ĐỀ THI THỬ ĐẠI HỌC NĂM HỌC 2012-2013 MÔN TOÁN ĐỀ 18 docx

Nguoithay.vn a/ c/ cos ax+b a cos ax+b dx : I b/ sin ax+b dx cos ax+b sin ax+b a sin ax+b dx d/ cos ax+b dx ln sin ax+b sin ax+b ln cos ax+b f ( x)dx R sin m x; cos n x ta ý : b/ c nguyên hàm II Tính tích phân sau : 2005) I a ( sin 2x sin x cos x b a I sin x sin x dx 3cos x t 3cos x 2cos x s inx I 3cos x dx KQ: ln 1 t2 ;s inxdx=- tdt cosx= 3 x sin 2x cos x dx cos x I 2005 dx t2 1 t t tdt 2; x 2 t 2t dt 2 t t 34 27 Nguoithay.vn b I sin x cos x dx cos x 0 cos x 2sin x cos x dx s inxdx cosx+1 cos x dt=-sinxdx, x=0 t cosx 2 f ( x)dx t dt t t 2 t 2 t dt t f ( x)dx I t=2;x= 1 dt t 2 t 2t ln t 2ln 1 2 Tính tích phân sau a - I 2006 sin 2x cos2 x 4sin x a I cos 3x dx sin x 2005 I b sin 2x cos x 4sin x 2 dx t2 2tdt 2sin x cos x 8sin x cos x dx 3sin xdx x t 1; x 2 f ( x)dx I b I tdt 31 t cos2 x 4sin x sin xdx tdt t KQ: KQ: 3ln cos2 x 4sin x t dx 2 t dt 31 cos 3x dx sin x Ta có : cos3x=4cos3 x 3cos x 4cos2 x cosx= 4-4sin x cosx= 1-4sin x cosx cos3x Cho nên : f ( x)dx dx 1+sinx 4sin x s inx dt=cosxdx,x=0 t s inx 2 I f ( x)dx t=1;x= t f ( x)dx 4t t dt t cosxdx t 2 dt dt t 4t 8t 2t 3ln t 2 3ln Tính tích phân sau Trang - a sin xdx 2005 I b I 2006 sin x cos x sin 2x sin x cos x.cos dx x KQ: ln a I b I sin xdx x sin x cos x.cos 2 sin x cos x sin 2x sin x cos x dx s inx+cosx ; 4 sinx+cosx d sinx+cosx x x d s inx+cosx sinx+cosx ln s inx+cosx sin x 4 ln 2 ln1 ln Tính tích phân sau a b I 2006 cos2x sin x cos x Cho nên : f ( x)dx cos2x dx 2sin 2x dx Vì : cos x cos2 x sin x cos2x sinx-cosx+3 cosx-sinx dx s inx-cosx+3 f ( x)dx sinx-cosx+3 f ( x)dx I t2 dt t3 t dt t3 KQ: KQ: sin x cos x dx t2 31 t t2 cosx+sinx cosx-sinx cosx+sinx dx t 32 ln 4 dt= cosx+sinx dx; x t cos2x I 2006 a I ln cosx-sinx dx V ln cosx Vì : s inx+cosx= sin x Cho nên : I s inx dx 1+cosx sin x cos x dx s inx+cosx dx sinx+cosx sin xdx sin x cos x cosx 2, x t dt t3 32 - Trang b I I cos2xdx= dt 4 cos xdx dt cos2x dx 2sin 2x t 2sin x x cos2x dx 2sin 2x ln t dt 41 t t 1; x t ln Tính tích phân sau : a C 2006 I b a I b I 4sin3 x dx cos x 4sin3 x dx cos x I 2006 KQ: sin3x sin3 3x dx cos3x cos2 x cosx s inxdx=4 cosx s inxdx=4 1 cosx 2 2 sin3x sin3 3x dx cos3x Ta có : sin 3x sin3 3x sin 3x sin 3x sin3xdx=- dt dt=-3sin3xdx t cos3x x t f ( x)dx 1 t dt 32 t sin 3x.cos 3x 2; x t 2 1 t 2t ln t 1 t dt 31 t 1 ln Tính tích phân