mẹo học công thức toán học
cụng thuc toan' hoc toan hoc MC LC i Mc lc 3.9 S hc 1.1 Ghi S ng thc lng giỏc nghch o 3.10 ng thc Tớch vụ hn 3.11 Tớch Phõn 3.12 S Phc i S 3.13 ng thc lng giỏc 2.1 S i s 3.13.1 ng thc lng giỏc c bn 2.2 Phộp Toỏn i S 2.3 Toỏn S Nguyờn 3.13.2 ng thc lng giỏc Tun hon, i xng v tnh tin 2.4 Toỏn Phõn S 3.13.3 ng thc Pytago 2.5 Toỏn S Phc 3.13.4 ng thc Tng v hiu ca gúc 2.6 Toỏn Ly tha 3.13.5 Cụng thc h bc 2.7 Toỏn Cn 3.13.6 Cụng thc gúc chia ụi 2.8 Toỏn Log 3.13.7 ng thc Bin tớch thnh tng 2.9 Hm s 3.13.8 ng thc lng giỏc nghch o 3.13.9 ng thc Dng s phc 10 Lng Giỏc 3.13.10 ng thc Tớch vụ hn 10 3.1 Gúc 3.13.11 ng thc Gii tớch 10 3.2 Cỏc hm s lng giỏc c bn 3.13.12 ng thc ng Dựng 10 3.3 Phộp Toỏn Lng Giỏc 3.13.13 ng thc gúc bi 11 3.3.1 ng thc lng giỏc c bn 3.13.14 Cỏc Hm lng giỏc nghch o 11 3.3.2 ng thc lng giỏc Tun hon, i xng v tnh tin Gii Tớ 11 3.3.3 ng thc Pytago 4.1 Phộp Toỏn Gii Tớch 11 3.3.4 ng thc Tng v hiu ca gúc 4.2 Toỏn o hm 11 Cụng thc h bc 4.2.1 Cụng c Toỏn o Hm 11 3.4.1 ng thc Bin tớch thnh tng 4.2.2 Hoỏn Chuyn o Hm 12 3.4.2 ng thc lng giỏc nghch o 3.4.3 ng thc Tớch vụ hn 3.4.4 ng thc Gii tớch 3.4.5 o Hm Lng Giỏc C Bn 3.4.6 Tng Hai Gúc 3.4.7 Hiu Hai Gúc 3.4.8 Tng Hai Hm 3.4.9 Hiu Hai Hm 3.4.10 ng thc gúc bi Cỏc Hm lng giỏc nghch o 3.5.1 Chui S 3.5.2 Tớch Phõn 3.5.3 S Phc 3.6 Cụng thc h bc 3.7 Cụng thc gúc chia ụi 3.8 ng thc Bin tớch thnh tng 3.4 3.5 Phộp Toỏn Tớ Phõn 5.0.3 Cụng thc tớch phõn 12 12 Ngun, ngi úng gúp, v giy phộp o bn v hỡnh nh 25 6.1 Vn bn 25 6.2 Hỡnh nh 25 6.3 Giy phộp ni dung 25 2.4 Toỏn Phõn S S hc 1.1 Ghi S S hc l mụn hc v s v cỏc phộp tớnh v s S l cỏch thc ngi ghi li s lng cỏc i tng nh cụng c sn xut, sỳc vt chn nuụi Cỏc dõn tc khỏc cú cỏch kớ hiu khỏc , mi kớ hiu thng c gi l mt ch s, hay mt s, ngy thng c gi l ký s Ngi ta ghộp cỏc ch s khỏc vo theo nhng quy c nht nh to thnh cỏc s Ngy cũn li ph bin l cỏch ghi s ca: Ngi Arp gi l S Rp (0, 1, 2, 3, 4, 5, 6, 7, 8, 9), 2.4 Toỏn Phõn S 1/ ab = b a a b +c= a+bc b a b c= abc b a b ìc= ac b a b /c ab c = c + a b = a+bc b c a b = abc b c ì a b = ac b c/ ab = bc a Ngi La-mó c gi l S La Mó (I, V, X, L, C, D, M), 10 a b + c d = ad+bc bd V nhiu cỏch ghi s khỏc 11 a b c d = adbc bd 12 a b ì c d = ac bd 13 a c b /d i S 2.1 S i s = ad bc 2.5 Toỏn S Phc (a + ib) + (c + id) = (a + c) + i(b + d) 2.2 Phộp Toỏn i S (a + ib) (c + id) = (a + c) i(b + d) (a + ib)x(c + id) = (ac bd) + i(ad + bc) 2.3 Toỏn S Nguyờn (a+ib) (c+id) = (a+ib)(c+id) (c+id)(c+id) = (acbd)+i(ad+bc) (c+id2 ) |Z1 |1 + |Z2 |2 = a + = a |Z1 |1 |Z2 |2 = a = a |Z1 |1 ì |Z2 |2 = Z1 ì Z2 (1 + ) a ì = |Z1 |1 /|Z2 |2 = a/0 = 00 (a + ib) + (a ib) = 2a Z1 Z2 (1 ) a + (a) = 10 (a + ib) (a ib) = i2b a (a) = 2a 11 (a + ib) ì (a ib) = a2 b2 a ì (a) = a2 a/(a) = a + a = 2a 12 (a+ib) (aib) = (a2 b2 ) (aib)2 j + j = 2j 10 a a = j + (j) = 11 a ì a = a2 j j = 12 a/a = j j = 2j 13 a + b = b + a j ì j = j 14 a + b + c = (a + b) + c = (a + b) + c j ì (j) = j 2 I S j j j j 2.7 Toỏn Cn =1 = 1 j = 10 j = 0=0 1=1 = j 10 a = aẵ a a = a2 ì a = a3 an = a a( n 2) m ( an )m = a n m n a = mn a ab = a b a a b = b a0 = 11 n a a1 = a 12 j = 11 j = i 12 j = 13 j n = n = 2m 14 j n = j n = 2m + 2.6 Toỏn Ly tha a1 = a an = an b = n a n b 13 j = 1 = 14 j = 1 = j 15 j = 1 1 = (an )m = am n (a)n = an Vi n = 2m 2.8 Toỏn Log (a)n = an Vi n = 2m + Nu b > with b , Vy cho mi s thc y,a,c am /an = amn logb (y a ) = a logb (y) (am )n = amn logb (ba ) = a 10 am + an = am (1 + an m) 11 a a = a (1 a m m n m logb (ac) = logb (a) + logb (c) logb (a/c) = logb (a) logb (c) n logb (a) = 12 a ì a = a m n m+n ( a )m b = for any d > 0, d 2.9 Hm s 13 (ab)m = am ì bm 14 logd (a) logd (b) am bm 15 (a + b)n = (a + b) ì (a + b) ã ã ã ì (a + b) Hm s i s l mt biu thc i s dựng mụ t tng quan gia hai i lng Hm s i s cú ký hiu toỏn n 16 (a b)n = (a b) ì (a b) ã ã ã ì (a b) n f (x) = x Vi a2 b2 = (a b)(a + b) a2 + b2 = a2 + 2ab + b2 = (a + b)2 2ab x a2 + b2 = a2 2ab + b2 = (a b)2 + 2ab f (x) 3.4 Cụng thc h bc Loi hm s Hm s c bit 3.3.3 ng thc Pytago Cỏc ng thc sau cú th d thy trờn vũng trũn n v: ng thc sau cng ụi hu ớch: a sin x + b cos x = 3.1 Lng Giỏc a2 + b2 ã sin(x + ) vi { Gúc = arctan(b/a), + arctan(b/a), neu a 0; neu a < Hai ng thng ct ti mt im to mt Gúc gia hai ng thng Ký hiu Gúc l A Gúc o bng n v o Cho thớ d A = 300 3.3.4 ng thc Tng v hiu ca gúc Cỏch chng minh nhanh cỏc cụng thc ny l dựng cụng thc Euler 3.2 Cỏc hm s lng giỏc c bn Cho bit tng quan gia Cnh v Gúc tam giỏc sin(x y) = sin(x) cos(y) cos(x) sin(y) vuụng cos(x y) = cos(x) cos(y) sin(x) sin(y) tan(x) tan(y) tan(x y) = tan(x) tan(y) cs(x + y) = cs(x) cs(y) 3.3 Phộp Toỏn Lng Giỏc cs(x) cs(x y) = cs(y) 3.3.1 ng thc lng giỏc c bn vi Cos(x) Sin(x) Sec(x) = cos(x) CoSec(x) = sin(x) sin(x) T an(x) = cos(x) cos(x) Cotan(x) = sin(x) 3.3.2 cs(x) = ex = cos(x) + sin(x) v = 3.4 Cụng thc h bc Gii cỏc phng trỡnh cụng thc bi cho cos2 (x) v sin2 (x), thu c: ng thc lng giỏc Tun hon, i xng v tnh tin + cos(2x) Cỏc ng thc sau cú th d thy trờn vũng trũn n cos(2x) v: sin2 (x) = ng thc sau cng ụi hu ớch: cos(4x) 2 sin (x) cos 2(x) = 23 sin 2(x) sin(3x) a sin x + b cos x = a2 + b2 ã sin(x + ) sin3 (x) = vi 32 cos(x) + cos(3x) cos (x) = { Cos(3x) = 4cos^3(x) - 3cos(x) Sin(3x) = 4sin^3(x) + eu a 0; arctan(b/a), n = 3sin(x) eu a < + arctan(b/a), n cos2 (x) = 3.4.1 ng thc Bin tớch thnh tng Dựng cụng thc tng v hiu gúc bờn trờn cú th suy cos (x) cos (y) = cos (x + y) + cos (x y) cos (x y) cos (x + y) sin (x y) + sin (x + y) sin (x) cos (y) = sin (x) sin (y) = cosh x = n=1 ) (x) sin x = cos n x n=1 ng thc s 3.4.4 ng thc Gii tớch Cỏc cụng thc gii tớch sau dựng gúc o bng radian lim x0 ng thc lng giỏc nghch o x2 1+ (n 21 )2 ng thc Bin tng thnh tớch 3.4.2 ( LNG GIC sin(x) = 1, x cos(x) = 0, x0 x d sin(x) = cos(x) dx lim arcsin(x) + arccos(x) = /2 arctan(x) + arccot(x) = /2 { eu x > /2, n arctan(x) + arctan(1/x) = eu x < /2, n ( ) x+y arctan(x) + arctan(y) = arctan xy ( ) xy arctan(x) arctan(y) = arctan + xy sin(arccos(x)) = x2 cos(arcsin(x)) = x2 x sin(arctan(x)) = + x2 3.4.5 o Hm Lng Giỏc C Bn Cỏc ng thc sau cú th suy t trờn v cỏc quy tc ca o hm: d cos(x) = sin(x) dx d tan(x) = sec2 (x) dx d cot(x) = csc2 (x) dx d sec(x) = sec(x) tan(x) cos(arctan(x)) = dx + x2 d csc(x) = csc(x) cot(x) x dx tan(arcsin(x)) = 1x d arcsin(x) = dx x2 x2 tan(arccos(x)) = x d arctan(x) = dx + x2 3.4.3 ng thc Tớch vụ hn Cỏc biu thc v tớnh tớch phõn cú th tỡm ti danh sỏch tớch phõn vi hm lng giỏc v danh sỏch tớch phõn Trong cỏc ng dng vi hm c bit, cỏc tớch vụ hn vi hm lng giỏc ngc sau cú ớch: sin x = x ( n=1 sinh x = x ( 1+ n=1 cos x = ( n=1 x2 n2 3.4.6 Tng Hai Gúc ) x2 n2 sin (x + y) = sin x cos y + cos x sin y cos (x + y) = cos x cos y sin x sin y ) x2 (n 12 )2 3.4.7 Hiu Hai Gúc ) sin (x y) = sin x cos y cos x sin y cos (x y) = cos x cos y + sin x sin y 3.5 Cỏc Hm lng giỏc nghch o 3.4.8 Tng Hai Hm ( sin x + sin y = sin Tng quỏt x+y ( cos x + cos y = cos ) ( cos x+y ) xy ( cos Nu T l a thc Chebyshev bc n thỡ ) cos(nx) = Tn (cos(x)) xy ) cụng thc de Moivre: cos(nx) + i sin(nx) = (cos(x) + i sin(x))n tan x + tan y = sin (x + y) cos x cos y cot x + cot y = sin (x + y) sin x sin y 3.4.9 Hm ht nhõn Dirilet D(x) s xut hin cỏc cụng thc sau: + cos(x) + cos(2x) + cos(3x) + ã ã ã + cos(nx) (( ) ) sin n + 21 x = sin(x/2) Hiu Hai Hm tan x tan y = sin (x y) cos x cos y Hay theo cụng thc hi quy: cot x cot y = sin (x y) sin x sin y sin(nx) = sin((n 1)x) cos(x) sin((n 2)x) ( sin x sin y = cos x+y ( cos x cos y = sin 3.4.10 cos(nx) = cos((n 1)x) cos(x) cos((n 2)x) ) x+y ( sin xy ) ( sin ng thc gúc bi ) xy 3.5 Cỏc Hm lng giỏc nghch o ) 3.5.1 Chui S Cỏc hm nghch o cú th c ký hiu l sin1 hay cos1 thay cho arcsin v arccos Vic dựng ký hiu m cú th gõy nhm ln vi hm m ca hm lng giỏc Cỏc hm lng giỏc nghch o cng cú th c nh Bi hai Cỏc cụng thc sau cú th suy t cỏc cụng ngha bng chui vụ hn: thc trờn Cng cú th dựng cụng thc de Moivre vi n = ( ) ( ) z5 ( 1ã3ã5 ) z7 arcsin z = z + 12 z3 + (1ã3 +) 2ã4ã6 + ããã 2ã4 (2n)! z 2n+1 = n=0 22n (n!)2 (2n+1) sin(2x) = sin(x) cos(x) arccos z cos(2x) = cos2 (x)sin2 (x) = cos2 (x)1 = 12 sin2 (x) tan(2x) = tan(x) tan2 (x) arctan z Cụng thc gớc kộp cú th dựng tỡm b ba Pytago arccsc z Nu (a, b, c) l b ba Pytago thỡ (a2 b2 , 2ab, c ) cng vy Bi ba Vớ d ca trng hp n = 3: sin(3x) = sin(x) sin3 (x) cos(3x) = cos3 (x) cos(x) arcsec z arccot z = = (z = z ( arcsin ) ( ) z7 1ã3 z + + (2ã4 +) 1ã3ã5 2ã4ã6 (2n)! z 2n+1 n=0 22n (n!)2 (2n+1) = z z3 + = n=0 ( ) z3 z5 z7 + n 72n+1 (1) z 2n+1 ããã |z| < + ããã) |z| < |z| < ( ) = arcsin z ( ) ( ) z5 ( 1ã3ã5 ) z7 = z + 21 z +( 1ã3 |z| > 2ã4 )+ 2ã4ã6 + ããã (2n)! z (2n+1) = n=0 22n (n!)2 2n+1 ( ) = arccos z ( ) ( ) z5 ( 1ã3ã5 ) z7 = (z + 21 z + ( 1ã3 2ã4 )+ 2ã4ã6 + ããã) (2n)! z (2n+1) = n=0 22n (n!)2 (2n+1) = = = z arctan (z z3 + z5 z7 + ã ã ã ) n 2n+1 (1) z n=0 2n+1 |z| < 3.5.2 LNG GIC 3.7 Cụng thc gúc chia ụi Tớch Phõn Chỳng cng cú th c nh ngha thụng qua cỏc biu ay x/2 cho x cụng thc trờn, ri gii phng thc sau, da vo tớnh cht chỳng l o hm ca cỏc trỡnh cho cos(x/2) v sin(x/2) thu c: hm khỏc x arcsin (x) = dz, z2 1 dz, z2 arccos (x) = x x arctan (x) = dz, + z2 arccot (x) = x arcsec (x) = x |x| < x sin = (x) = cos(x) x R dz, z2 + z>0 x>1 dz, |z| z tan (x) sin(x/2) = = cos(x/2) cos x + cos x (1) Nhõn vi mu s v t s + cos x, ri dựng nh lý Pytago n gin húa: ( ) tan x2 = 1cos x (1+cos x)2 x>1 (1cos x)(1+cos x) (1+cos x)(1+cos x) = S Phc = Cụng thc trờn cho phộp m rng hm lng giỏc nghch o cho cỏc bin s phc|phc: ( ( )) arcsin(z) = i log i z + z ( arccos(z) = i log z + arctan(z) = i log ( iz + iz ( ) tan x2 = (1cos x)2 (1cos2 x) ) z2 ) Cụng thc h bc Gii cỏc phng trỡnh cụng thc bi cho cos2 (x) v sin2 (x), thu c: + cos(2x) sin2 (x) = cos(2x) cos(4x) sin (x) cos 2(x) = cos3 (x) = 32 cos(x) + cos(3x) (1cos x)(1cos x) (1+cos x)(1cos x) = cos x sin x (x) = sin(x) cos(x) = + cos(x) sin(x) Nu t = tan 23 sin 2(x) sin(3x) sin3 (x) = Suy ra: tan cos2 (x) = sin x + cos x Tng t, li nhõn vi mu s v t s ca phng trỡnh (1) bi cos x, ri n gin húa: = 3.6 + cos(x) Dn n: dz, |z| z arccsc (x) = 3.5.3 |x| < cos (x) (x) , thỡ: Phng phỏp dựng t thay th nh trờn hu ớch [[gii tớch chuyn cỏc t l thc cha sin(x) v cos(x) thnh [[hm ca t Cỏch ny giỳp tớnh [[o hm ca biu thc d dng 3.11 Tớch Phõn 3.8 ng thc Bin tớch thnh tng Dựng cụng thc tng v hiu gúc bờn trờn cú th suy cos (x) cos (y) = cos (x + y) + cos (x y) cos (x y) cos (x + y) sin (x y) + sin (x + y) sin (x) cos (y) = sin (x) sin (y) = cosh x = ( n=1 x2 1+ (n 21 )2 (x) sin x = cos n x n=1 3.11 Tớch Phõn Chỳng cng cú th c nh ngha thụng qua cỏc biu thc sau, da vo tớnh cht chỳng l o hm ca cỏc hm khỏc [[ng thc Bin tng thnh tớch x arcsin (x) = 3.9 ng thc lng giỏc nghch o dz, z2 |x| < 1 dz, z2 |x| < 1 arccos (x) = arcsin(x) + arccos(x) = /2 x arctan(x) + arccot(x) = /2 { eu x > /2, n arctan(x) + arctan(1/x) = eu x < /2, n ( ) x+y arctan(x) + arctan(y) = arctan xy ( ) xy arctan(x) arctan(y) = arctan + xy sin(arccos(x)) = x2 cos(arcsin(x)) = x2 x sin(arctan(x)) = + x2 cos(arctan(x)) = + x2 x tan(arcsin(x)) = x2 x2 tan(arccos(x)) = x 3.10 ng thc Tớch vụ hn x arctan (x) = dz, + z2 arccot (x) = x arcsec (x) = x sin x = x ( n=1 sinh x = x ( n=1 cos x = n=1 x n2 ( ) x2 (n 12 )2 z>0 arccsc (x) = x x>1 dz, |z| z x>1 3.12 S Phc Cụng thc trờn cho phộp m rng hm lng giỏc nghch o cho cỏc bin [[s phc: ( ( )) arcsin(z) = i log i z + z ( ) arccos(z) = i log z + z ( iz + iz ) 3.13 ng thc lng giỏc ) x2 1+ 2 n dz, z2 + x R dz, |z| z i arctan(z) = log Trong cỏc ng dng vi [[hm c bit, cỏc [[tớch vụ hn sau cú ớch: ) ) Trong [[toỏn hc, cỏc ng thc lng giỏc l cỏc [[phng trỡnh cha cỏc [[hm lng giỏc, ỳng vi mt di ln cỏc giỏ tr ca bin s Cỏc [[ng thc ny hu ớch cho vic rỳt gn cỏc biu thc cha hm lng giỏc Vớ d vic [[tớnh tớch phõn vi cỏc hm khụng phi l lng giỏc: cú th thay chỳng bng cỏc hm lng giỏc v dựng cỏc ng thc lng giỏc n gin húa phộp tớnh 3.13.1 ng thc lng giỏc c bn tan(x) = sin(x) cos(x) cotg(x) = cos(x) = sin(x) tan(x) ]] tanx.cosx = 3.13.2 LNG GIC vi cs(x) = ex = cos(x) + sin(x) v ng thc lng giỏc Tun hon, i xng v tnh tin = Cỏc ng thc sau cú th d thy trờn vũng trũn n v]]: 3.13.5 Cụng thc h bc ng thc sau cng ụi hu ớch: a sin x + b cos x = a2 + b2 ã sin(x + ) vi { arctan(b/a), + arctan(b/a), = eu a 0; n eu a < n Gii cỏc phng trỡnh cụng thc bi cho cos2 (x) v sin2 (x), thu c: cos2 (x) = + cos(2x) sin2 (x) = cos(2x) sin2 (x) cos2 2(x) = 3.13.3 ng thc Pytago sin3 (x) = cos(4x) 23 sin 2(x) sin(3x) Cỏc ng thc sau cú th d thy trờn [[vũng trũn n 32 cos(x) + cos(3x) v: cos3 (x) = ng thc sau cng ụi hu ớch: a sin x + b cos x = a2 + b2 ã sin(x + ) vi { arctan(b/a), + arctan(b/a), = 3.13.4 eu a 0; n eu a < n ng thc Tng v hiu ca gúc Xem thờm [[nh lý Ptolemaios 3.13.6 Cụng thc gúc chia ụi ay x/2 cho x cụng thc trờn, ri gii phng trỡnh cho cos(x/2) v sin(x/2) thu c: cos sin (x) = (x) = + cos(x) cos(x) Dn n: Cỏch chng minh nhanh cỏc cụng thc ny l dựng (x) sin(x/2) cos x tan = = [[cụng thc Euler cos(x/2) + cos x sin(x y) = sin(x) cos(y) cos(x) sin(y) cos(x y) = cos(x) cos(y) sin(x) sin(y) tan(x) tan(y) tan(x y) = tan(x) tan(y) Nhõn vi mu s v t s + cos x, ri dựng nh lý Pytago n gin húa: ( ) tan x2 = 1cos x (1+cos x)2 (1cos x)(1+cos x) (1+cos x)(1+cos x) cs(x + y) = cs(x) cs(y) cs(x y) = cs(x) cs(y) (1) = sin x + cos x = 10 d sin(x) = cos(x) dx 3.13.13 ng thc gúc bi Bi hai Cỏc ng thc sau cú th suy t trờn v cỏc quy tc ca [[o hm: d cos(x) = sin(x) dx LNG GIC Cỏc cụng thc sau cú th suy t cỏc cụng thc trờn Cng cú th dựng cụng thc de Moivre]] vi n = d tan(x) = sec2 (x) dx sin(2x) = sin(x) cos(x) d cot(x) = csc2 (x) dx cos(2x) = cos2 (x)sin2 (x) = cos2 (x)1 = 12 sin2 (x) d sec(x) = sec(x) tan(x) dx tan(2x) = d csc(x) = csc(x) cot(x) dx Cụng thc gớc kộp cú th dựng tỡm b ba Pytago]] Nu (a, b, c) l b ba Pytago thỡ (a2 b2 , 2ab, c ) cng vy d arcsin(x) = dx x2 tan(x) tan2 (x) Bi ba d arctan(x) = dx + x2 Cỏc biu thc v tớnh tớch phõn cú th tỡm ti danh sỏch Vớ d ca trng hp n = 3: tớch phõn vi hm lng giỏc]] v danh sỏch tớch phõn vi hm lng giỏc ngc]] sin(3x) = sin(x) sin3 (x) 3.13.12 ng thc Thng Dựng cos(3x) = cos3 (x) cos(x) sin (x + y) = sin x cos y + cos x sin y Tng quỏt sin (x y) = sin x cos y cos x sin y cos (x + y) = cos x cos y sin x sin y cos (x y) = cos x cos y + sin x sin y ( ) ( ) x+y xy sin x + sin y = sin cos 2 ( ) ( ) x+y xy sin x sin y = cos sin 2 ) ( ) ( xy x+y cos cos x + cos y = cos 2 ( ) ( ) x+y xy cos x cos y = sin sin 2 Nu T l a thc Chebyshev]] bc n thỡ cos(nx) = Tn (cos(x)) [[cụng thc de Moivre: :\cos(nx)+i\sin(nx)=(\cos(x)+i\sin(x))^n \, Hm '''ht nhõn Dirichlet]] Dn(x) s xut hin cỏc cụng thc sau: + cos(x) + cos(2x) + cos(3x) + ã ã ã + cos(nx) (( ) ) n + 21 x sin(x/2) tan x + tan y = sin (x + y) cos x cos y = tan x tan y = sin (x y) cos x cos y Hay theo cụng thc hi quy: cot x + cot y = sin (x + y) sin x sin y sin(nx) = sin((n 1)x) cos(x) sin((n 2)x) cot x cot y = sin (x y) sin x sin y cos(nx) = cos((n 1)x) cos(x) cos((n 2)x) sin 11 3.13.14 Cỏc Hm lng giỏc nghch o 4.2.2 Hoỏn Chuyn o Hm Chui S Cỏc hm nghch o cú th c ký hiu l Hoỏn Chuyn Laplace sin1 hay cos1 thay cho arcsin v arccos Vic dựng ký df hiu m cú th gõy nhm ln vi [[hm m ca hm s = dt lng giỏc Cỏc hm lng giỏc nghch o cng cú th c nh = df s dt ngha bng chui vụ hn: arcsin z = z+ = arccos z = = (z = arctan z arccsc z arcsec z arccot z 4.1 Hoỏn Chuyn Fourier ( 1ã3 ) z5 ( 1ã3ã5 ) z7 + ( 2ã4 +) 2ã4ã6 + ããã df (2n)! z 2n+1 j|z|=< n=0 22n (n!)2 (2n+1) dt ( ) z3 z ( arcsin ) ( ) z7 1ã3 z + + (2ã4 +) 1ã3ã5 2ã4ã6 (2n)! z 2n+1 n=0 22n (n!)2 (2n+1) = z z3 + = n=0 ( ) z3 z z + (1)n z 2n+1 2n+1 ããã |z| < 1 df = j dt + ã ã ã ) |z| < Hoỏn Chuyn Z Z = 90o 1 df = = ( ) Z dt = arcsin z ) ( ) ( ) ( z 1ã3ã5 z = z + 12 z +( 1ã3 |z| > 2ã4 )+ 2ã4ã6 + ããã (2n)! Cụng thc tng quỏt z (2n+1) = n=0 22n (n!)2 2n+1 df ( ) = s = j = 90o = arccos z dt ( ) ( ) ( ) 1ã3 z 1ã3ã5 z = (z + 21 z + ( 2ã4 )+ 2ã4ã6 df + ã1ã ã ) |z| >1 (2n+1) (2n)! z = = = = n=0 22n (n!)2 (2n+1) dt s j = = = z arctan (z z3 + z5 z7 + ã ã ã ) (1)n z2n+1 n=0 2n+1 Gii Tớch Phộp Toỏn Gii Tớch |z| < Thớ d di L = sL = jL = L90o dt dv = sC = jC = 90o dt C dv 1 C = = = 90o dt sC jC C C Phộp Toỏn Tớch Phõn 4.2 4.2.1 Toỏn o hm Cụng Thc Toỏn o Hm Chỳng cng cú th c nh ngha thụng qua