《总习题四高等数学同济大学第六本.docx》由会员分享，可在线阅读，更多相关《总习题四高等数学同济大学第六本.docx（56页珍藏版）》请在三一办公上搜索。
1、总习题四高等数学同济大学第六本 总习题四 求下列不定积分(其中a, b为常数): 1. dxe-exx-x; =ee2xx 解 2. dxe-ex-x-1dx-1e2x-1de=x12ln|e-1e+1xx|+C. (1-x)x3dx; 1(x-1)2 解 3. (1-x)x6263dx=-dxdx-1(x-1)3dx=1x-1+12(1-x)21+C. a-x(a0); 113 解 x626a-xdx=3(a3)2-(x3)2d(x)=16a3ln|x+ax-a3333|+C. 4. dx; x+sinx 解 dx=d(x+sinx)=ln|x+sinx|+C. x+sinxx+sinx 5
2、. lnlnxxdx1+cosx11+cosx; 11dx=lnxlnlnx-lnx+C 解 6. lnlnxxdx=lnlnxdlnx=lnxlnlnx-lnxlnxx. sinxcosx1+sin4x4dx; sinx1+sin4 解 sinxcosx1+sinxdx=xdsinx=1211+(sin2x)2d(sin2x)=12arctansin2x+C. 7. tan4xdx; 解 tanxdx= =134sincos42xxdtanx=tan22xsin1tan1322xdtanxtantan24xx+1dtanx=(tanx-1+x+1)dtanx =tan3x-tanx+arct
3、antanx+c=tan3x-tanx+x+c. 8. sinxsin2xsin3xdx; 解 sinxsin2xsin3xdx=-(cos3x-cosx)sin3xdx 2 =-cos3xsin3xdx+cosxsin3xdx 22 =cos3xd(cos3x)+(sin4x+sin2x)dx 64 = 9. 解 112611111cos23x-116cos4x-18cos2x+C. dxx(x+4)dxx(x+4)6; 11x65=(-4xx+4)dx=14ln|x|-124ln(x+4)+C. 6dx(a0); 10. a-xa+xdx=du=adx+dx 解 222222a-xa-xa
4、-xa-xa+xa+x1x =aarcsin 11. dxx(1+x)xa-a-x22+C. ; 解 dxx(1+x)=211+(x)2dx=2ln(x+1+(x)+C=2ln(2x+1+x)+C. 12. xcos2xdx; 解 xcos2xdx=(x+xcos2x)dx=x2+xdsin2x 244 =x2+4114xsin2x-14111sin2xdx=14x+214xsin2x+18cos2x+C. 13. eaxcosbxdx; 解 因为 eaxcosbxdx=cosbxdea1ax=1aeaxcosbx+beaaxsinbxdx22 =ea1axcosbx+ba2sin22bxde
5、(1aeax=1aeaxcosbx+ba2ba2eaxsinbx-baeaxcosbxdx, 所以 e = 14. axcosbxdx=a2axa+bcosbx+eaxsinbx)+C1a+bdx1+ex22eax(acosbx+bsinbx)+C. ; 解 dx1+ex令1+e=ux1uxdln(u-1)=221u-12du=(1u-1-1u+1)du. =ln|u-1u+1dx|+c=ln1+e-11+e+1x+c. 15. x2x-1dx令x=sect22; 解 x2x-121sec2ttantsecttantdt=costdt=sint+C = 16. 解 x-1xdx+C. ; (a
6、-x)dx225/2令x=asint25/2(a-x)21(acost)25acostdt = = =1a441costan4tdt=1a31a4(tant+1)dtant 13a13a43t+x24tant+C 1a42+23xa-x2+C. (a-x) 17. dxx41+x2; 解 dxx4令x=tant21+xcostsin431tan4tsectsec2tdt =t4dt=cossin224ttdsint13sin3 =(1sint-1sin3t)dsint=-t+1sint+C =-(1+x)3x32+1+xx2+C. 18. xsin 解 xsinxdx; 令x=txdxtsin
7、t2tdt=2tsintdt 2 =-2t2dcost=-2t2cost+2cost2tdt =-2t2cost+4tdsint=-2t2cost+4tsint-4sintdt =-2t2cost+4tsint+4cost+C =-2xcosx+4xsinx+4cosx+C. 19. ln(1+x2)dx; 解 ln(1+x2)dx=xln(1+x2)-x =xln(1+x2)-2(1-11+x22x1+x2dx )dx =xln(1+x2)-2x+2arctanx+C. 20. 解 sincossincos122323xxxxdx; sin2dx=xcosxdtanx=(tanx-2tanx
8、tan2x+1)dtanx =tan2x-ln(tan21x+1)+C. 21. arctanxdx; 解 arctan =xarctan =xarctanxdx=xarctanx-(1-1x-x)dx11+xdx 1+xx-x+arctanx+C =(x+1)arctan 22. 1+cosxsinx1+cosxsinxx-x+C. dx; 2cosx2x2dx=2cscx2dx2=2ln|cscx2-cotx2|+C 解 dx=2sinx2. cos 23. x382(1+x)x38dx; 解 (1+x)2dx=14(1+x8)21dx411x4=+arctanx+C421+x84. 提示
