Toán 8 Phương pháp đổi biến số34691

PH D ng 1: NG PHÁP I BI N S u() = a ; u() = b t x = u(t) b  a   f (x).dx =  f[u(t)].u (t).dt Ví d 1: Tính tích phân sau :  a)  x dx  dx = Cos t.dt t x = Sin t t  [ Khi /2    x dx = =  t=  t = /2 x=0 x= 1  Sin t.cos t.dt = /2  b) (1  cos 2t).dt = (t+ dx   x2 /2  cos t cos t.dt = 0  /2 sin 2t  sin   ) = ( + )= 2 2   ) 2  dx = (tan2 t+ 1).dt ; t  ( ; t x = tant Khi /4  I= c)  1 x x =  tan t =0 x =  tan t = (tan t  1).dt = (tan t  1) dx Khi  /3 dx 1 x =  Cos t.dt  /3   Sin t Khi x=1 x= e  dt =(t)  t =0  t = /4  /4 =   dx = Cos t.dt t  (  t= x=0 x= /4 t x = Sin t   ; ] 2  t = /3 Cos t.dt  Cos t /3  dt =(t)  /3 =   t =  t = /6 D ng 2: t : t = v(x) => dt = v/(x).dx Khi bi n đ i f(x) thành m t bi u th c có d ng g[v(x)].v/(x) i c n : x = a  t = v(a) x = b  t = v(b) ThuVienDeThi.com   ; ) 2 b v(b) a v(a)  f (x).dx =  g[v(x)].v(x).dx Ví d 2: Tích tích phân sau :  x a)  2x dx t t =3+2x2 => dt = 4xdx => Khi  1 b) x  t=  t =9 x=1 x= x  2x dx = e -x dx  t.dt = t3 = (27 53 ) t t = x3 => dt = 3x.dx =>  1 x = 1 x=1 Khi -x  x e dx = 1 e3 dx  x ln x c) 1 t e e3 6x  2 3x  2x  dx 6x   3x2  2x  1 1 = e1 + e= (e ) 3 e dx x =ln3 t t= 3x2 2x+7 => dt=(6x2).dx Khi 1  t=  t =3 x=e x = e3 dx dt  x ln x =  t =ln t e d)  1 t t =lnx => dt = Khi  t=  t =8 x=0 x=1 dt =ln t t dx =  =ln8 ln7 Bài t p : Bài 1: Tính tích phân : a)  (3x  2)5 dx = dt =x.dx  t=  t = 1  e dt = e t dt =x.dx (3x  2)6 3.6 = (2)6 =  18 18 ThuVienDeThi.com b)  x(x  1) dx t t =x2+1 => dt = 2xdx => Khi 1  x(x  1) dx = 2 c)  (2x  1)7 dx =  t=1  t =2 x=0 x=1 t7  t dt = (2x  1)8 2.8 = = 27  14 14 58 38  16 16 x d) x  3.dx t t =x2 +3 => dt = 2xdx => 1 x = 1 x=0 Khi  x x  3.dx = 1 /  e)  Sinx dx  3Cosx  Sinx dx =  3Cosx /  f) dt =x.dx  t=4  t =3 t3 = ( 27 8) t t = 1+3cosx  dt =3sinx.dx =>  x=0  t= x = /2  t = Khi / t.dt = dt =x.dx dt  t = ln t 1 1 = ln1+ ln4= ln4 3  Cosx Sinx.dx /  t t = 1+cosx  dt =sinx.dx => dt=sinx.dx Khi x=0  t= x = /2  t = 1  Cosx Sinx.dx =2  t.dt =2 /  g) e tanx Cos x dx t t = tanx => dt = Khi /  e tanx Cos2 x x=0 x =/ dx =  e t dt =et t3 4 = + 3 dx cos2 x  t=  t =1 1 =e1 ThuVienDeThi.com dt =sinx.dx e2 dx e x ln x  h) e2 t t =lnx => dt = dx x Khi  t=  t =2 x=e x = e2 dx dt = =ln t =ln2   e x ln x t Bài 2: Tích tích phân sau :  2ln x dx x e  a) t t =6+2lnx => dt =2 Khi e   2ln x dx = x / b) =    Cosx sin x dx +  /  = t.dt = Cosx  Cos3 x.dx =  /  t=  t =8 x=1 x=e Cosx sin x.dx +  / / t4 = (16 64 ) Cosx(1  Cos2 x).dx =  /  Cosx sin x dx  /  / /  Cosx sin x dx  Cosx.sin x.dx / dt dx => = dx x x t t= cosx => dt = sinx.dx => dt =sinx.dx ic n: Khi x = /2  t = x=0  t =1 x = /2  t=0 I=  / c) d) t.dt   t.dt =2  t.dt =2 2dx   / 12 3.Cos  5 3x x.dx x  144 = tan3x 3 /9  /12 t3 =   = (tan tan )= ( 1) 9 t t =x2+144  dt =2x.dx=> Khi x = 5 x=9  t = 169  t = 225 ThuVienDeThi.com dt =x.dx x.dx = x  144  5 ln  e) 225 ex dx  ex x=0 x =ln3 dt ex dx =  =ln t x t 5e  169 t t = 5+ex Khi ln 225 dt  t = ln t 169 f)  x x  1.dx = (ln225ln169) =ln15ln13  dt =ex.dx  t=  t =8 =ln8ln6 x  => x=t2 +1 => dx =2t.dt t t= ic n: Khi  t=  t = x=1 x=5 2  x x  1.dx =  (t  1).t.