Đang tải... (xem toàn văn)
Tài liệu tham khảo các dạng bài tập liên quan đến các vấn đề trong tích phân. Đây là các dạng bài tập tích phân được trình bày theo hình thức tiếng Anh.
Chapter 5e Integral of Irrational Function 8/ 2 5 * 6 7 9 11 12 * 13 * 14 * 15 17 * 18 * .19 .20 22 22 23 25 27 .28 .29 .31 32 .33 36 37 37 38 38 39 39 39 40 41 42 42 43 44 * .44 45 46 47 47 1 48 49 50 51 52 52 55 56 58 .58 59 8/ ( ) 2 dx I x q ax bx c = − + + ∫ ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 dx 1 dt 1 Put x q dx , t t x q t x q ax bx c 1 1 1 2q 1 ax bx c a q b q c a q b q c t t t t t a b 2aq 1 aq bq c a t b 2aq t aq bq c t t t − = ⇒ = − = − − + + + + = + + + + = + + + + + ÷ ÷ ÷ ÷ + = + + + + = + + + + + ∫ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 dt dx t 1 1 x q ax bx c a t b 2aq t aq bq c t t dt dt t 1 a t b 2aq t aq bq c t aq bq c t b 2aq a t − ⇒ = − + + + + + + + − = = − + + + + + + + + + + ∫ ∫ ∫ ∫ 2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 dt Put aq bq c A, b 2aq 1 a t t aq bq c aq bq c aq bq c dt b 2aq b 2aq 1 a t A 2A A 4A b 2aq b 2q 4Aa b 2aq a Put t y dy dt, N 2A A 4A 4 aq bq c = − + + = + ÷ + + ÷ + + + + + + ÷ = − + + ÷ + + − ÷ ÷ + + − + + = ⇒ = = − = ÷ + + ∫ ∫ ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 dy Ady I If N 0 N n 4Aa b 2aq , y N y N A 4Aa b 2aq 4a a.q b.q c b 4abq 4 aq 4ac b b c b A 0 A m a.q b.q c 0 a q 0 2a a 4 b 4ac b a q a 0 2a 4 = − = − ≥ ⇒ = ⇒ ≥ + + + ≥ + ⇔ + + ≥ + + ⇔ ≥ > ⇒ = ⇒ + + ≥ ⇔ + + − ≥ ÷ − ⇔ + + ⇔ > ÷ ∫ ∫ ( ) ( ) ( ) 2 2 2 2 2 m.dy I m.ln y y n y n B 1 b 2q y t 2A x q 2 aq bq c ⇒ = − = − + + ÷ + + = + = + ÷ − + + ∫ 3 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 b 2aq b 2aq b 2aq a a y n t t t 2A A 4A aq bq c aq bq c t aq bq c t b 2aq a aq bq c t aq bq c t b 2aq a t t aq bq c + + + ÷ + = + + − = + + ÷ ÷ + + + + ÷ + + + + + = + + + + + + + ÷ = ÷ + + ÷ ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 t aq bq c t b 2aq a * t 1 2q 1 1 1 a q b q c a q b q c t t t t t ax bx c ax bx c y n x q aq bq c 1 1 x q t t x q + + + + + = + + + + + = + + + + ÷ ÷ ÷ ÷ + + = + + ⇒ + = − + + = + ⇒ = ÷ − ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 ln y y n dx I 1 x q ax bx c aq bq c 1 b 2aq 1 ax bx c ln aq bq c x q x q 2 aq bq c aq bq c + + ÷ ⇒ = = − − + + + + + + + ÷ ÷ = − + + + + ÷ ÷ − − + + + + ÷ ÷ ∫ 4 ( ) 2 dx * I x 1 1 x = − − ∫ ( ) 2 2 2 2 2 dx 1 * I Put : x 1 t x 1 1 x dt 1 1 2t dx 1 x 1 1 t t t = − = − − − + ⇒ = − = − + = − ÷ ∫ ( ) ( ) ( ) 2 2 1 1 2 dt dt t I 1 2t 1 2t t. t because1 x 0 x 1 x 1 0 t 0 t t d 2t 1 1 2t 1 1 2 1 x I 1 2t 1 C 1 2 2 x 1 1 x 1 2t 1 2 − + ⇒ = − = − − − − − > ⇔ < ⇒ − < ⇒ < ⇔ = − − − − − − = − = − = − − − = − − − = − + − + − − − + ∫ ∫ ∫ ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 3x 4 VD : dx x 6x 8 1 x 3 x 6x 8 Put x 3 t x t 3 dx dt 3 3 t 4 3x 4 dx dx x 6x 8 1 t d 1 t 3t.dt dt 3 13 13arcsin t 2 1 t 1 t 1 t d 1 t 2t.dt 3 1 t 13arcsin t 3 x 6x 8 13arcsin x 3 C + − + − = − − − + − − = ⇒ = + ⇒ = + + + = − + − − − = + = − + − − − − = − = − − + = − − + − + − + ∫ ∫ ∫ ∫ ∫ ∫ For evaluating integral 2 R x, ax bx c + + ÷ ∫ we can make a trigonometric change of variables: 5 ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 b c b ax bx c a x 2a a 4a b b 4ac a x a u d If b 4ac 0, a 0 2a 4a b 4ac b a x a u d If b 4ac 0, a 0 2a 4a b 4ac b a x a d u If b 4ac 0, a 0 2a 4a ÷ + + = + + − ÷ ÷ − ÷ = + − = − − > > ÷ ÷ − ÷ = + + = + − < > ÷ ÷ − ÷ = − − + = − − − > < ÷ ÷ 2 2 2 2 2 2 2 2 2 d R x, ax bx c R u, u d Put u sin t R x, ax bx c R u, d u Put u d.sin t R x, ax bx c R u, u d Put u d.tgt + + = − = ÷ ÷ + + = − = ÷ ÷ + + = + = ÷ ÷ ∫ ∫ ∫ ∫ ∫ ∫ * ( ) ( ) 3 2 x 2 3 dx 1 I sin arctan C 3 3 x 4x 7 + = = + ÷ ÷ + + ∫ ( ) ( ) ( ) ( ) ( ) ( ) 2 2 3 3 2 2 2 2 2 3 2 3 dx du * I x 4x 7 x 2 3 Put u x 2 I x 4x 7 u 3 3 3 Put u 3.tan t du , u 3 3 tan t 1 cos t cos t x 2 3 3.cos t.dt sin t 1 u 3 1 I sin arctan sin arctan 3 3 3 3 3 cos t. 3 = + + = + + = + ⇒ = + + + = ⇒ = + = + = + ⇒ = = = = ÷ ÷ ÷ ∫ ∫ ∫ 6 ( ) ( ) ( ) 2 2 3 2 2 2 b x dx 4a 2a * I sin arctan with a 0, 4ac b 0 a. 4ac b 4ac b ax bx c a ÷ + ÷ ÷ = = > − ≥ ÷ − − ÷ + + ÷ ∫ ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 3 2 2 2 2 3 2 2 2 2 2 2 2 dx b 4ac b * I a 0, 4ac b 0 ax bx a a x 2a 4a ax bx c b 4ac b 1 du Put u x , m I , 2a a. a 4a u m m.dt m Put u m.tan t du , u m m tan t 1 cos t cos t − = > − ≥ + + = + + ÷ ÷ ÷ + + − = + = ⇒ = ÷ ÷ + = ⇒ = + = + = ∫ ∫ ( ) ( ) 3 3 2 2 2 3 2 2 2 2 2 2 2 2 1 du 1 m.dt 1 cos t.dt 1 sin t I . a. a a. a a. a a. a.m cos t. m m u m cos t. cos t b x 1 u 4a 2a sin arctan sin arctan m 4ac b a. 4ac b 4ac b a. a. 4a a ⇒ = = = = + ÷ ÷ + ÷ ÷ = = ÷ ÷ − − − ÷ ÷ ÷ ÷ ∫ ∫ ∫ ( ) ( ) ( ) ( ) c 1 2 3 c 2b b 2a 2a 2 2 c b x dx 4a 2a I lim .sin arctan m a. 4ac b ax bx c b c 4a 4a 2a . lim sin arctan m a. 4ac b a. 4ac b +∞ →+∞ − − →+∞ + ÷ ÷ ÷ ⇒ = = ÷ − + + ÷ + ÷ ÷ ÷ = = ÷ − − ÷ ∫ 7 ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 n 2 2 2 2 n n 2 2 2 2 2 2 2 n n n 2 2 2 dx b 4ac b * I a 0, 4ac b 0 ax bx a a x 2a a ax bx c b 4ac b 1 du Put u x , m I , 2a a a u m m.dt m Put u m.tan t du , u m m tan t 1 cos t cos t 1 du 1 m.dt I m a a u m cos t. cos t − = > − ≥ + + = + + ÷ ÷ ÷ + + − = + = ⇒ = ÷ ÷ + = ⇒ = + = + = ⇒ = = + ∫ ∫ ∫ ( ) n 2 n n 1 n cos t .dt 1 m a − − = ÷ ∫ ∫ ( ) ( ) ( ) n 2 2 n 1 n n 2 2 2 2 cos t .dt dx 1 * I a 0, 4ac b 0 m a ax bx c 4ac b b u m , u x , t arctan 2a m a − − = = > − ≥ + + − = = + = ÷ ÷ ∫ ∫ ( ) 2 2 2 2 2 2 