Integrallashning asosiy usullari differensial ostiga kiritish usuli. O‘rniga qo‘yish (o‘zgaruvchini almashtirish) usuli. Bo‘laklab intеgrallash usuli

INTEGRALLASHNING ASOSIY USULLARI Differensial ostiga kiritish usuli. O‘rniga qo‘yish (o‘zgaruvchini almashtirish) usuli. Bo‘laklab intеgrallash usuli 7.2.1. Aniqmas integralda x o‘zgaruvchidan boshqa u  u(x) o‘zgaruvchiga o‘tish orqali  f (x)dx integralni jadval integraliga keltirib integrallash usuliga differensial ostiga kiritish usuli deyiladi. Bu usulda f (u)du  d ( f (u)) formulaga asoslangan quyidagi almashtirishlar keng qo‘llaniladi: du  d(u  a), ( ), 1 d au b a du   ( ), 2 1 2 udu  d u cosudu  d(sin u), sin udu  d(cosu), (ln ), 1 du d u u  ( ), cos 1 2 du d tgu u  (arcsin ) 1 1 2 du d u u   , ( ) 1 1 2 du d arctgu u   , a,b o‘zgarmas sonlar. 303 1 misol. Integrallarni differensial ostiga kiritish usuli bilan toping: 1)  ; 16 9 2 x dx 2) ; 2 e xdx x 3)  ; 1 2 3 dx x arctg x 4)   . sin cos cos sin dx x x x x 1)             C u arctg u du x d x x dx 4 4 1 3 1 3 4 1 16 (3 ) (3 ) 3 1 16 9 2 2 2 2 . 4 3 12 1 C x arctg  2)         . 2 1 2 1 2 1 ( ) 2 2 1 2 2 2 e xdx e d x e du e C e C x x u u x 3)          . 4 1 4 ( ) 1 4 4 3 3 2 3 C arctg x C u dx arctg xd arctgx u du x arctg x 4)              ln | | ln |sin cos | . sin cos (sin cos ) sin cos cos sin u C x x C u du x x d x x dx x x x x 7.2.2. Aniqmas integralda integral ostidagi funksiyaning bir qismini u  u(x) o‘zgaruvchi bilan almashtirish orqali  f (x)dx integralni integrallash qulay bo‘lgan  f (u)du integralga keltirib integrallash usuliga o‘rniga qo‘yish (yoki o‘zgaruvchini almashtirish) usuli deyiladi. Bu usul   f (x)dx  f ((t))(t)dt (2.1) formulaga asoslanadi. Ayrim hollarda t (x) o‘rniga qo‘yish tanlashga to‘g‘ri keladi. U holda (2.1) formula o‘ngdan chapga qo‘llaniladi, ya’ni    f ((x))(x)dx f (t)dt . 2 misol. Integrallarni o‘rniga qo‘yish usuli bilan toping: 1) x x 3dx;   2) 1 cos sin 2 ; 2   x xdx 3) dx x x x   ln 1 ln ; 4)  . 4 2 2 dx x x 1) x  3  t o‘rniga qo‘yishni bajaramiz. U holda 3, 2 . 2 x  t  dx  tdt Shu sababli  x x  3dx   (t  3) t  2tdt  2 (t  3t )dt  2 4 2 ( 3) 2 ( 3) . 5 2 3 6 5 2 6 2 5 3 5 3 4 2 C x x C t t   t dt   t dt           304 2) 2 2 1 cos x  t deymiz. U holda sin 2 2 , 1 cos . 2 x   tdt t   x Bundan (1 cos ) . 3 2 3 1 cos sin 2 ( 2 ) 2 2 3 3 2 C x C t   x xdx   t  t dt         3) 2 1 ln x  t bo‘lsin. Bundan ln 1, 2 , 2 tdt x dx x  t   t  1 ln x.U holda                                C t t dt t t t t dt t t tdt dx x x x 1 1 ln 2 1 2 1 1 2 1 1 2 1 2 ln 1 ln 2 2 2 2 . 1 ln 1 1 ln 1 2 1 ln ln C x x x         4) x 2sin t, dx 2costdt, 4 x 2cost 2     deymiz. Bunda . 