bai tap tich phan co loi giai

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<p>Gio n ph o trng pt cp 2-3 akiaCau1: Tnh cac tch phan sau:a/2 231x 2xI dx;xb/x 440J (3x e )dx. Giai:a/ Ta co:222111 2 2I dx ln|x| (ln2 1) (ln1 2) ln2 1.x x x| ` | ` + + + . , . ,b/ Ta co:4x2403J x 4e (24 4e) (0 4) 28 4e.2| ` . ,Cau2: Tnh tch phan: 1 520xI dx.x 1+Giai:T5 3 2 2x x (x 1) x(x 1) x. + + +Ta c: 113 4 2 2200x 1 1 1 1 1I x x dx x x ln(x 1)] ln2 .4 2 2 2 4 x 1| `] + + + ] ] + . ,Cau3: Tnh / 20sinxdx.cosx sinx+Giai:Ta co:sinx cosx sinx (A B)cosx (A B)sinxA Bcosx sinx cosx sinx cosx sinx + + | ` + + + + . ,ong nhat ang thc, ta c:A B 01A B .A B 1 2+ ' Vay:/ 2/ 2 / 200 0sinx 1 cosx sinx 1 1dx dx x ln(cosx sinx) .cosx sinx 2 2(cosx sinx 2 2 4 ]] + ]]+ +] ] Cau4: Tnh tch phan :2 220 2xI dx.1 xGiai:Trang1Trng pt cp 2-3 akia gio n ph o 12at x = sint, khi o: dx = costdt .oi can:vi x=0 t =0 2x= t2 4 ' Lai co: 2 2 2 22 2x dx sin t.costdt sin t.costdt sin tcostdt 1(1 cos2t)dt.cost cost 21 x 1 sin t Khi o: / 4/ 4001 1 1 1I (1 cos2t)dt t sin2t .2 2 2 8 4 | ` . ,Cau5: Tnh tch phan : 2/ 322dxIx x 1Giai:at 21 costx , khi o: dx dtsint sin t oi can: x=1 t =22x= t33 ' Khi o: / 2 / 22/ 2/ 3/ 3 / 321costdtsin tdt t161sint 1sin t Cau6: Tnh tch phan : 0aa xI dx, (a 0)a x+ &gt;Giai:at x a.cos2t, khi o: dx 2a.sin2tdt. oi can: x=-a t =2x=0 t4 ' Lai co:a x a a.cos2tdx ( 2a.sin2tdt) cott ( 2a.sin2tdt)a x a a.cos2t+ + 24a.cos t.dt 2a(1 cos2t)dt. + Trang2Gio n ph o trng pt cp 2-3 akiaDo o: / 2/ 2/ 4/ 41I 2a (1 cos2t)dt 2a t sin2t a 12 4 | ` | ` + . , . ,.Cau7: Tnh tch phan : / 32/ 6cosdxIsin x 5sinx 6 +Giai:at x = sint, khi o: dt = cosxdxoi can: 1x=t =6 23x= t3 2 ' Ta co: 2 2cosdx dt dt(t 2)(t 3) sin x 5sinx 6 t 5t 6 + +A B [(A B)t 2A 3B]dtdtt 3 t 2 (t 2)(t 3)+ | ` + . ,T o: A B 0 A 12A 3B 1 B 1+ ' ' Suy ra:2cosxdx 1 1dt.t 3 t 2 sin x 5sinx 6| ` + . , Khi o: 3/ 23/ 21/ 2 1/ 21 1 t 3 3(6 3)I dt ln lnt 3 t 2 t 2 5(4 3) | ` . , Cau8:: Tnh tch phan : 7 33 20x dxI1 x+Giai:at 3 2 3 2t x 1 t x 1, + + khi o:223t dt3t dt 2xdx dx .2x oi can: x=0 t =1 x= 7 t 2 ' Ta co: 3 3 23 43 2x dx x .3t dt3t(t 1)dt 3(t t)dt.2xt1 x +Khi o: 22 5 241 1t t 141I 3 (t t)dt 3 .5 2 10| ` . ,Trang3Trng pt cp 2-3 akia gio n ph o 12Cau9:: Tnh tch phan : 120081I x sinxdx Giai:Viet lai I ve di dang:0 12008 20081 0I x sinxdx x sinxdx. + (1)Xet tch phan 020081J x sinxdx. at x t dx dt khi o:223t dt3t dt 2xdx dx .2x oi can: {x=-1 t =1 x=0 t 0Khi o: 0 12008 20081 0I ( t) sin( t)dt x sinxdx. (2)Thay (2) vao (1) ta c I = 0.Cau10:: Tnh tch phan : / 2 44 40cos xI dx.cos x sin x+Giai:att x dx