n n m m m n m a x a x a b x b x b x p x q x r ... ... ) ( ) ( ) ( 1 1 0 1 1 0 + + + + = = - - ) ( ) ( ) ( ) ( ) ( ) ( 1 x p x q x m x p x q x r n n m n m + = = - k k k k k n m q px x c x b q px x c x b q px x c x b a x a a x a a x a x p x q ) ( ... ) ( ) ( ... ) ( ) ( ) ( 2 2 2 2 2 2 1 1 2 2 1 + + + + + + …

$ko`ra, integrallanishi jihatidan bir-biridan farq qiladigan quyidagi to`rt xil sodda (elementar) ratsional kasrlarni integrallash masalasiga keltiriladi: ; bu yerda a0,b0,c0, a,p va q lar berilgan sonlar: 2(n(n, p 2–4q int(a0/(x-a),x)=int(a0/(x-a),x); shape \* mergeformat 2) . > int(a0/(x-a)^n,x)=int(a0/(x-a)^n,x); shape \* mergeformat 3) > int((b0*x+c0)/(x^2+p*x+q),x)=int((b0*x+c0)/(x^2+p*x+q),x); 4) bu yerda . oxirgidan ko`rinadiki, agar ko`rinishdagi integrallarni ololsak, masala haldir. agar bu integralda m=1 desak, bo`ladi. endi, bo`lgan holni qaraylik. oxirgi integralni bo`laklab integrallaymiz: demak, rekkurrent (qaytma) formulani olamiz. endi, m=2,3,…,n qiymatlarni berish natijasida jn ni topamiz. shunday qilib, berilgan ratsional kasrning, yuqoridagi usullar bilan, integralini topa olamiz, ya`ni ratsional kasr integrali elementar funksiyadan iborat bo`lar ekan. > restart; > ik2:=int(1/(t^2+a^2)^m,t); > m:=1:value(ik2); > m:=2:value(ik2); > m:=3:value(ik2); > m:=4:ik2:=value(ik2); > m:=4:a:=1:ik2:=value(ik2); n=2,3 bo`lganda integralini formula asosida topish. 1) > restart; > ik3:=int((a*(t-p/2)+b)/(t^2+a^2)^m,t); > m:=2:ik3:=value(ik3); > m:=3:ik3:=value(ik3); 2) > a:=4:b:=5:p:=6:q:=25:a:=sqrt(4*q-p^2)/2: > with(student): > changevar(t=x+p/2, (ik3, t), x); 3) n=2,3 bo`lganda bevosita topish. > restart; > ik4:=int((4*x+5)/(x^2+6*x+25)^2,x); …$

$2,x,4); > k:=solve({kp0,kp1,kp2,kp3,kp4},{a1,a2,a3,a4,a5}); > a1:=0; a4:=-2/3; a3:=2/3; a5:=-2/3; a2:=1;y1(x); > int(y1(x),x); topilgan koeffitsientlar asosida berilgan kasirni sodda kasirlarga ajratilgan ko`rinishin yozamiz: bebosita sodda kasirlarga ajratish: > factor(x^5+x^2); > with(genfunc): rgf_pfrac((x^6+1)/(x^5+x^2), x); > int((x^6+1)/(x^5+x^2), x)=int((x^6+1)/(x^5+x^2),x); ba`zi bir trigonometrik ifodalarni integrallash 1. integralda r o`z argumentlarining ratsional funksiyasi bo`lsin. u holda, bu integralda umumiy trigonometrik almashtirish deb ataluvchi almashtirish yordamida ratsional funksiya integraliga kelinadi. haqiqatdan ham, ekanligini e`tiborga olsak, , bu yerda - ratsional funksiya. 1-misol. bu almashtirish yordamida integrallar jadvalidagi 16-formulani keltirib chiqarish mumkin: 17-formula esa ekanligidan va 16-formuladan kelib chiqadi. 2-misol. integralni toping. 1) o`zgaruvchini almashtirish yordamida integralni topish. > restart; > with(student): ia13:=changevar(x=2*arctan(t),int(1/(1+sin(x)+cos(x)),x),t); > ia13:=value(%); > ia13:=changevar(t=tan(x/2), (ia13, t),x); 2) bevosita integrallash. > restart; > int(1/(1+sin(x)+cos(x)),x)=int(1/(1+sin(x)+cos(x)),x); shape \* mergeformat 2. = , r sinx ga nisbatan toq ratsional funksiya. bu yerda ham umumiy almashtirishdan yoki qulayroq bo`lgan cosx=t dan foydalanish mumkin. 3-misol. bu integral sinx ga nisbatan toq: = …$