2 arcsin x t  U holda                dt ctgt t C t dt dt t t dt t t dx x x 2 2 2 2 2 2 2 sin sin 1 sin sin 4 cos                             C x x x C x x ctg 2 arcsin 2 sin arcsin 2 1 sin arcsin 2 arcsin 2 arcsin 2 C x x x      2 arcsin 4 2 . Ba’zan bajarilgan o‘rniga qo‘yishdan so‘ng shunday integral hosil bo‘ladiki, bu integralni boshqa o‘rniga qo‘yish orqali soddalashtirish yoki jadval integraliga keltirish lozim bo‘ladi. 3 misol. (8 1) 4 1 2 2 x x dx integralni toping. t x 1  o‘rniga qo‘yishni bajaramiz. U holda 2 t dt dx   va  (8 1) 4 1 2 2 x x dx . (8 ) 4 1 4 1 8 2 2 2 2 2                 t t tdt t t t dt Keyingi integralda 2 2 4  t  z o‘rniga qo‘yishdan foydalanamiz. Bundan , 8 4 2 2 tdt  zdz  t  z  . U holda     2 2 (8 t ) 4 t tdt . 2 2 1 ( 4) 4 2 2 C z arctg z dz z z zdz           305 z ni x orqali ifodalaymiz: . 1 4 1 4 4 2 2 2 x x x z t       Demak,  (8 1) 4 1 2 2 x x dx . 2 4 1 2 1 2 C x x arctg     7.2.3. Aniqmas integralda integral ostidagi ifodani udv ko‘paytma shaklida ifodalash va     udv uv vdu (2.2) formulani qo‘llash orqali  f (x)dx integralni integrallash qulay bo‘lgan  vdu integralga keltirib topish usuliga bo‘laklab integrallash usuli deyiladi. Bo‘laklab integrallash usuli bilan topiladigan integrallarni asosan uch guruhga ajratish mumkin: P x arctgxdx  ( ) , P(x)arcctgxdx,   P(x)ln xdx,  P(x)arcsin xdx ,  P(x) arccos xdx (bu yerda P(x)  ko‘phad) ko‘rinishdagi 1-guruh integrallari. Bunda dv  P(x)dx deb olish va qolgan ko‘paytuvchilarni u orqali belgilash qulay;  P(x)e dx, kx  P(x)sin kxdx,  P(x)cos kxdx ko‘rinishdagi 2-guruh integrallari. Ularni topishda u  P(x) va qolgan ko‘paytuvchilarni dv deb olish maqsadga muvofiq; e sin kxdx, kx  e kxdx kx cos  ko‘rinishdagi 3-guruh integrallari (2.2) formulani takroran qo‘llash orqali topiladi. 4 misol. Integrallarni bo‘laklab integrallash usuli bilan toping: 1)  arctgxdx; 2) xdx  2 ln ; 3) sin 2 ; 2  x xdx 4) e cos xdx. x   1)  arctgxdx integral 1- guruhga kiradi. U holda             dx x x xarctgx dx dv v x x dx arctgx u du arctgxdx 2 2 1 , , 1 , ln1 . 2 1 1 (1 ) 2 1 2 2 2 dx xarctgx x C x d x xarctgx          306 2) 1- guruh xdx  2 ln integraliga (2.2) formulani ketme-ket ikki marta qo’llaymiz:            x x xdx dx dv v x x dx x u du x xdx ln 2 ln , ln , 2ln , ln 2 2 2 ln 2 ln 2 ln 2 ln 2 . , ln , , 2 2 x x x x dx x x x x x C dx dv v x x dx x u du              3)  x sin 2xdx 2 integral 1- guruhga kiradi. U holda             x x x x xdx xdx dv v x u du xdx x xdx cos2 cos2 2 1 2 cos2 sin 2 , , 2 , sin 2 2 2 2            x x x x x xdx xdx dv v x u du dx sin 2 2 1 sin 2 2 1 cos2 2 1 2 sin 2 cos2 , , , 2 cos2 . 4 1 sin 2 2 1 cos2 2 1 2   x x  x x  x  C 4)  e xdx x   cos integral uchinchi guruh integrali bo’lgani sababli (2.2) formulani takroran qo‘llaymiz:                 x xdx dv v e u du e dx I e xdx x x x sin cos , , , cos e x   e xdx  x x        sin sin 1             x xdx dv v e u du e dx x x cos sin , ,         e x   e x   e x x x x             cos cos 1 sin 1 I x x e x 2 2 2 sin cos           Bundan . sin cos 2 2 C x x I e x            307 Ko‘rsatilgan uch guruh bo‘laklab integrallanadigan barcha integrallarni o‘z ichiga olmaydi. Masalan,  x xdx 2 sin integral yuqorida keltirilgan integral guruhlariga kirmaydi, lekin uni bo‘laklab integrallash usuli bilan topish mumkin:                ln |sin | . , sin , sin 2 2 xctgx ctgxdx xctgx x C