cỏc biu thc sau, da vo tớnh cht chỳng l o hm ca cỏc hm khỏc o Hm Ca Hm S Lng Giỏc o Hm Ca Hm S ng Cong o Hm Ca Hm S c Bit o Hm Bc N x arcsin (x) = x x arccos (x) = arctan (x) = dz, z2 dz, |x| < 1 z2 dz, x R + z2 arccot (x) = x |x| < z2 dz, +1 z>0 12 PHẫP TON TCH PHN x x + sin 2ax+C = + sin ax cos ax+C 4a 2a x ( ) x x x x2 cos2 ax dx = + sin 2ax+ cos 2ax+C arccsc (x) = dz, x > 4a 8a 4a |z| z x xn sin ax n n x cos ax dx = xn1 sin ax dx a a 5.0.3 Cụng thc tớch phõn cos ax (ax)2k (1)k dx = ln |ax| + +C x 2k ã (2k)! sin ax dx = cos ax + C k=1 a cos ax cos ax a sin ax x x dx = dx (for n = 1) n1 sin ax dx = sin 2ax+C = sin ax cos ax+Cxn (n 1)x n1 xn1 4a 2a ( ax ) dx x x tan = ln + +C x sin ax dx = sin 2ax cos 2ax + C cos ax a 4 4a 8a ( ) dx sin ax n2 dx x x x = + (for n > 1) n n1 n2 sin 2ax cos 2ax+C x2 sin2 ax dx = cos ax a(n 1) cos ax n cos ax 4a 8a 4a dx ax = tan +C sin((b1 b2 )x) sin((b1 + b2 )x) ax|b1 | = a |b2 |)2 sin b1 x sin b2 x dx = +C1 + cos (for 2(b1 b2 ) 2(b1 + b2 ) dx ax n1 = cot +C sin ax cos ax n n n2 cos ax a sin ax dx = + sin axdx (for n > 0) na n x dx x ax ax = tan + ln cos +C dx ax + cos ax a a = ln tan +C sin ax a x dx x ax ax = cot + ln sin +C n2 dx cos ax dx cos ax a a + (for n > 1) = sinn ax a(1 n) sinn1 ax n sinn2 ax cos ax dx ax = x tan +C + cos ax a sin ax x cos ax x sin ax dx = +C a2 a ax cos ax dx = x cot +C n cos ax a x n n n1 x sin ax dx = cos ax+ x cos ax dx (for n > 0) a a sin(a1 a2 )x sin(a1 + a2 )x cos a1 x cos a2 x dx = + +C (for |a1 | a2 2 2(a1 a2 ) 2(a1 + a2 ) a (n 6) nx x sin dx = (for n = 2, 4, ) a 1 a 24n2 2 tan ax dx = ln | cos ax|+C = ln | sec ax|+C a a sin ax (ax)2n+1 dx = (1)n +C n n1 tan ax dx = tan ax tann2 ax dx (for n = 1) x (2n + 1) ã (2n + 1)! n=0 a(n 1) sin ax a cos ax sin ax dx q dx = + dx = (px+ ln |q sin ax+p cos ax|)+C (for p2 + xn (n 1)xn1 n1 xn1 q tan ax + p p +q a ( ax ) dx 1 dx = tan +C = ln | sin ax| + C sin ax a tan ax a ( ax ) ( ax ) x dx x x dx = tan + ln cos +C = + ln | sin ax + cos ax| + C + sin ax a a tan ax + 2a ( ax ) ( ax ) x x dx dx x = cot + ln sin +C = + ln | sin ax cos ax| + C sin ax a a tan ax 2a ( ax ) sin ax dx tan ax dx x = x + tan +C = ln | sin ax + cos ax| + C sin ax a tan ax + 2a x tan ax dx cos ax dx = sin ax + C = + ln | sin ax cos ax| + C a tan ax 2a cosn1 ax sin ax n n n2 dxn=> 0)ln |sec ax + tan ax| + C cos ax dx = + cos ax dx sec ax (for a na n secn1 ax sin ax n cos ax x sin ax n sec ax dx = + secn2 ax dx (for n = x cos ax dx = + +C a(n 1) n1 a2 a arcsec (x) = dz, |z| z x>1 cos2 ax dx = 13 secn x dx = n2 n2 sec x dx [2] n1 secn2 x tan x n1 + sin ax cosn ax dx = cosn+1 ax+C a(n + 1) (for n = 1) sinn1 ax cosm+1 ax n sin ax cos ax dx = + sinn2 ax c dx x + + a(n m) n m = x tan + C sec x + sinn+1 ax cosm1 ax m n m sin ax cos ax dx = + sinn ax cosm a(n + m) n+m csc ax dx = ln |csc ax cot ax| + C a dx = ln |tan ax| + C sin ax cos ax a csc x dx = cot x + C dx dx = + (for n n ax n1 ax n2 ax cscn1 ax cos ax n sin ax cos a(n 1) cos sin ax cos n n2 csc ax dx = + csc ax dx (for n = 1) a(n 1) n1 dx dx = + (for n n1 n2 sin ax cos ax a(n 1) sin ax sin ax cos ax cot ax dx = ln | sin ax| + C a sin ax dx = +C (for n = 1) n ax n n1 n2 cos a(n 1) cosn1 ax cot ax dx = cot ax cot ax dx (for n = 1) a(n 1) ( ax ) 1 sin2 ax dx = sin ax+ ln tan + +C dx tan ax dx cos ax a a = + cot ax tan ax + sin2 ax dx sin ax dx = (for n = tan ax dx dx cosn ax a(n 1) cosn1 ax n cosn2 ax = cot ax tan ax sinn1 ax sinn2 ax dx sinn ax dx ( ax ) = + (for n = 1) dx = ln tan +C cos ax a(n 1) cos ax cos ax sin ax a sinn ax dx sinn+1 ax nm+2 sinn ax dx ( ) dx = ( = tan ax +C cosm ax a(m 1) cosm1 ax m1 cosm2 ax (cos ax sin ax)2 2a ( sin)n1 ax n1 sinn2 ax dx sinn ax dx dx sin x cos x dx = + ( = 2(n 2)cosm ax a(n m) cosm1 ax n m cosm ax (cos x + sin x)n n (cos x + sin x)n1 (cos x + sin x)n2 sinn ax dx sinn1 ax n1 sinn2 ax dx cos ax dx x = (for = + ln |sin ax + cos ax| + C cosm ax a(m 1) cosm1 ax m cosm2 ax cos ax + sin ax 2a cos ax dx cos ax dx x = +C (for n = 1) n = ln |sin ax cos ax| + C sin ax a(n 1) sinn1 ax cos ax sin ax 2a cos2 ax dx 1( ax ) sin ax dx x = cos ax + ln tan +C = ln |sin ax + cos ax| + C sin ax a cos ax + sin ax 2a ( ) cos2 ax dx cos ax dx sin ax dx x = + (for n = = ln |sin ax cos ax|+C sinn ax n a sinn1 ax) sinn2 ax cos ax sin ax 2a cos ax dx ax ax cosn ax dx cosn+1 ax nm2 cosn ax dx = tan2 + ln tan +C = sin ax(1 + cos ax) 4a 2a sinm ax m1 a(m 1) sinm1 ax sinm2 ax ax ax cos ax dx cosn1 ax n1 cosn ax dx cosn2 ax dx = cot2 ln tan +C = + (fo sin ax(1 + cos ax) 4a 2a sinm ax sinm ax a(n m) sinm1 ax n m ( ax ) (ax n ) sin ax dx cos cosn1 ax n1 cosn2 ax dx = cot2 + + ln tan + ax dx +C= ( cos ax(1 + sin ax) 4a 2a sinm ax a(m 1) sinm1 ax m sinm2 ax ( ax ) (ax ) sin ax dx = tan2 + ln tan + +C cos ax(1 sin ax) 4a 2a 2sin ax4 tan ax dx = a (ln | sec ax+tan ax|sin ax)+C tann ax dx sin2 ax + C sin ax cos ax dx = = tann1 (ax)+C (for n = 1) 2a a(n 1) sin ax cos(a1 + a2 )x cos(a1 a2 )x tann ax dx sin a1 x cos a2 x dx = +C (for |a=1 | = |a2 |) tann+1 ax+C (for n = 1) 2(a1 + a2 ) 2(a1 a2 ) cos ax a(n + 1) cotn ax dx n n+1 sin ax cos ax dx = sin ax+C (for n = 1)2 cotn+1 ax+C (for n = 1) = a(n + 1) a(n + 1) sin ax n m 14 PHẫP TON TCH PHN x x c (for n = 1) arccot dx = x arccot + ln(c2 + x2 ) c c c 2 x c +x x cx x arccot dx = arccot + sin x dx = c c c c c x x3 x cx2 c3 2 cos x dx = cos x dx = cos x dx = sin c x arccot c dx = arccot c + ln(c +x ) c c n+1 c x xn+1 x c x dx n x arccot dx = arccot + ( n = 1) tan x dx = c n+1 c n+1 c + x2 c a2 (ax + b)n+1 2 a (n 6) nx (ax + b)n dx = ( n = 1) 2 dx = x cos (for n = 1, 3, ) a(n + 1) 2 a a 24n dx = ln |ax + b| x x ax + b a arcsin dx = x arcsin + c2 x2 c c a(n + 1)x b ( ) x(ax+b)n dx = (ax+b)n+1 ( n {1, 2}) x x c x x 2 a (n + 1)(n + 2) x arcsin dx = arcsin + c x c c x x b dx = ln |ax + b| 2 x x x + 2c x ax + b a a 2 arcsin + c x x arcsin dx = c c x b dx = + ln |ax + b| ( n (ax + b) a (ax + b) a x sin x dx = n+1 xn+1 sin1 x a(1 n)x b x dx = ( n {1, 2}) (ax + b)n a (n 1)(n 2)(ax + b)n1 ) ( ) xn x2 nxn1 sin1 x (ax + b)2 x2 n2 + +n x sin x dx dx = 2b(ax + b) + b2 ln |ax + b| n1 ax + b a ( ) x2 b2 dx = ax + b 2b ln |ax + b| x x (ax + b)2 a3 ax + b arccos dx = x arccos c x2 c c ( ) ( ) x2 2b b2 dx = ln |ax + b| + x x c2 x x (ax + b)3 a3 ax + b 2(ax + b)2 x arccos dx = c x2 arccos c c ( x2 1 2b 2 dx = + x x x x + 2c n n3 2 (ax + b) a (n 3)(ax + b) (n 2)(a + b)n2 x arccos dx = c x arccos c c dx ax + b x x c = ln arctan dx = x arctan ln(c2 + x2 ) x(ax + b) b x c c a dx ax + b x c2 + x2 x cx = + ln x arctan dx = arctan x2 (ax + b) bx b x c c ( ) 1 dx ax + b 3 x x cx c x = a + ln arctan + ln c2 + x2 x2 arctan dx = x2 (ax + b)2 b2 (ax + b) ab2 x b3 x c c 6 n+1 x dx x xn+1 x c x dx = arctan 2 xn arctan dx = arctan ( n = 1) x +a a a c n+1 c n+1 c2 + x2 dx x ax x x x = arctanh = ln ( |x| < |a|) arcsec dx = x arcsec + ln |x x2 1| x2 a2 a a 2a a + x c c c|x| x xa dx ( ) = arccoth = ln ( |x| > |a|) 2 x arcsec x dx = x arcsec x x2 x a a a 2a x + a 2ax + b dx n arctan ( 4acb2 > 0) = 2 ax2 + bx + c x arcsec x dx = 4ac b 4ac b ( n+1 ( arcsec x n1 xn1 x2 n+1 x dx 2ax + b 2ax = artanh = ln 2 ax2 + bx + c b 4ac b 4ac b 4ac 2ax ( ))) dx +(1 n) xn1 arcsec x + (1 n) xn2 arcsec x dx = ( 4ac b2 = 0) ax + bx + c 2ax + b cotn ax dx = tan1n ax+C cos ax a(1 n) 15 x3 r2n+1 dx = x4 r dx = dxr dx a a+r = r a ln = r a arsinh ax + bx x+ c x x 3 mx + n m 2an bm r dx 2axr + b a2+ r dx = ln ax2 + bx + c + arctan = > 0) + 2a r (a4acb ln ax2 + bx + c 2a a 4ac b2 4ac b x x mx + n m 2an bm r5 dx 2axr5+ b a2 r3 2 dx = ln ax + bx + c + artanh = (+4acb 0) a + r + a4 r a |c|) xr a xr a c c x2 r dx = ln (x + r) 8 x cx c+x x x xr5 a2 xr3 a4 xr a6 ( x (0, c)) arsech dx = x arsech c arctan x r dx = ln (x + r) c c xc 24 16 16 x x x + x2 + c2 r a r arcsch dx = x arcsch +c ln ( x (0, c)) x3 r dx = c c c Chỳ ý: bi ny quy c x>0 r7 a2 r 3 x r dx = x b dx = ln ax2 + bx + c ax + bx + c 2a 2a ln cx dx = x ln cx x r2n+5 a3 r2n+3 2n + 2n + x3 r a2 xr3 a4 xr a6 + + ln (x + r) 16 16 5 (ln x)2 dx = x(ln x)2 2x ln x + 2x x r a xr a xr 3a xr 3a + + + ln (x + r) 16 64 128 128 r 2a r a r n n x5 r dx = + (ln cx) dx = x(ln cx) n (ln cx)n1 dx r9 2a2 r7 a4 r5 x5 r3 dx = + (ln x)i dx r2n+7 2a2 r2n+5 a4 r2n+3 = ln | ln x| + ln x + 2n+1 x r dx = + ln x i ã i! 2n + 2n + 2n + i=2 x4 r3 dx = 16 PHẫP TON TCH PHN ecx (cx 1) c2 dx x dx ) ( ( n = 1) = + x 2x (ln x)n (n 1)(ln x)n1 n (ln x)n1 + x2 ecx dx = ecx c c c n xn ecx dx = xn ecx xn1 ecx dx ( ) c c ln x m m+1 ( m = 1) x ln x dx = x m + (m + 1)2 ecx dx (cx)i = ln |x| + x i ã i! i=1 ( ) cx e dx ecx ecx xm+1 (ln x)n n = + c dx ( n = 1) m n m n1 x (ln x) dx = x (ln x) dx xn ( m =n 1) xn1 xn1 m+1 m+1 ecx ln x dx = ecx ln |x| Ei (cx) c n n+1 (ln x) dx (ln x) ecx = ( n = 1) ecx sin bx dx = (c sin bx b cos bx) x n+1 c + b2 ecx ecx cos bx dx = (c cos bx + b sin bx) c + b2 ln x dx ln x = ( m = 1)cx n ecx sinn1 x n(n 1) xm (m 1)xm1 (m 1)2 xm1 e sin x dx = (c sin xn cos x)+ ecx sinn c2 + n2 c + n2 ecx cosn1 x n(n 1) cx n e cos x dx = (c cos x+n sin x)+ ecx cosn + n2 c c + n n n n1 (ln x) dx (ln x) n (ln x) dx = + ( 2m = 1) xm (m 1)xm1 m xm xecx dx = ecx 2c 2 xà e(xà) /2 dx = (1 + erf ) m m+1 m x dx x m+1 x dx = + ( n = 1) n n1 n1 (ln x) (n 1)(ln x) n1 (ln x) ( ) x2 n1 e dx = ex j=0 c2j x2j+1 + (2n x2 1)c2n2 xe 2n dx (n > 0), dx = ln | ln x| ã ã ã ã ã (2j 1) 2j ! x ln x c2j = = 2j+1 j! 22j+1 i i ax2 dx (n 1) (ln x) i e dx = = ln | ln x| + (1) a xn ln x i ã i! i=1 (2n)! ( a )2n+1 2 x2n ex /a dx = n! 1 dx sinh cx dx = cosh cx = ( n = 1) n n1 c x(ln x) (n 1)(ln x) cosh cx dx = sinh cx c x x sin(ln x) dx = (sin(ln x) cos(ln x)) sinh2 cx dx = sinh 2cx 4c x cosh2 cx dx = sinh 2cx + 4c x cos(ln x) dx = (sin(ln x) + cos(ln x)) n1 n n1 sinh cx dx = sinh cx cosh cx sinhn2 cx dx (n cn n n+2 sinhn cx dx = sinhn+1 cx cosh cx sinhn+2 cx dx ecx dx = ecx c(n + 1) n+1 c cx n1 acx dx = a ( a > 0, a = 1) coshn cx dx = sinh cx coshn1 cx+ coshn2 cx dx (n c ln a cn n xecx dx = 17 coshn cx dx = n+2 sinh cx coshn+1 cx c(n + 1) n+1 dx cx = ln sinh cx c sinh(ax+b) cos(cx+d) dx = a c cosh(ax+b) cos(cx+d)+ a2 + c2 a +c cosh(ax+b) sin(cx+d) dx = a c sinh(ax+b) sin(cx+d) a2 + c2 a + c2 dx sinh cx = ln sinh cx c cosh cx + cosh(ax+b) cos(cx+d) dx = a2 a c sinh(ax+b) cos(cx+d)+ 2 +c a +c Assume (x2 > a2 ) , for (x2 < a2 ) , see next section: dx cosh cx = ln sinh cx c cosh cx + dx cosh cx n2 = n1 sinhn cx c(n 1) sinh cx n a c n+2 cosh sinh(ax+b) cx dx sin(cx+d) (n< dx0,=n =2 1)2 cosh(ax+b) sin(cx+d) a +c a + c2 dx cosh cx = ln sinh cx c sinh cx dx = arctan ecx cosh cx c xs dx = dx s s dx a ( n==s1) a arccos x x dx sinh cx n2 dx ( n = 1) dx x+s = + n = = ln n1 n2 cosh cx c(n 1) cosh cx n cosh cx s 2 a x a ( ) n n1 n2 cosh cx n1 cosh cx coshNotecx x+s x x+s that ln = sgn(x) arcosh = ln + dx = dx a( m = n) a