sau a I = sin x sin x cot gx dx sin x b I = 3 Trang 4 0 a I = sin( x) dx x) 4 d I = cos x( sin x cos x)dx sin x dx c I = sin( 2 3 sin x sin x cot gx dx sin x sin x cot xdx s inx s inx 3 - cot xdx sin x 3 cot x cot xdx sin( b I = sin( 2 x) dx x) d cosx+sinx cosx+sinx cosx-sinx dx cosx+sinx ln cosx+sinx 2 2 sin x dx c I = cos2x 1 cos2x+ cos4x dx 8 dx 12 cos4x 2cos 2x dx 40 1 x sin 2x sin 4x 32 16 4 d I = cos x( sin x cos x)dx Vì : sin x cos4 x sin 2 x Cho nên : I 2 12 sin x cos2xdx= cos2xdx- sin 2 x cos xdx 20 sin x 2 sin x 0 Tính tích phân sau a I = sin xdx b I = sin x cot gx dx 2 d */I = ( cos x tg x cot g x 2dx c I = sin x )dx 2 0 a I = sin xdx 2 cos x sinxdx=- 2cos x cos x d cosx cosx+ cos3 x cos x 0 15 - Trang b I = sin x cot gx dx dx sin x 2tdt cot x t t cot x x 2tdt t I dt 2t 2 1 t 3 t anx-cotx dx t anx-cotx dx Vì : tanx-cotx= Cho nên : x sinx cosx ;2 t anx-cotx dx t anx-cotx dx ln sin x 4 d I = ( cos x cos2x sin2x 2cot x t anx-cotx0;x 3 ; 3 cot x 3 I ln sin x sin x cos x sinxcosx 2x ; 6 cosx sinx 2tdt tg x cot g x 2dx c I = 3; x t dx sin x ; cos2x dx sin2x ln sin x )dx (1) x t dx dt , x t ;x t I cos t sin t dt sin t 3 cost dt sin x cosx dx 2 2I I Tính tích phân sau a tan xdx (Y-HN-2000) b cos2x dx (NT-2000) c sinx+cosx+2 4 Trang - cos x dx (NNI-2001) sin x sin x dx ( GTVT-2000) cos x e a tan xdx Ta có : f ( x) tan x sin x cos x d 2sin x dx (KB-03) sin x sin x dx cos x f cos x 1 cos x cos x cos x ó: I 1 dx cos x cos x f ( x)dx t anx+ tan x dx cos x tan x tan x x 4 3 12 2 12 12 * Chú ý : Ta cịn có cách phân tích khác : tan x tan x tan x 1 f ( x) tan x tan x 3 tan x tan x I tan x tan x dx 4 b dx cos x 3 tan x t anx+x I tan x tan x tan x 3 3 1 dx cos x tan x 1 dx 12 cos2x dx sinx+cosx+2 Ta có : f ( x) sinx+cosx+9 I cos x sin x cos2x sinx+cosx+9 cosx+sinx f ( x)dx sinx+cosx+2 cosx-sinx cosx+sinx sinx+cosx+9 cosx-sinx dx cosx+sinx=t-2.x=0 t t=3;x= cosx-sinx dx f ( x)dx s inx+cosx+2 dt 2 I t2 dt t3 1 t t2 sin t cost sin t cost sin t cost+9 dt 2 2, t t dt t3 t2 dt t3 1 2 1 2 sin t cost cost sin t dt sin t cost+9 - 2 2 f ( x) Trang c cos x dx sin x Ta có : f ( x) cot x I sin x cos6 x sin x 3sin x 3sin x sin x sin x sin x 2 dx dx 3 dx sin x sin x 4 1 x cot x 3cot x 3x sin x 4 sin x dx cos6 x cos x dx cos6 x d 4 1 tan x dx cos x 2 4 e sin x dx cos x 2sin x dx sin x f 23 12 1 dx cos x cos x cos2 x dx sin x tan x