$qo`yish) qulayroq, chunki cosxdx=d(sinx) dir. 4-misol. bu integral cosx ga nisbatan toq: = = - 1) o`zgaruvchini sinx=t almashtirish yordamida integralni topish. > restart; > with(student): > it5:=changevar(sin(x)=t,int(cos(x)^3/sin(x)^4,x),t); > it5:=value(%); > it5:=changevar(t=sin(x), (it5, t),x); 2) bevosita integrallash. > restart; > int(cos(x)^3/sin(x)^4,x)=int(cos(x)^3/sin(x)^4,x); shape \* mergeformat integralda r- o`z argumentlarining ratsional funksiyasi bo`lib, sinx va cosx larga nisbatan juft funktsiya bo`lsa: quyidagich almashtirish qilamiz. , 5-misol. da juft bo`lgani uchun integralni tgx=t almashtirish yordamida topamiz. 1) o`zgaruvchini tgx=t almashtirish yordamida integralni topish. > restart; > with(student): > it8:=changevar(tan(x)=t,int(1/(1+cos(x)^2),x),t); > it8:=value(%); > it8:=changevar(t=tan(x), (it8, t),x); 2) bevosita integrallash. > restart; > int(1/(1+cos(x)^2),x)=int(1/(1+cos(x)^2),x); shape \* mergeformat 6-misol. da sinx va cosx larga nisbatan juft funktsiya bo`lgani uchun integralni tgx=t almashtirish yordamida topamiz. ekanini etiborga olib, = = 1) o`zgaruvchini tgx=t almashtirish yordamida integralni topish. > restart; > with(student): > it9:=changevar(tan(x)=t,int(1/(sin(x)^2- 4*sin(x)*cos(x)+ 5*cos(x)^2),x),t); shape \* mergeformat > it9:=value(%); > it9:=changevar(t=tan(x), (it9, t),x); 2) bevosita integrallash. …$

$sol. | |= = =|t=tgx|= 1) o`zgaruvchini tgx=t almashtirish yordamida integralni topish. > restart; > with(student): > it10:=changevar(tan(x)=t,int(tan(x)^3,x),t); shape \* mergeformat > it10:=value(%); > it10:=changevar(t=tan(x), (it10, t),x); 2) bevosita integrallash. > restart; > int(tan(x)^3,x)=int(tan(x)^3,x); shape \* mergeformat 9-misol. tgx=t almashtirish yordamida integralni topish > restart; > with(student): > it12:=changevar(tan(x)=t,int(tan(x)^5,x),t); > it12:=changevar(t=tan(x), (it12, t),x); 10-misol. bu integralni topishda yuqoridagilardan farqli quyidagicha almashtrish qilamiz: bu holda = 1) o`zgaruvchini almashtirish yordamida integralni topish. > restart; > with(student): > it13:=changevar(2+3*sin(2*x)=t, int(cos(2*x)/(2+3*sin(2*x))^(2/3),x),t); > it13:=changevar(t=2+3*sin(2*x), (it13, t),x); 2) bevosita integrallash. > restart; > int(cos(2*x)/(2+3*sin(2*x))^(2/3),x)= int(cos(2*x)/(2+3*sin(2*x))^(2/3),x); shape \* mergeformat 3. sinus va kosinus toq darajali bo`ganda integrallash a) bu yerda t=cosx. b) bu yerda t=sinx. > restart; > it3m:=int(sin(x)^(2*m+1),x)=int(sin(x)^(2*m+1),x); shape \* mergeformat > m:=1:it3m; shape \* mergeformat > m:=2:it3m; shape \* mergeformat 11-misol.. integralni hisoblang. yechish. va ekanligini hamda almashtirish kiritib, quyidagini hosil hilamiz: > restart; > it5m:=int(sin(x)^3*cos(x)^4,x)=int(sin(x)^3*cos(x)^4,x); shape \* mergeformat 4. sinus va kosinus juft …$

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"ratsional funksiyani (kasrni) integrallash. ba`zi bir trigonometrik ifodalarni integrallash" haqida

1576157184.doc n n n m m m n m a x a x a b x b x b x p x q x r ... ... ) ( ) ( ) ( 1 1 0 1 1 0 + + + + = = - - ) ( ) ( ) ( ) ( ) ( ) ( 1 x p x q x m x p x q x r n n m n m + = = - k k k k k n m q px x c x b q px x c x b q px x c x b a x a a x a a x a x p x q ) …

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