dt2 oi can: x=0 t =2x= t 02 ' Khi o: 40 / 2 / 2 4 44 4 4 44 4/ 2 0 0cos ( t)( dt)sin tdt sin x2I dx.cos t sin t cos x sin xcos ( t) sin ( t)2 2 + + + Do o:/ 2 / 2 4 44 40 0cos x sin x2I dx dx I .2 4 cos x sin x + + Cau11:: Tnh tch phan:1/ 21/ 21 xI cosx.ln dx.1 x | ` + . , Trang4Gio n ph o trng pt cp 2-3 akiaGiai:0 1/ 21/ 2 01 x 1 xI cosx.ln dx cosx.ln dx1 x 1 x | ` | ` + + + . , . , . (1)Xet tnh chat 01/ 21 xJ cosx.ln dx1 x | ` + . ,at x t dx dt oi can: 1 1x=-t =2 2x=0 t 0 'Khi o: 0 1/ 2 1/ 21/ 2 0 01 t 1 t 1 xI cos( t).ln dt cost.ln dt cosx.ln dx1 t 1 t 1 x+ | ` | ` | ` + + . , . , . , (2)Thay (2) vao (1) ta c I = 0.Cau12:: Tnh tch phan:1 4x1x dxI2 1+Giai:Bien oi I ve dang:0 1 4 4x x1 0x dx x dxI2 1 2 1 ++ + (1)Xet tch phan 0 4x1x dxJ2 1+at x = t dx = dtoi can: {x=-1 t =1 x=0 t 0. Khi o:0 1 1 4 4 t 4 xt t x1 0 0( t) dt t .2.dt x .2 .dxJ2 1 2 1 2 1 + + + (2)Thay (2) vao (1) ta c:1 1 1 1 4 x 4 4 x4x x x0 0 0 0x .2 .dx x dx x (2 1)dx 1I x dx .5 2 1 2 1 2 1+ + + + + Cau13: Tnh tch phan:/ 2 nn n0cos xdxIcos x sin x+Giai:att x dx dt2 Trang5Trng pt cp 2-3 akia gio n ph o 12oi can: x=0 t =2x= t 02 'Khi o: n0 / 2 / 2 n nn n n nn n/ 2 0 0cos t ( dt)sin tdt sin x2I dx.cos t sin t cos x sin xcos t sin t2 2 | ` . , | ` | ` + + + . , . , Do o:/ 2 / 2 n nn n0 0cos x sin x2I dx dx I .2 4 cos x sin x + + Cau14:: Tnh tch phan:20xsinxdxI .4 cos xGiai:Bien oi I ve dang:2 20 0 0xsinxdx xsinxdxI xf(sinx)dx.4 (1 sin x) 3 sin x + at x t dx dt oi can: {x=t =0 x=0 t Khi o: 02 2 2 20 0 0( t)sin( t)dt ( t)sintdt sintdt tsintdtI4 cos ( t) 4 cos t 4 cos t 4 cos t 2 2 20 0 0d(cost) d(cost) d(cost)I 2I4 cos t 4 cos t cos t 4 20 0d(cost) 1 cost 2 ln9I . ln .2 2 4 cost 2 8 cos t 4 + Cau15:: Tnh tch phan:230I x.cos xdx Giai:at x 2 t dx dt oi can: {x=2t =0 x=0 t 2 Khi o: 0 23 32 0I (2 t).cos (2 t)( dt) (2 t).cos tdt 2 2 23 30 0 02 cos tdt tcos tdt (cos3t 3cost)dt I2 + Trang6Gio n ph o trng pt cp 2-3 akia2012I sin3t 3sint 0 I 0.2 3| ` + . ,Cau16: Tnh tch phan:/ 201 sinxI ln dx.1 cosx+ | ` + . ,Giai:att x dx dt2 oi can: x=0 t =2x= t 02 'Khi o: 0 / 2/ 2 0 01 sin t1 cost 1 sint2I ln ( dt) ln dt ln dt1 sint 1 cost1 cos t2 | ` | `+ + + | ` | `. , + + | ` . , . , + . , . , / 201 sinxln dx I 2I 0 I 0.1 cosx+ | ` + . ,Cau17:: Tnh tch phan:/ 40I ln(1 tgx)dx. +Giai:att x dx dt4 oi can: x=0 t =4x= t 04 'Khi o: 0 / 4 / 4/ 4 0 01 tgt 2I ln[1 tg( t)dt ln(1 )dt ln dt4 1 tgt 1 tgt + + + + / 4 / 4 / 4/ 400 0 0[ln2 ln(1 tgt)]dt ln2 dt ln(1 tgt)dt ln2.t I + + ln2 ln22I I .4 8 Trang7Trng pt cp 2-3 akia gio n ph o 12Cau 18:Tnh tch phan: 221ln(1 x)I dx.x+Giai:at:21u