(2t.dt) =  (2t  2t ).dt =( g) dx  2 4 x dx 4 2  Khi x =4 x=2  t=2  t =1 4t.dt =4  dt =4t =4(21)=4 t 1 x x => x=42t2 => dx =4t.dt 2 t t= ic n: 2t 2t + ) = / h)   4Sinx.Cos x.dx  dt =4Cos x.dx => t t = 1+4Sin x Khi /  1  4Sinx.Cos x.dx = 4 i) x=0  t= x = /6  t =  e  t.dt = 1 x x dx t t= Khi x => dt = x=1 x  t= t3 dt =cosx.dx = ( 27 1) dx => 2dt= ThuVienDeThi.com x dx = 26 24 +  t =2 x=4  x e x dx =2  e t dt =2et  Bài 3: a)  Sin x.dx = =2e2 2e /  / sin x.sin x.dx =  (1  cos x).sin x.dx 0 t t = cos x  dt =sinx.dx Khi x=0  t=1 x = /2  t =    2  Sin x.dx = (1  t )(dt) = (1  t ).dt =(t  0 /2  b) t3 1 ) =1 = 3 /2 Sin x.Cos2 x.dx =  /2  (1  cos Sin x.Cos x.sin x.dx = x)Cos x.sin x.dx t t = cos x  dt =sinx.dx Khi x=0  t=1 x = /2  t = /2    1 =  = 15 / / c) (1  cos x)Cos x.sin x.dx = (1  t ).t ( dt) = (t  t ).dt =(  Sin x.dx =  (1  cos x) sin x.dx 0 t t = cos x  dt =sinx.dx Khi x=0  t=1 x = /2  t = /  Sin x.dx =  (1  t / 3 2t t  ) / ) (dt) =  (1  t ) dt =  (1  2t  t ).dt 2 =(t  d) 2 = 8/15 / dx dx dx =  Cosx =   =   x  x  / sin(  x)  / 2.sin(  ).cos(  ) /4 4 ThuVienDeThi.com t t5  ) / dx x x    / 2.tan(  ).cos (  ) 4  x t t= tan(  ) => dt=   x 2.cos2 (  )  = x = /3 Khi ic n x = /4 tan /  12   t = tan  t = tan  12 tan  tan(  / 8) dx dt dt = =ln t =   t  t x  x tan(  /12)   / 2.tan(  ).cos (  ) tan tan 12 4   =ln(tan ) ln(tan ) 12 / / / Sin x Sin x e)  dx dx = tan x .dx =   2 Cos x Cos x Cos x Cos x 0 dx Cos x x=0  t= x = /4  t =  dt = t t = tan x Khi /  tan x / f)  Cos x  t dt  dx = tan5 x.dx Ta có : t3 = tan5x = (tan3xtanx)(tan2x + ) + tanx /  tan5 x.dx = /  (tan x  tanx)(tan x  )dx  / Tính K =   tan x.dx = K J (tan x  tanx)(tan x  1)dx = /  (tan x  tanx).d(tanx) tan x tan x  / 1  ) =  = 4 4 =( / J = /  /4 tan x.dx =    /4 d(Cosx) = ln  ln2 = (฀ ln Cosx ) Cosx 2 ThuVienDeThi.com / tan5 x.dx =   V y: 1  ln2 /4 / g) I=   Sin x.Cos x.dx ; Xét J= Sin x.Cos x.dx 0 /4  I+J= Sin x.Cos2 x.(Sin x  Cos2 x).dx = = Sin 2x dx = /4  I–J= /4 = /4   /4 = 0 Sin 2x.d(Sin2x) = V y I=  (  ) 32 24 x  1.dx t t=  x Sin x.Cos x.dx    1  Cos4x  dx =  x  Sin4x  Sin x.Cos x.(Cos x  Sin x).dx  h)  /4  /4 3 Sin 2x 24 J = /   32 Sin 2x.Cos2x.dx   /4 = 24  (  ) 32 24 x  => x=t3 +1 => dx =3t2.dt ic n: Khi 3  t=  t =  x x  1.dx =  x=1 x=3 (t  1).t.