tan x d t anx tan x tan x d tan x 0 tan x tan x 2sin x dx cos2x d sin x sin x dx cos x 2 sin x dx cos2x 4 1 dx cos x cos x 1 tan x dx cos x t anx+ tan x tan x t anx- tan x cos2x dx 1 3 sin x sin x sin x 15 d cos2x cos2x ln cos2x ln sin x ln 2 Tính tích phân sau : 2 a sin x cos xdx b sin x dx s inx+ 3cosx c I J cos2x dx cosx- s inx cos6 x cos x d cosx cos x cos x.s inxdx 1 cos7 x cos5 x Trang K 2 a sin x cos xdx cos x dx s inx+ 3cosx sin 3x dx 2cos3x 0 35 - ln b 3sin 3x dx cos 3x sin 3x dx 2cos3x sin x cos x dx s inx+ 3cosx c Ta có : I J Do : 20 x 2sin cos x+ x x tan d tan x ln tan 2 sin x 3cos x dx s inx+ 3cosx I 3J 16 20 sin x 6 sin x ln dx x x tan 6 ln (1) 3cosx sin x 3cosx dx s inx+ 3cosx ln d tan 1 x x tan 2cos 2 6 - ln cos 3x 16 dx 201 cosx s inx+ 2 sin x I d cos 3x cos 3x I 3J s inx- 3cosx dx cosx- s inx 0 ln I 3J I t x dt dx cos t+3 x ;t 4 0.x cos2t dt sint+ 3cost dt sin t+3 J ln 16 ln 16 t 6 cos 2t+3 K I J (2) I ln J 2 Tính tích phân sau a dx sin x -99) b 10 c 10 4 sin x cos x sin x cos x dx (SPII-2000) d a dx s inx+cosx dx sin x b 4 s inx+cosx dx cos x s inxsin x+ dx tan x -LN-2000) -2000) dx 4 dx s inx+cosx - Trang tan t x x 2cos 1 t t2 dt 2 tan u f (t )dt t u2 2du I 2u u1 u2 2 2dt t t 0, x t 2 ; t tan u 2 2 du 2du 2 tan u cos u tan u u2 u1 u1 2dt t 2t du; t cos 2u 2dt 2dt ;x t2 dx dt t2 2t t2 t 1 x tan dx; 2 dx I dt arctan 2 arxtan sin10 x cos10 x sin x cos x dx c Ta có : sin10 x cos10 x sin x cos4 x sin x cos2 x cos4 x sin x cos6 x sin x cos2 x sin x cos2 x sin x cos4 x sin x cos2 x sin x sin x cos2 x I x 6 sin cos x+ sin x sin x+ s inxsin x+ f ( x)dx ln sin x 2 I x 6 s inx ln x s inxsin x+ I sin x 32.8 sin x cos x+ cosx sinx sin x ln 15 64 6 * 6 dx ln s inx = cosx-sinxco x s inxsin x+ 6 ln cosx-sinxco x ln sin x+ 6 * Chú ý : Ta cịn có cách khác Trang 10 1 cos4x+ cos8x 32 f ( x) cosx sinx sin x 15 32 dx s inxsin x+ Ta có : x 1 cos4x cos8x sin x 16 32 1 15 cos4x+ cos8x dx 32 32 d 15 32 cos 2 x - f(x)= s inxsin x+ dx cot x sin x I sin x s inx+ cosx 2 s inx 6 2d cot x cot x ln cot x ln cot x 6 Tính tích phân sau s inxcos3 x dx (HVBCVT-99) cos x 2 a b cos x cos 2 xdx ( HVNHTPHCM-98) c sin x dx cos x sin x s inxcos3 x a dx cos x -01) cos x (sin x)dx cos x dt dx cos x d -95) 2sin x cos xdx sin xdx t cos x I 1 t 22 t cos x t 1; x dt 1 dt 21 t t 2; x ln t t 2 t ln 2 b cos x cos 2 xdx cos2x cos4x