ln(1 x) du dx1 xdx1 dvvxx + +' ' Khi o:22 211 11 1 1 1 1I ln(x 1) dx ln3 ln2 dxx x(x 1) 2 x 1 x| ` + + + + + + + . , 211 3ln3 ln2 (ln|x| ln(x 1)) ln3 3ln2.2 2 + + + +Cau 19:Tnh tch phan: 12 2x0(x x)e dx +Giai:12 2x0(x x)e dx +.at22xu x xdv e dx +' ( )2xdu 2x 1 dx1v e2 +' I = 112x 2 2x 21001 1e (x x) (2x 1)e dx e I2 2+ + I1 = 12x0(2x 1)e dx +, at2xu 2x 1dv e dx + ' 2xdu 2x 1dx1v e2 + ' I1 = 1 112x 2x 2 2x00 01 1 1e (2x 1) e dx (3e 1) e2 2 2+ </p> <p>= ( )2 2 21 13e 1 (e 1) e2 2 .VayI = 22 21 ee e2 2 Cau 20:Tnh tch phan:3 05 x1x .e dxGiai:I =3 05 x1x .e dx.att = x3 dt = 3x2dx, x = 0 t = 0,x = 1 t = 1 Trang8Gio n ph o trng pt cp 2-3 akia I = 0 1t t11 01 1 1( t).e dt t.e dt I3 3 3| ` . , . ViI1 = 1t0te dt . attu tdv e dt ' tdu dtv e ' I1 = 1 11t t t00 0e .t e dt e e 1 .VayI =11 1I3 3 Cau 21:Tnh tch phan: / 220I (x 1)sinxdx. +Giai:at:2du 2xdx u (x 1)v cosx dv sinxdx +' ' Khi o:/ 2 / 2/ 2200 0I (x 1)cosx 2 xcosxdx 1 2 xcosxdx + + + (1)Xet tch phan / 20J xcosxdx. at:u x du dxdv cosxdx v sinx ' ' Khi o:/ 2/ 2 / 20 00J xsinx sinxdx cosx 12 2 + (2)Thay (2) vao (1) ta c:I 1 2 1 1.2 | ` + . ,Cau 22:Tnh tch phan: 1x0xe dxGiai:1x0xe dx.att =x t2 = x2tdt = dx x = 1 t = 1,x = 0 t = 0 I = 1 12 t 3 t10 0t e 2tdt 2 t e dt 2I .at3tu tdv e dt' 2tdu 3t dtv e' I1 = 11t 3 t 2200e .t 3 e .t dt e 3I .ViI2 = 1t 20e .t dt.at2tu tdv e dt' tdu 2tdtv e ' Trang9Trng pt cp 2-3 akia gio n ph o 12 I2 = 11t 2 t300e .t 2 e tdt e 2I1 .viI3 = 1t0e tdt.attu tdv e dt ' tdu dtv e ' I3 = 1 11t t t00 0e .t e dt e e e (e 1) 1 VayI = 2I1 = 2(e 3I2) = 2e 6I2= 2e 6(e 2I3) = 12I3 4e = 12 4e Cau 23:Tnh tch phan: 2x 20I e sin xdx.Giai:Bien oi I ve dang:2x 2 2x0 01I e sin xdx e (1 cos2x)dx2 (1) Xet tch phan:22x 2x1001 e 1I e dx e2 2 2 (2) Xet tch phan:2x20I e cos2xdxat:2x 2xdu 2sin2xdxu cos2x1v e dv e dx2 ' ' Khi o:22x 2x 2x200 01 e 1I e cos2x e sin2xdx e sin2xdx2 2 2 + + (3) Xet tch phan:2x2, 10I e sin2xdxat: 2x 2xdu 2cos2xdxu sin2x1v e dv e dx2 ' ' Khi o:22x 2x2, 1 200I1I e sin e cos2xdx I .2 1 442 4 43(4)Thay (4) vao (3), ta c:2 22 2 2e 1 e 1I I I .2 2 4 4 (5)Thay (2), (5) vao (1), ta c: 2 221 e 1 e 1 1I [ ( )] (e 1).2 2 2 4 4 8 Trang10Gio n ph o trng pt cp 2-3 akia I1 = 1 11t t t00 0e .t e dt e e 1 .Vay I = 2Cau 24:Lap cong thc truy hoi tnh:/ 2nn0I sin x.dx (n N) Giai: at: n 1 n 2u sin x du (n 1)).sin x.dx dv sinx.dx v cosx. n 1 / 2n 0 n 2 n n n 2n 1I sin x.cosx] (n 1).(I I ) I In </p> <p> + </p> <p>Cau 25:Lap cong thc truy hoi tnh:/ 2nn0I cos x.dx (n N) Giai: at: n 1 n 2u cos x du (n 1).cos x.dx dv cosx.dx v sinx. n 1 / 2n 0 n 2 n n n 2n 1I cos x.sinx] (n 1).