xs , sinhm cx sinhm cx c(n m) sinhm1 cx n m x where the positive value of arcosh a is to be taken coshn+1 cx nm+2 coshn cx coshn cx dx = + dx ( m = 1) m2 sinhm cx m1 c(m 1) sinhm1 cx cx sinh x dx coshn cx n1 coshn1 cx coshn2scx = s + dx ( m = 1) dx = m2 sinhm cx c(m 1) sinhm1 cx m sinh cx x dx = sinhm cx sinhm1 cx m1 sinhm2 scx s( m = n) dx = + dx n cx coshn cx cosh c(m n) coshn1 cx m n x dx = s5cx 3s mn+2 sinhm cx sinhm+1 cx sinhm + dx ( n = 1) dx = coshn cx n1 c(n 1) coshn1 cx coshxn2 dxcx = 5s sinhm cx m1 sinhm1 cx sinhm2scx + dx ( n = 11) dx = dx coshn cx c(n 1) coshn1 cx n coshn2x cx = s2n+1 (2n 1)s2n1 1 2m2 x sinh cx dx = x cosh cx sinh cx x2m dx x2m1 2m x dx c c = + 2n+1 2n1 2n1 s 2n s 2n s 1 x cosh cx dx = x sinh cx cosh cx x dx xs a x+s c c = + ln s 2 a cx dx = ln | cosh cx| c x x+s x dx = + ln s3 s a coth cx dx = ln | sinh cx| c x3 s 3 x+s x4 dx = + a xs + a4 ln s 8 a n n1 n2 cx dx = cx+ cx dx ( n = 1) c(n 1) xs a2 x x+s x dx = + a ln s a cothn cx dx = cothn1 cx+ cothn2 cx dx s ( n = 1) c(n 1) x 1x x+s x dx = + ln 2 s bx) s( b = sc ) a sinh bx sinh cx dx = (b sinh cx cosh bxc cosh cx sinh b c2 ( ) 2m nm1 n m x2( x dx nm cosh bx cosh cx dx = (b sinh bx cosh cxc sinh cx cosh bx) = (1) ( b2 = c2a)2(nm) s2n+1 2(m + i) + i s2( b c2 i=0 1 x dx cosh bx sinh cx dx = (b sinh bx sinh cxc cosh bx cosh = cx) ( b2 = c2 ) b c s a s sinh n2 cx dx 18 [ ] dx x x3 = s5 a4 s s3 [ ] dx x x3 x5 = + s7 a s s3 s5 [ ] dx x x3 x5 x7 = + s9 a s s3 s5 s7 x dx x3 = s5 a2 3s3 [ ] x dx 1 x3 x5 = s7 a s3 s5 [ ] x dx 1x x5 x7 = + s9 a6 s3 s5 s7 ( ) x xu + a2 arcsin (|x| |a|) u dx = a xu dx = u3 (|x| |a|) x a x x2 u dx = u3 + (xu+a2 arcsin ) (|x| |a|) a u dx a+u = u a ln (|x| |a|) x x x dx = arcsin (|x| |a|) u a x dx 1( x) = xu + a2 arcsin (|x| |a|) u a 1( x ) u dx = xu sgn x arcosh (for |x| |a|) a x dx = u (|x| |a|) u Tớ phõn hm hp (Integrals involving) ax2 + bx + c R = Assume (ax + bx + c) cannot be reduced to the following expression (px + q)2 for some p and q dx = ln aR + 2ax + b R a (for a > 0) x R b dx = R a 2a PHẫP TON TCH PHN dx R x 2bx + 4c dx = R3 (4ac b2 )R x b 2n+1 2n1 R (2n 1)aR 2a ( ) dx cR + bx + 2c = ln xR x c ( ) dx bx + 2c = arsinh xR c |x| 4ac b2 Sdx = dx = dx R2n+1 2S 3a dx 2S = S a ( ) 2b arcoth Sb (for b > 0, ax > 0) ( ) dx = 2b artanh Sb (for b > 0, ax < 0) xS ( ) S arctan (for b < 0) b b ( ( )) S b arcoth (for b > 0, ax > 0) S ( b )) ( S S (for b > 0, ax < 0) dx = S b artanh b x ( ( )) S b arctan S (for b < 0) b ( n1 ) xn x n dx = x S bn dx S a(2n + 1) S ( ) n n n1 x Sdx = x S nb x Sdx a(2n + 3) ( ( ) ) 1 S dx dx = + n a xn S b(n 1) xn1 xn1 S sin ax dx = cos ax + C a sin2 ax dx = x x sin 2ax+C = sin ax cos ax+C 4a 2a sin3 ax dx = cos 3ax cos ax +C 12a 4a x2 x x sin2 ax dx = sin 2ax cos 2ax + C (for a > 0, 4acb > 0) 4a 8a ( ) x x x dx x2 sin2 ax dx = sin 2ax cos 2ax+C = ln |2ax + b| (for a > 0, 4ac b2 = 0) 4a 8a 4a R a sin((b1 b2 )x) sin((b1 + b2 )x) 2ax + b dx sin0,b1|2ax x sin b+2 xb|dx +C (for |b1 = arcsin (for a < 0, 4acb2 < < = b2 2(b4ac) 2(b1 + b2 ) b2 ) R a b 4ac 4ax + 2b dx sinn1 ax cos ax n n = sin ax dx = + sinn2 ax dx (for n > R3 (4ac b2 )R na n ( ) 4ax + 2b 8a dx ax dx = + = ln tan +C R5 3(4ac b2 )R R2 4ac b2 sin ax a ( ) cos ax n2 dx dx 2ax + b dx dx = + (for n > 1) = + 4a(n 1) n n1 n2 2n+1 2n1 2n1 sin ax n R (2n 1)(4ac b ) R R a(1 n) sin ax sin ax 2ax + b dx = arsinh R a 4ac b2 19 sin ax x cos ax cos ax dx ax + C = x tan +C a a + cos ax a 2kn 2k+1n ax cos ax dx xn2k xn n xn12k n! +ax+ C n n1 k+1 = x n!cot (1) (1)k 2+2k x sin ax dx = cos ax+ x cos ax dx = cos cos ax a 1+2k a a a (n 2k)! a (n 2k k=0 k=0 sin(a1 a2 )x sin(a1 + a2 )x a2 cos a1 x cos a2 x dx = + +C (for |a1 | nx a3 (n2 6) 2(a1 a2 ) 2(a1 + a2 ) x2 sin2 dx = (for n = 2, 4, ) 2 a a 24n 1 tan ax dx = ln | cos ax|+C = ln | sec ax|+C 2n+1 a a sin ax (ax) dx = (1)n +C x (2n + 1) ã (2n + 1)! n n1 n=0 tan ax dx = tan ax tann2 ax dx (for n = 1) a(n 1) sin ax sin ax a cos ax dx = + dx dx q xn (n 1)xn1 n1 xn1 = (for p2 + (px+ ln |q sin ax+p cos ax|)+C ( ) q tan ax + p p + q a dx ax = tan +C dx x 1 sin ax a = + ln | sin ax + cos ax| + C ( ) ( ) tan ax + 2a x dx x ax ax +C = tan + ln cos dx x 1 + sin ax a a = + ln | sin ax cos ax| + C ( ) ( ) tan ax 2a x dx x ax ax = cot + ln sin +C sin ax a a x tan ax dx = ln | sin ax + cos ax| + C ( ) tan ax + 2a sin ax dx ax = x + tan +C sin ax a tan ax dx x = + ln | sin ax cos ax| + C tan ax 2a cos ax dx = sin ax + C a sec ax dx = ln |sec ax + tan ax| + C a x x cos2 ax dx = + sin 2ax+C = + sin ax cos ax+C 4a 2a sec2 x dx = tan x + C cosn1 ax sin ax n n n2 cos ax dx = + cos ax dx (for n > 0) na n secn2 ax tan ax n n sec ax dx = + secn2 ax dx (for n = cos ax x sin ax a(n 1) n1 x cos ax dx = + +C a2 a ( ) secn2 x tan x secn x dx = + x3 x x n1 2 x cos ax dx = + sin 2ax+ cos 2ax+C n2 n2 [3] sec x dx 4a 8a 4a n1 x sin ax dx = xn sin ax n x cos ax dx = a a n xn1 sin ax dx = 2k+1n 2kn n2k1 xn2k n! dx x x n! (1)k =2+2k (1)k 1+2k cos ax+ x tan + C (n a (n 2k)! sec x + a 2k 1)! k=0 k=0 dx x = x cot + C sec x csc ax dx = ln |csc ax + cot ax| + C a (for n = 1) csc2 x dx = cot x + C cos ax (ax)2k dx = ln |ax| + (1)k +C x 2k ã (2k)! k=1 cos ax a cos ax sin ax dx = dx xn (n 1)xn1 n xn1 ( ax ) dx = ln tan + +C cos ax a cscn1 ax csc ax n cscn ax dx = + cscn2 ax dx (for n sin ax n2 dx dx a(n 1) n = + (for n > 1) n n1 n2 cos ax a(n 1) cos ax n cos ax sin x2 dx dx