cos2x+cos4x+cos4x.cos2x 2 1 1 cos2x+cos4x+ cos6x+cos2x cos2x+ cos4x+ cos6x 4 8 Ta có : f ( x) cos2 x cos2 x I 1 cos2x+ cos4x+ cos6x dx 8 1 sin x sin x sin x x 16 16 48 c sin x dx cos x sin x Vì : d sin x cos6 x d sin x cos6 x sin xdx 6sin5 x cos x 6cos5 x sin x dx 6sin x cos x sin x cos4 x 3sin x sin x cos2 x sin x cos2 x dx sin xdx sin x dx cos x sin x 3sin 2x cos 2xdx d sin x cos6 x 6 d sin x cos x sin x cos6 x ln sin x cos6 x - ln Trang 11 d dx cos x cos x dx cos x t anx+ tan x tan x d t anx Tính tích phân sau b sin x cos xdx (NNI-96) a sin11 xdx ( HVQHQT-96) 0 c cos x cos xdx (NNI-98 ) d cos2x dx -97 ) a sin11 xdx Ta có : sin11 x sin10 x.sinx= 1-cos2 x sinx= 1-5cos2 x 10cos3 x 10cos4 x 5cos5 x cos6 x sinx 1-5cos x 10cos3 x 10cos x 5cos5 x cos6 x s inxdx Cho nên : I 5 cos7 x cos6 x 2cos5 x cos x cos3 x cosx 118 21 b sin x cos xdx sin x cos x cos2x cos2x 2 1 cos2x 2cos x cos 2 x 1 2cos x cos 2 x cos2x-2cos 2 x cos3 x 1 1+cos4x 1+cos4x cos2x-cos 2 x cos3 x cos2xcos2x 8 2 1 cos2x-cos4x+cos4x.cos2x 16 3cos x cos6x-cos4x 32 cos6x+cos2x cos2x-cos4x+ 16 I 3cos x cos6x-cos4x dx 32 1 sin x sin x sin x x 32 64 32.6 32.4 d cos xdx cos2x dx 0 cosx dx cosxdx cosxdx 2 s inx s inx Trang 12 1 2 - b b f ( x)dx f (b x)dx x x b -x=t , suy x=b-t dx=-dt , b f ( x)dx f (b x)dx Vì tích phân khơng f (b t )dt b 0 : Tính tích phân sau a/ 4sin xdx s inx+cosx b/ dx f/ sin x cos x dx sin x cos3 x t n 4sin xdx a/ I s inx+cosx dt x x dx, x 0 2 4sin f ( x)dx 4cosx f (t )dt t t sinx+cosx dx ;x 2 t sin I sin x dx sin x cos6 x e/ xm x dx s inx+cosx 2 5cos x 4sin x d/ c/ log t anx dx t b b f (b t )( dt ) V b t t t cos I cos t dt cost+sint t dt f (t )dt 2 2I I 2 b/ I cos x dx tan x 5cos x 4sin x s inx+cosx 4 s inx+cosx s inx+cosx dx s inx+cosx dx dx - Trang 13 I 5cos x 4sin x s inx+cosx 5sin t cos t dx cost+sint dt 5sin x 4cosx s inx+cosx dx 2 2 2I s inx+cosx dx tan x dx cos x I 2 c/ log t anx dx dt , x dx t x x t ;x t f ( x)dx log t anx dx log tan Hay: f (t ) log 1 tan t tan t f (t )dt I t dt log 2 tan t dt t dt log 2 log t log tdt dt 2I t I sin x dx (1) sin x cos6 x d/ I sin sin cos6 t t cos6 x dx I (2) cos6 x sin x d t t cos6 x sin x dx cos6 x sin x 2I dx x2 I n e/ xm x dx -x suy x=1-t Khi x=0,t=1;x=1,t=0; dt=-dx 0 I 1 t m n t n (1 t ) m dt t ( dt ) xn (1 x)m dx 4sin x dx cosx s inxcos3 x dx cos x 4 3 x5 x Trang 14 dx -97 ) cosx+2sinx dx cos x 3sin x (XD-98 ) x s inx dx ( HVNHTPHCM-2000 ) cos x x sin x dx ( AN-97 ) cos x - s inx+2cosx dx 3sin x cosx -2000) Ta phân tích : s inx dx 1+cosx -2000 ) sin x cos x dx sin x cos3 x x sin x dx cos x I ln -2000 ) 10 asinx+bcosx+c dx a 's inx+b'cosx+c' asinx+bcosx+c dx a 's inx+b'cosx+c' A B a ' cosx-b'sinx a 's inx+b'cosx+c' C a 's inx+b'cosx+c' -S - Tính I : I A B a ' cosx-b'sinx a 's inx+b'cosx+c' C dx a 's inx+b'cosx+c' Ax+Bln a 's inx+b'cosx+c' C dx a 's inx+b'cosx+c' Tính tích phân sau : a s inx-cosx+1 dx s inx+2cosx+3 c b s inx+7cosx+6 dx 4sin x 3cos x a s inx-cosx+1 dx Ta có : f ( x) s inx+2cosx+3 cosx+2sinx dx ( XD-98 ) cos x 3sin x d I = sinx-cosx+1 sinx+2cosx+3 cos x sin x A 3sin x 3cos x B cosx-2sinx sinx+2cosx+3 f ( x) A 2B 2A B B 3A C C I I d s inx+2cosx+3 dx 5 s inx+2cosx+3 ln 10 5 J 42 dx s inx+2cosx+3 C sinx+2cosx+3 1 Thay vào (1) A A B s inx+ 2A+B cosx+3A+C s inx+2cosx+3 dx 5 ln s inx+2cosx+3 10 J - Tính tích phân J : - Trang 15 dx t ; x t 0, x cos x 2 2dt 2dt f ( x)dx 2 2t t t 2t t 2 t t dt t tan x dt t du t cos 2u tan u 2du cos 2u cos 2u f (t )dt u2 du j= u b u2 u1 cosx+2sinx dx; 4cos x 3sin x f ( x) I I I 3cos x 4sin x 5 cos x 3sin x ;B dx sin x s inx cot x dx sin x 2 u1 ; t 2dt tan u t ln 10 5 A tan u1 u2 u1 tan u2 B 3cos x 4sin x 4cos x 3sin x ln cos x 3sin x x 5 4 ln 10 3cosx 4sin x dx 3sin x cos x sin x sin x dx sin x s inx-cosx dx sin x 15sin 3x cos 3xdx 2 s inxcosx a 2cos x b sin x ln s inx dx cos x Trang 16 dx a,b 2 C 4cos x 3sin x 2 (3) ;C=0 cos5 x sin x dx u2 3 2 du cosx+2sinx 4cos x 3sin x A tan u J tan xdx 0 10 cos4x.cos2x.sin2xdx - tan x 11 dx ( KA-08) cos2x 12 cos x cos xdx (KA-09 ) 14 0 16 x sin x dx sin x cos x 18 sin 3x sin3 x dx cos3x cos x 4sin x dx (KA-06) sin 2004 x dx sin 2004 x cos 2004 x -06) 21 sin x sin2 x dx dx 20 sin x -05) 19 x sin x x cosx dx (KA-2011 ) x sin x cosx x sin x dx (KB-2011) cos x 15 17 dx (KB-08) sin x s inx+cosx 13 sin x s inxsin x+ -05) -06) -06) - Trang 17 ... 10cos3 x 10cos x 5cos5 x cos6 x s inxdx Cho nên : I 5 cos7 x cos6 x 2cos5 x cos x cos3 x cosx 118 21 b sin x cos xdx sin x cos x cos2x cos2x 2 1 cos2x 2cos x cos 2 x 1 2cos x cos 2 x cos2x-2cos... cos4x.cos2x.sin2xdx - tan x 11 dx ( KA-08) cos2x 12 cos x cos xdx (KA-09 ) 14 0 16 x sin x dx sin x cos x 18 sin 3x sin3 x dx cos3x cos x 4sin x dx (KA-06) sin 2004 x dx sin 2004 x cos 2004 x -06) 21 sin