(3t dt) = 2  3t 3t + ) (3t  3t ).dt =( 3 12 3 45 2+ 32= 14  Tính tích phân b ng ph ng pháp đ i bi n s f(x) có ch a d ng sau :   a2  x2 đ t x= a.Sin t v i t  [ ; ] 2 ho c đ t x = a.Cos t v i t  [0 ; ]   a  x ho c (a2 + x2 ) đ t x = a.tan t v i t  ( ; ) 2 a  2 x a đ t x= v i t  [0;]\  Cos t = Bài : a) dx  1 x t x = tant   ) 2  dx = (tan2 t+ 1).dt ; t  ( ; ThuVienDeThi.com x =  tan t =0 x =  tan t = Khi /4  I= dx  b) Khi /4 dx  =  x2  dt =(t)    4Sin t / =   2Cos t.dt  2Cos t c)  x  x dx  t=  t = /4 x=0 x= 2Cos t.dt  /4  dx = 2Cos t.dt t x = 2Sin t  x2 /4 (tan t  1).dt = (tan t  1)  t =0  t = /4 t x = 2Sin t /4   /4 dt = t = /4 0  dx = 2.Cos t.dt Khi x /6 2  x dx =  t=  t = /6 x=0 x=  8.Sin t  4Sin t.Cos t.dt /6  = 16 Sin t Cos t Cos t.dt =16 /6 /6  Sin 2t.dt =  /  = 2(t  Sin 4t) =   4 d) x  dx t t=x2 4 => dt= 2x.dx => x2  ic n: Khi 4 x  x 4 dx =  t = 12 x=4 x= 4/3  12 1 0 dt t dt =x.dx  t = 4/3 4/3 = t 12 =  12 = e)  x  x dx =  x  x x.dx ThuVienDeThi.com (1  Cos 4t).dt t t=1+x2 => dt= 2x.dx => Khi ic n: 2  x  x x.dx = ( = f) t5  x 2 dt =x.dx ; x2 =t1  t=  t =2 x=0 x=1 2 1 t 5/ 2 3/2 = = (  (t  1) t.dt (t  t ).dt 1 1 5/ t3 ) =(  x dx 25  t3 ) 1 2 + 23 )(  )= 15 15 t x = 2Sin t  dx = 2.Cos t.dt 1 x = 1 x= Khi x /4   x dx = 1  t = /6  t = /4 8.Sin t  4Sin t.Cos t.dt  / /4 /4 / Sin 2t.dt =  (1  Cos 4t).dt = 16  Sin t Cos t Cos t.dt =16   /  /  / /4    5 = 2(  sin) 2(  sin )= + = 2(t  Sin 4t)  4 6 4 g)  x.dx 1 x ic n:  = 2 1 x   3/  t 1 x ic n: x.dx  x2 = +1 t t=1+x2 => dt= 2x.dx => = 3/ = t Khi 2  dt t x=0 x=1 = t dt =x.dx  t=  t = 3/4 x=0 x = 1/2 dt x.dx Khi x.dx h) t t=1x2 => dt= 2x.dx =>  dt =x.dx  t=  t =2 = 1 ThuVienDeThi.com Bài : e a) x dx  (ln x) Khi e x  (ln x)2 2 b) = Cos t.dt  Sin t x.dx = 1 x 0 /6  dt = /6 /6  t = x=0 2 x=   Cos t.dt  Cos t Khi 2  /6 t x2 = Sin t  2x.dx = Cos t.dt => x.dx= cost.dt  x4  x.dx   t =  t = /6 x=1 x= e /6 dx dx = Cos t.dt x  t ln x = Sin t  t = /6 Cos t.dt =  Sin t  /6  Cos t.dt = Cos t /6    = 12 dt = c)  t x.dx  2x x2 = Sin t  2 x.dx = Cos t.dt => x.dx= Khi  x.dx  2x = 2 /  2 cost.dt  t = x=0 x= 1  t = /4 Cos t.dt  Sin t = 2 /4  Cos t.dt = Cos t 2 ThuVienDeThi.com /4  dt =  2 ... Khi  t=  t =8 x=0 x=1 dt =ln t t dx =  =ln8 ln7 Bài t p : Bài 1: Tính tích phân : a)  (3x  2)5 dx = dt =x.dx  t=  t = 1  e dt = e t dt =x.dx (3x  2)6 3.6 = (2)6 =  18 18 ThuVienDeThi.com... Khi 1  x(x  1) dx = 2 c)  (2x  1)7 dx =  t=1  t =2 x=0 x=1 t7  t dt = (2x  1 )8 2 .8 = = 27  14 14 58 38  16 16 x d) x  3.dx t t =x2 +3 => dt = 2xdx => 1 x = 1 x=0 Khi  x x  3.dx... Khi ln 225 dt  t = ln t 169 f)  x x  1.dx = (ln225ln169) =ln15ln13  dt =ex.dx  t=  t =8 =ln8ln6 x  => x=t2 +1 => dx =2t.dt t t= ic n: Khi  t=  t = x=1 x=5 2  x x  1.dx =  (t  1).t.(2t.dt)