ax =x +C = tan +C csc x + cos x2 + sin x2 + cos ax a sin x2 dx ax dx = x+C = cot +C csc x cos x2 sin x2 cos ax a x dx x ax ax cot ax dx = ln | sin ax| + C = tan + ln cos +C a + cos ax a a x ax ax x dx n n1 cot ax dx = cot ax cotn2 ax dx (for n = 1) = cot + ln sin +C a(n 1) cos ax a a 20 PHẫP TON TCH PHN sin2 ax dx sin ax dx = (for n = cosn ax a(n 1) cosn1 ax n cosn2 ax dx tan ax dx = sinn ax dx sinn1 ax sinn2 ax dx cot ax tan ax = + (for n = 1) cos ax a(n 1) cos ax ( ) dx ax = ln tan +C sinn ax dx sinn ax dx sinn+1 ax nm+2 cos ax sin ax a = ( m m1 ( cos ax a(m 1) cos ax m1 cosm2 ax dx ) = tan ax +C (cos ax sin ax)2 2a sinn ax dx sinn1 ax n1 sinn2 ax dx ) ( = + ( m cos ax cosm ax dx dx a(n m) cosm1 ax n m sin x cos x = 2(n 2) (cos x + sin x)n n (cos x + sin x)n1 (cos x + sin x)n2n1 sin ax n1 sinn ax dx sinn2 ax dx = (for cos ax dx x cosm ax a(m 1) cosm1 ax m cosm2 ax = + ln |sin ax + cos ax| + C cos ax + sin ax 2a cos ax dx x cos ax dx = +C (for n = 1) = ln |sin ax cos ax| + C sinn ax a(n 1) sinn1 ax cos ax sin ax 2a cos2 ax dx 1( ax ) sin ax dx x = cos ax + ln tan +C = ln |sin ax + cos ax| + C sin ax a cos ax + sin ax 2a ( ) sin ax dx x cos2 ax dx cos ax dx = ln |sin ax cos ax|+C = + (for n = cos ax sin ax 2a sinn ax n a sinn1 ax) sinn2 ax cos ax dx ax ax = tan2 + ln tan +C cosn ax dx cosn+1 ax nm2 cosn ax dx sin ax(1 + cos ax) 4a 2a = m m1 sin ax m1 a(m 1) sin ax sinm2 ax cos ax dx ax ax = cot2 ln tan +C cosn ax dx cosn1 ax n1 cosn2 ax dx sin ax(1 cos ax) 4a 2a = + (fo m ( ( ax sin )ax sinm ax a(n m) sinm1 ax n m sin ax dx ) ax = cot + + ln tan + +C cos ax(1 + sin ax) 4a 2a 2cosn4ax dx cosn1 ax n1 cosn2 ax dx = ( ( ) ( m) m1 sin ax dx ax sin ax a(m 1) sin ax m sinm2 ax ax = tan + ln tan + +C cos ax(1 sin ax) 4a 2a sin ax tan ax dx = (ln | sec ax+tan ax|sin ax)+C a sin ax cos ax dx = cos ax + C 2a tann ax dx = tann1 (ax)+C (for n = 1) cos((a1 a2 )x) cos((a1 + a2 )x) sin ax (for a(n sin a1 x cos a2 x dx = +C |a1 | = 1) |a2 |) 2(a1 a2 ) 2(a1 + a2) tann ax dx 1 = tann+1 ax+C (for n = 1) n n+1 sin ax cos ax dx = sin ax+C (for n = 1) cos2 ax a(n + 1) a(n + 1) cotn ax dx 1 n n+1 = cotn+1 ax+C (for n = 1) sin ax cos ax dx = cos ax+C (for n =sin 1) a(n + 1) ax a(n + 1) n ax dx sinn1 ax cosm+1 ax n cot n2 n m = m ax dx tan1n + sin ax cos (for ax+C m, n > 0) (for n = 1) sin ax cos ax dx = ax cos a(1 n) a(n + m) n+m x m2 x sinn+1 ax cosm1 ax m n n m arcsin dx = xax arcsin n x>2 0) sin ax cos ax dx = + sin axccos dx c +(for cm, a(n + m) n+m ( ) dx x x c2 x x = ln |tan ax| + C x arcsin dx = arcsin + c x2 sin ax cos ax a c c dx dx x n = 1)x3 x x2 + 2c2 = + (for n n1 x2 arcsin dx = arcsin + c x2 sin ax cos ax a(n 1) cos ax sin ax cosn2 ax c c dx dx = + (for n = 1) ( n+1 n1 n2 sinn ax cos ax x sin x a(n 1) sin ax sin ax cos axxn sin1 x dx = n+1 dx = + cot ax tan ax dx tan ax + sin ax dx = +C (for n = 1) cosn ax a(n 1) cosn1 ax ) ( ax ) xn x2 nxn1 sin1 x n2 1 sin2 ax dx +n x sin x dx = sin ax+ ln tan + +C + n1 cos ax a a 21 x x dx cx dx = x arccos c2 x2 = ln c c sinh cx c ( ) x x c x x dx cosh cx x arccos dx = arccos c2 x2 = ln c c sinh cx c sinh cx 2 x x x + 2c x dx sinh cx x2 arccos dx = arccos c2 x2 = ln c c sinh cx c cosh cx + x x c dx cosh cx arctan dx = x arctan ln(c2 + x2 ) = ln c c sinh cx c cosh cx + x c2 + x2 x cx dx x arctan dx = arctan = arctan ecx c c cosh cx c 3 x x x cx c n2 dx cosh cx dx x2 arctan dx = arctan + ln c2 + x2 (n = 1) = n c c 6 n1 n2 sinh cx c(n 1) sinh cx n sinh cx x xn+1 x c xn+1 dx sinh cx dx n2 xn arctan dx = arctan (ndx = 1) = + (n = 1) c n+1 c n+1 c2 + x2 n n1 cosh cx c(n 1) cosh cx n coshn2 cx x x x arcsec dx = x arcsec + ln |x x2 1| coshn1 cx coshn2 cx coshn cx n1 c c c|x| dx = dx ( + sinhm cx sinhm cx c(n m) sinhm1 cx n m ) 1( x arcsec x x2 x arcsec x dx = coshn cx nm+2 coshn+1 cx coshn cx + dx dx = m m1 sinh cx m1 c(m 1) sinh cx sinhm2 cx n x arcsec x dx = ( n+1 ( n1 1 coshn cx coshn1 cx n1 coshn2 cx 21 x arcsec x x x n+1 n dx = + dx m sinh cx c(m 1) sinhm1 cx m sinhm2 cx sinhm cx sinhm1 cx sinhm2 cx m1 ( ))) + dx = dx ( n coshn cx +(1 n) xn1 arcsec x + (1 n) xn2 arcsec x dx cosh cx c(m n) coshn1 cx m n sinhm cx sinhm+1 cx mn+2 sinhm cx dx = + dx n n1 x x c cosh cx n1 c(n 1) cosh cx coshn2 cx arccot dx = x arccot + ln(c2 + x2 ) c c sinhm cx sinhm1 cx m1 sinhm2 cx 2 dx = + dx x c +x x cx n n1 cosh cx x arccot dx = arccot + c(n 1) cosh cx n coshn2 cx c c 1 3 x x x cx c x sinh cx dx = x cosh cx sinh cx x2 arccot dx = arccot + ln(c2 +x2 ) c c c c 6 n+1 1 n+1 x cosh cx dx = x sinh cx cosh cx x x x c x dx xn arccot dx = arccot + (n = 1) c c c n+1 c n+1 c2 + x2 1 cx dx = ln | cosh cx| sinh cx dx = cosh cx c c 1 coth cx dx = ln | sinh cx| cosh cx dx = sinh cx c c n n1 x cx dx = cx+ tanhn2 cx dx (n = 1) sinh2 cx dx = sinh 2cx c(n 1) 4c n n1 x coth cx dx = coth cx+ cothn2 cx dx (n = 1) cosh cx dx = sinh 2cx + c(n 1) 4c 1 n1 sinh (b sinh cx cosh bxc cosh cx sinh bx) ( sinhn cx dx = sinhn1 cx cosh cx sinhn2 cx dxbx sinh (ncx >dx 0) = b c2 cn n 1 n+2 n+2 cosh bx (b sinh bx cosh cxc sinh cx cosh bx) sinhn cx dx = sinhn+1 cx cosh cx sinh cxcosh dx cx dx (n = < 0,2 n =21) b c c(n + 1) n+1 1 n1 n n1 n2 cosh (b sinh bx sinh cxc cosh bx cosh cx) cosh cx dx = sinh cx cosh cx+ cosh cx dxbx sinh(ncx>dx 0) = b c2 cn n n+2 a c n+2 coshn cx dx = sinh cx coshn+1 cx cosh cx dx (n 0), a c cosh(ax+b) sin(cx+d) dx = sinh(ax+b) sin(cx+d) cosh(ax+b) cos(cx+d) a + c2 a + ãc3 ã ã ã ã (2j 1) (2j) ! c2j = = a c 2j+1 j! 22j+1 cosh(ax+b) cos(cx+d) dx = sinh(ax+b) cos(cx+d)+ cosh(ax+b) sin(cx+d) a + c2 a + c2 ax2 x x e dx = arsinh dx = x arsinh x + c a c c (2n)! ( a )2n+1 2 x x x2n ex /a dx = arcosh dx = x arcosh x2 c2 n! c c x c x (|x| < |c|) ln cx dx = x ln cx x artanh dx = x artanh + ln |c2 x2 | c c x x c 2 arcoth dx = x arcoth + ln |x c | (|x| > |c|) (ln x)2 dx = x(ln x)2 2x ln x + 2x c c cx n n x c+x (ln cx) dx = x(ln cx) n (ln cx)n1 dx x x (x (0, c)) arsech dx = x arsech c arctan c c xc dx (ln x)i = ln | ln x| + ln x + x x x + x2 + c ln xc)) i ã i! arcsch dx = x arcsch +c ln (x (0, i=2 c c c dx x dx = + (n = 1) n n1 n1 ecx dx = ecx (ln x) (n 1)(ln x) n (ln x) c ) ( ln x m m+1 cx cx (m = 1) x ln x dx = x a dx = a (a > 0, a = 1) m + (m + 1)2 c ln a xm+1 (ln x)n n ecx xm (ln x)n dx = xm (ln x)n1 dx (m = xecx dx = (cx 1) m+1 m+1 c ( ) (ln x)n dx (ln x)n+1 x 2x = (n = 1) x2 ecx dx = ecx + x n+1 c c c ln x dx ln x 1 n cx n n cx n1 cx = (m = 1) x e dx = x e x e dx m m1 xm1 x (m 1)x (m 1) c c (ln x)n n (ln x)n1 dx (ln x)n dx ecx dx (cx)i = + ( m = 1) = ln |x| + m m1 x (m 1)x m1 xm x i ã i! i=1 m ( ) x dx xm+1 m+1 xm dx cx e dx ecx ecx = + ( n = 1) n n1 = n1 + c dx (n = 1)(ln x) (n 1)(ln x) n1 (ln x)n1 xn n1 x xn1 dx cx = ln | ln x| cx e ln x dx = e ln |x| Ei (cx) x ln x c (n 1)i (ln x)i dx ecx cx = ln | ln x| + (1)i e sin bx dx = (c sin bx b cos bx) n x ln x i ã i! c + b2 i=1 ecx dx ecx cos bx dx = (c cos bx + b sin bx) = (n = 1) c + b2 x(ln x)n (n 1)(ln x)n1 n(n 1) ecx sinn1 x n2 x sin(ln = x dx (sin(ln x) cos(ln x)) (c sin xn cos x)+ ecxx)sindx ecx sinn x dx = c2 + n2 c + n2 x ecx cosn1 x n(n 1) cx n cos(ln x)cos dxn2 = x (sin(ln x) + cos(ln x)) e cos x dx = (c cos x+n sin x)+ ecx dx 2 2 c +n c +n (ax + b)n+1 cx2 (ax + b)n dx = (n = 1) xecx dx = e a(n + 1) 2c 2 xà dx e(xà) /2 dx = (1 + erf ) = ln |ax + b| ax + b a sinh(ax+b) cos(cx+d) dx = PHẫP TON TCH PHN 23 x(ax+b)n dx = a(n + 1)x b (ax+b)n+1 + 1)(n + 2) a2 (n (n [1] {1,http://aas.aanda.org/index.php?option=com_article& 2}) access=standard&Itemid=129&url=/articles/aas/pdf/ 1998/10/h0596.pdf x x b dx = ln |ax + b| ax + b a a [2] Stewart, James Calculus: Early Transcendentals, 6th Edition omson: 2008 x b dx = + ln |ax + b| (ax + b) a (ax + b) a [3] Stewart, James Calculus: Early Transcendentals, 6th Edition omson: 2008 x a(1 n)x b dx = (n {1, 2}) (ax + b)n a2 (n 1)(n 2)(ax + b)n1 ( ) x2 (ax + b)2 dx = 2b(ax + b) + b2 ln |ax + b| ax + b a ( ) x b2 dx = ax + b 2b ln |ax + b| (ax + b)2 a ax + b ( ) x 2b b2 dx = ln |ax + b| + (ax + b)3 a3 ax + b 2(ax + b)2 ( ) x2 1 2b b2 dx = + (n {1, 2, 3}) (ax + b)n a (n 3)(ax + b)n3 (n 2)(a + b)n2 (n 1)(ax + b)n1 dx ax + b = ln x(ax + b) b x dx a ax + b = + ln x (ax + b) bx b x ( ) 1 dx ax + b = a + ln x2 (ax + b)2 b2 (ax + b) ab2 x b3 x x dx = arctan x2 + a2 a a dx x ax = arctanh = ln (|x| < |a|) x2 a2 a a 2a a + x dx x xa = arccoth = ln (|x| > |a|) x2 a2 a a 2a x + a dx 2ax + b arctan (4acb2 > 0) = 2 ax + bx + c 4ac b 4ac b2 dx 2ax + b 2ax + b b2 4ac = artanh = ln (4acb2 < 0) ax2 + bx + c b2 4ac b2 4ac b2 4ac 2ax + b + b2 4ac dx = (4ac b2 = 0) ax + bx + c 2ax + b x b dx dx = ln ax + bx + c 2 ax + bx + c 2a 2a ax + bx + c 2ax + b m 2an bm mx + n arctan (4acb2 > 0) dx = ln ax2 + bx + c + 2 ax + bx + c 2a a 4ac b 4ac b2 2ax + b mx + n m 2an bm artanh (4acb2 < 0) dx = ln ax2 + bx + c + ax2 + bx + c 2a a b2 4ac b2 4ac mx + n m 2an bm dx = ln ax2 + bx + c (4acb2 = 0) ax2 + bx + c 2a a(2ax + b) 2ax + b (2n 3)2a dx dx = + n 2 n1 2 (ax + bx + c) (n 1)(4ac b )(ax + bx + c) (n 1)(4ac b ) (ax + bx + c)n1 bx + 2c b(2n 3) x dx dx = (ax2 + bx + c)n (n 1)(4ac b2 )(ax2 + bx + c)n1 (n 1)(4ac b2 ) (ax2 + bx + c)n1 x2 dx b dx = ln 2 x(ax + bx + c) 2c ax + bx + c 2c ax + bx + c 24 NGUN, NGI ểNG GểP, V GIY PHẫP CHO VN BN V HèNH NH Ngun, ngi úng gúp, v giy phộp cho bn v hỡnh nh 6.1 Vn bn Cụng thc toỏn hc Ngun: https://vi.wikibooks.org/wiki/C%C3%B4ng_th%E1%BB%A9c_to%C3%A1n_h%E1%BB%8Dc?oldid=148841 Ngi úng gúp: Tnt1984, Doón Hiu, RadiX, AmieKim v ngi vụ danh 6.2 Hỡnh nh Tp_tin:Angle_acute.png Ngun: https://upload.wikimedia.org/wikipedia/commons/1/1e/Angle_acute.png Giy phộp: CC-BY-SA-3.0 Ngi úng gúp: No machine-readable source provided Own work assumed (based on copyright claims) Ngh s u tiờn: No machinereadable author provided Eddideigel assumed (based on copyright claims) Tp_tin:Angle_obtuse.png Ngun: https://upload.wikimedia.org/wikipedia/commons/c/c1/Angle_obtuse.png Giy phộp: CC-BY-SA3.0 Ngi úng gúp: No machine-readable source provided Own work assumed (based on copyright claims) Ngh s u tiờn: No machine-readable author provided Eddideigel assumed (based on copyright claims) Tp_tin:Cos.svg Ngun: https://upload.wikimedia.org/wikipedia/commons/b/b6/Cos.svg Giy phộp: Public domain Ngi úng gúp: self-made; graphed in GNUPlot edited in Illustrator Ngh s u tiờn: Self: Commons user Keytotime Tp_tin:Cot.svg Ngun: https://upload.wikimedia.org/wikipedia/commons/7/7f/Cot.svg Giy phộp: Public domain Ngi úng gúp: No machine-readable source provided Own work assumed (based on copyright claims) Ngh s u tiờn: No machine-readable author provided Keytotime assumed (based on copyright claims) Tp_tin:Csc.svg Ngun: https://upload.wikimedia.org/wikipedia/commons/5/5b/Csc.svg Giy phộp: Public domain Ngi úng gúp: ? 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