mavzu. Bir o‘zgaruvchi funksiyasining integral hisobi -masala. Aniqmas integrallarni toping: 1) dxxxxxx       ( 1)( 5) ; 2) dxxxx     cos sin 3cos; 3) dxxxxx        6 3 3

2-MAVZU. BIR O‘ZGARUVCHI FUNKSIYASINING INTEGRAL HISOBI 5-masala. Aniqmas integrallarni toping: 1) dx x x x x x       ( 1)( 2 5) 4 7 5 2 2 ; 2) dx x x x     1 cos 2 sin 3cos ; 3) dx x x x x        3 3 2 6 3 3. ( 3) 3 ; 4)    . (1 ) 12 5 3 4 2 dx x x x Yechish. 1) Integral ostidgi funksiya to‘g‘ri kasrdan iborat. Kasrning maxrajidagi 2 5 2 x  x  kvadrat uchhad ko‘paytuvchilarga ajralmaydi, chunki 4 0. 4 2  q    p U holda kasrni ( 1)( 2 5) 1 2 5 4 7 5 2 2 2            x x Bx C x A x x x x x ko’rinishda yozib olamiz. Tenglikning chap va o‘ng tomonlarini umumiy maxrajga keltiramiz va suratlarni tenglashtiramiz: 4 7 5 ( 2 5) ( )( 1). 2 2 x  x   A x  x   Bx  C x  A, B, C koeffitsiyentlarni topamiz:            : 5 5 . : 4 , 1:16 8 , 0 2 x A C x A B x A 11 Bundan A  2, B  2, C  5. Shunday qilib,                        2 5 ( 2 5) 2ln | 1| 2 5 2 5 1 2 ( 1)( 2 5) 4 7 5 2 2 2 2 2 x x d x x dx x x x x x dx dx x x x x x . 2 1 2 3 2ln | 1| ln | 2 5 | ( 1) 2 ( 1) 3 2 2 2 C x x x x arctg x d x              2) Integralda almashtirishlar bajaramiz: 3 . 1 cos 1 sin 3 1 cos 3 3cos 1 sin 1 cos 2 sin 3cos dx x I1 C x x dx dx x x x dx x x x                    1 I integralni universal trigonometrik o‘rniga qo‘yish orqali ratsionallashtiramiz:                x arctgt t dt dx t t x t t x x t tg dx x x I , 1 2 , 1 1 , cos 1 2 , sin 2 1 cos 1 sin 2 2 2 2 1           2 2 2 2 1 2 1 1 1 1 2 1 t dt t t t t                2 2 2 2 2 1 (1 ) 1 2 1 1 2 t d t t t tdt dt dt t t t . 2 2ln cos 2 2 ln1 2 ln |1 | 2 2 x x tg x tg x  t   t  tg     Demak, . 2 2ln cos 2 3 1 cos 2 sin 3cos C x x dx x tg x x x         3) 6 x  3  t belgilash kiritamiz, chunki EKUK(2,3,6)  6. Bundan 3, 6 . 6 5 x  t  dx  t dt U holda              t dt t t t t dx x x x x 5 3 2 4 3 3 2 6 6 3 3. ( 3) 3           t dt t t t dt t t 6 ( 1) 1 1 6 4 4 2 3 ( 3) . 5 6 ( 3) 7 6 5 6 7 6 6 5 6 7 6 5 7  t  t  t  C  x   x   x  C 12 4). Integral ostidagi funksiyani standart shaklda yozib olamiz: 1 . 3 2 4 1 12 17         x x Demak, . 3 2 , 4 1 , 12 17 m   n  p  Bundan 1. 1     p n m Chebishevning uchinchi o‘rniga qo‘yishidan foydalanamiz: 4 3 1 4 1 1 x  x t yoki ( 1) 1. 4 3 1 x t   Bundan , 1 3 1 4 4           x x t ( 1) , 3 4 x  t  12 ( 1) . 2 3 5 dx t t dt     U holda               dx  t t t t t dt x x x 3 2 3 5 2 3 3 3 1 17 2 12 5 3 4 2 12 ( 1) ( ( 1) ) ( 1) (1 )            t t dt t dt 2 2 4 5 3 2 3 17 2 12 ( 1) 12 . 1 5 12 5 12 3 5 4 4 5 C