(I I ) I In </p> <p> + </p> <p>Cau 26:Lap cong thc truy hoi tnh:/ 2 / 2n nn n0 0I x .cosx.dx vaJ x .sinx.dx. Giai: at: n n 1u x du n.x .dx. dv cosx.dx v sinx nnn n n 1I x sinx nJ nJ (1)220 | ` . , Tng t: n n 1J 0 nI (2) + T (1) va (2) n nn n 2 n n 2I n(n 1)I . I n(1 n)I2 2 | ` | ` + + . ,. ,Trang11Trng pt cp 2-3 akia gio n ph o 12 Tng t co :n 1n n 2J n(1 n)J n2 | ` + . ,Cau 27:Lap cong thc truy hoi tnh:1n xn0I x .e .dx Giai: at: n n 1u x du nx .dx x xdv e .dx v e . n x 1n 0 n 1 n 1I [x .e ] nI e nI Cau 28:Lap cong thc truy hoi tnh:1 1 nn xn n x0 0xI dx hay I x .e .dxe Giai: at: n n 1u x du nx .dx </p> <p>x xdv e .dx v e . n x 1n 0 n 1 n 11I [ x .e ] nI nIe + +Cau 29:Lap cong thc truy hoi tnh:en *n1I ln x.dx (n Z ) Giai: at: n n 11u ln x du n.ln x, dxx </p> <p>dv dx v x. n en 1 n 1 n n 1I [x.ln x] n.I I e nI . Cau 30:Chng minh rang:Neu f(x) lien tuc va la ham chan tren oan [a ; a] th:a aa 0I f(x)dx 2 f(x)dx. Giai:Bien oi I ve dang:a 0 aa a 0I f(x)dx f(x)dx f(x)dx + (1) Trang12Gio n ph o trng pt cp 2-3 akiaXet tnh phan 0aJ f(x)dx. at x t dx dt oi can: {x=-a t =a x=0 t 0. Mat khac v f(x) la ham chan f(t) = f(t)Khi o: 0 a a aa 0 0 0J f( t)dt f(t)dt f(t)dt f(x)dx (2)Thay (2) vao (1) ta c a0I 2 f(x)dx </p> <p>Cau 31:Chng minh rang Neu f(x) lien tuc va la chan tren R th :x0f(x)dxI f(x)dx vi R vaa 0.a 1 + &gt;+ Giai:Bien oi I ve dang:0x x x0f(x)dx f(x)dx f(x)dxIa 1 a 1 a 1 ++ + + Xet tnh phan 01 xf(x)dxIa 1+at x t dx dt oi can: {x=0 t =0 x=- t . Mat khac v f(x) la ham chan f(t) = f(t).Khi o: 0 t t1 t t t0 0f( t)dt af(t)dt af(t)dtIa 1 a 1 a 1 + + + Vay:t xt x x0 0 0 0af(t)dt f(x)dx (a 1)f(x)dxI f(x)dx.a 1 a 1 a 1 + + + + + Cau 32:Chng minh rang Neu f(x) lien tuc tren0;2] ] ] th: / 2 / 20 0f(sinx)dx f(cosx)dx. Giai:att x dx dt2 oi can: x=0 t =2x= t 02 'Trang13Trng pt cp 2-3 akia gio n ph o 12Khi o: / 2 0 / 2 / 20 / 2 0 0f(sinx)dx f(sin( t)dt f(cost)dt f(cosx)dx2 </p> <p>Cau 33:Chng minh rang Neu f(x) lien tuc va f(a + b x) = f(x) th b ba aa bI xf(x)dx f(x)dx.2+ Giai:atx a b t dx dt + oi can: {x=a t =b x=b t a Khi o: a bb aI (a b t)f(a b t)( dt) (a b t)f(t)dt + + + b b b b ba a a a a(a b)f(t)dt tf(t)dt (a b) f(t)dt xf(x)dx (a b) f(t)dt I + + + b ba aa b2I (a b) f(t)dt I f(x)dx.2+ + Cau 34:Chng minh rang Neu f(x) lien tuc va f(a + b x) = f(x) th baI f(x)dx 0. Giai:atx a b t dx dt + oi can: {x=a t =b x=b t a Khi o: a b bb a aI f(a b t)( dt) f(t)dt f(x)dx I 2I 0 I 0. + Cau 35: Tnh tch phan sau:1x1J e 1dx. Giai:Xet dau cua ham so y = ex 1Ta co: y = 0 xe 1 0 x 0 Nhan xet rang: xx 0 e 1 y 0 &gt; &gt; &gt;;xx 0 e 1 y 0 </p>