2 b x dx 1 dt t 1 2a n 2 : I arctan a m a.m ax bx c 4ac b 4ac b a. 4a 4a 2 2ax b arctan 4ac b 4ac b ÷ ÷ + ÷ ÷ = = = = = ÷ ÷ + + ÷ − − ÷ ÷ ÷ + ÷ ÷ = ÷ ÷ − − ∫ ∫ ( ) ( ) 2 2 2 3 2 3 2 2 3 2 3 dx 1 1 1 cos 2t n 4 : I cos t .dt .dt 2 a .m a .m ax bx c 1 sin 2t 1 u 1 u t arctan .sin 2.arctan 2 m 2 m 2a .m 2a .m b x u 0, x u 2a + = = = = + + = + = + ÷ ÷ ÷ ÷ = − ⇒ = = +∞ ⇒ = +∞ ∫ ∫ ∫ 8 ( ) ( ) ( ) ( ) 2 2 3 2 0 b/2a 3/2 3/2 2 2 2 2 2 3 3/2 2 dx 1 u 1 u . arctan .sin 2.arctan m 2 m 2a .m ax bx c 1 sin 1 sin . . 2 2 2 2 4ac b 4ac b 2a . 2a . 4a 2a 4a sin . 2 2 4ac b +∞ +∞ − ⇒ = + ÷ ÷ ÷ + + π π π π = + = + ÷ ÷ − − ÷ ÷ π π = + ÷ − ∫ ( ) ( ) 3 n 3/2 2 n 2 2 i 1 c c.n b * lim 1 sin arctan 2 4ac b a i.c b.n.i.c n c →+∞ = →+∞ ÷ ÷ = − ÷ ÷ − + + ∑ ( ) ( ) [ ] ( ) ( ) ( ) i i i 3 2 i i 3 n n 2 3/2 3 n n 2 2 2 i 1 i 1 2 c c 2 n 3/2 n n 2 i 1 c c 1 c i.c Put f x , x 0, c , x , x n n ax bx c c.n c 1 I lim lim n a i.c b.n.i.c n c a i.c b.n.i.c n c n c 1 lim lim n i.c i.c a b c n n →+∞ →+∞ = = →+∞ →+∞ →+∞ →+∞ = →+∞ →+ = ∈ ∆ = = + + ⇒ = = + + + + ÷ ÷ = = ÷ + + ÷ ÷ ÷ ∑ ∑ ∑ ( ) ( ) c n i 3/2 c 2 i 1 0 2 dx f x . x lim ax bx c b 1 sin arctan 2 4ac b →+∞ = ∞ ∆ = + + ÷ ÷ = − ÷ ÷ − ∑ ∫ ( ) p m n x a bx dx+ ∫ with m, n, p is rational number. The Russian mathematicant Trebushep prove that the upper integral only can be expressed in elementary function in 3 follow cases: 1/ p is an interger, when that, put s x t= with s is the least common multiple of m, n. 2/ m 1 n + is an interger, put s s a bx t+ = with s is the denominator of p. 3/ m 1 p n + + is an interger, put n s ax b t − + = with s is the denominator of p. 9 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 10 1 1 2 4 10 4 4 3 3 10 10 9 10 2 8 9 8 9 4 4 dx Ex : I x x 1 x x 1 So p 10 is a interger, we have case 1/ Put x t dx 4t .dt t 1 1 d t 1 d t 1 4t .dt I 4 dt 4 t t 1 t 1 t 1 t 1 4 4 1 4 8 t 1 9 t 1 2 x 1 9 x 1 − − ÷ = = + ÷ ÷ + = − = ⇒ = + − + + ÷ ⇒ = = = − ÷ + + + + − − = + = + + + + + ∫ ∫ ∫ ∫ ∫ ∫ ( ) ( ) ( ) ( ) ( ) ( ) p m n p a a 1 m.a n.a a 1 p p i a m 1 1 n.a.i a m 1 1 i n.a p i i i p i p p i 1 i 1 * I x . a b.x .dx where p is a interger Put x t dx a.t .dt I a t . a b.t .t .dt I a t . C b.t .a .dt a C t .b a .dt Put n.a.i a m 1 1 c − − + − + + − − − = = = + = ⇒ = ⇒ = + = = ÷ ÷ ÷ ÷ + + − = ⇒ ∫ ∫ ∑ ∑ ∫ ∫ i c 1 i p i p p p i c i p i p i 1 i 1 C t .b a I a C t .b a .dt a. c 1 + − − = = = = ÷ ÷ + ∑ ∑ ∫ ( ) ( ) ( ) ( ) ( ) ( ) i n.i m 1 i p i i n.i m 1 i p i p p p p p m n i 1 i 1 1 i n.i m 1 i p i i i p i 1 p p p p p m n 1 i 1 i 1 0 0 C x .b a C x .b a I x . a b.x .dx a. a n.i m 1 n.i m 1 C x .b a C .b a I x . a b.x .dx n.i m 1 n.i m 1 + + − + + − = = + + − − = = = + = = + + + + ⇒ = + = = + + + + ∑ ∑ ∫ ∑ ∑ ∫ 10