9、: 已知递推公式 24. 解 dx(x+a)22n114=122a(n-1)(x+a)x22n-1+(2n-3)dx(x+a)22n-1. x8x+3x+2x8114dx; 14x884x+3x+2dx=x+3x+2dx4令x=t144t22t+3t+2dt =(1-41413t+2t+3t+2142)dt=14(1-4t+2+1t+1)dt =t-ln|t+2|+ln|t+1|+C 144 =x+ln4x+1x+244+C. 25. dx16-x4; 解 dx16-x114=1(4-x)(4+x)|+12arctanx2x222dx=18(14-x2+14+x2)dx =(ln|841322
10、+x2-x)+C =ln|2+x2-x|+116arctan+C. 26. dx; 1+sinx 解 dx=1+sinx =(sinxcos2sinxsinxsinx(1-sinx)1-sin2xdx=sinx-sincos22xxdxx-1+1cos2x)dx=secx-x+tanx+C. 27. dx; 1+cosx 解 dx=1+cosxxx+sinxx+sinxx+sinx2cos2x2dx=12xcos2x2dx+12sinxcos2x2dx =xdtan+tandx 22 =xtan 28. e 解 esinxxx2-tan3x2dx+tanx2dx=xtanx2+C. xcosx
11、cosx-sinx2cos3xdx; sinxsinxx-sinx2cosxdx=xecosxdx-esinxtanxsecxdx =xesinxdsinx-esinxdsecx =xdesinx-secxesinx+secxdesinx =xesinx-esinxdx-secxesinx+secxesinxcosxdx =xesinx-secxesinx+C. 3 29. xx(x+3x)dx; 3 解 xx(x+3x)tt+1dx令x=t61156tdt=6(-632tt+1)dtt(t+t)t2 =6ln+C=lnx(6x+1)6+C. 30. 解 dx(1+e)dxx2; 令1+e=t
12、x(1+e)x21t21t-1dt=(11-)dtt-1tt21 =ln(t-1)-lnt+C t1 =x-ln(1+ex)+ 31. 解 ee4x3x11+ex+C. +e2xx-e+1dx; e+ee2xx-x-2xee4x3x+e2xx-e+1dx=-1+edx=11+(e-ex-x)2d(e-ex-x) =arctan(ex-e-x)+C =arctan(2shx)+C. 32. xe(exx2+1)x2dx; 解 xex(e+1)dx=x(e+1)x2d(e+1)=-xdx1e+1x x =- =- =- =xe+1x+1e+1xdx=-xe+1x+1e(e+1)xxdexe+1xe
13、+1xxx+(1ex-1e+1xx)dexx+lne-ln(e+1)+C xxxee+1-ln(e+1)+C. 33. ln2(x+1+x2)dx; 解 ln2(x+1+x2)dx=xln2(x+1+x2)-xln2(x+1+x2)dx =xln2(x+1+x2)-2ln(x+1+x2)x1+x2dx =xln2(x+1+x2)-2ln(x+1+x2)d1+x2 =xln2(x+1+x2)-21+x2ln(x+1+x2)+21+x2ln(x+1+x2)dx =xln2(x+1+x2)-21+x2ln(x+1+x2)+2dx =xln2(x+1+x2)-21+x2ln(x+1+x2)+2x+C.
14、 34. lnx(1+x)23/2dx; 解 因为 1(1+x)lnx(1+x)23/223/2dx令x=tant1secx1+x23tsec2tdt=costdt=sint+C=x1+x2+C, 所以 dx=lnxd()=xlnx1+x2-x1+x21xdx =xlnx1+x2-ln(x+1+x)+C2. 35. 1-x2arcsinxdx; 解 1-x2arcsinxdx11令x=sinttcos2tdt=12(t+tcos2t)dt =t2+tsin2t=t2+tsin2t-sin2tdt 44444 =t2+tsin2t+cos2t+C 448111111 =(arcsinx)2+4x
15、arccosx1-x23112x1-x2arcsinx-14x+C1. 2 36. dx; 解 xarccosx1-x23dx=xarccosx2x1-x2dx=-xarccosxd1-x22 =-x21-x2arccosx+1-x2(x2arccosx)dx =-x21-x2arccosx+1-x2(2xarccosx-x211-x2)dx =-x21-x2arccosx+2x1-x2arccosxdx-x2dx =-x21-x2arccosx-x3-arccosxd(1-x2)3 33 =-x21-x2arccosx-x3-31232319(1-x)arccosx-2323122(1-x3
16、23x+292)dx =-x21-x2arccosx-x3-31(1-x)arccosx-2x+C3 =-131-x(x+1)arccosx-22x(x+6)+C. 37. dx; 1+sinxdx=dsinx=(-)dsinx 解 1+sinxsinx(1+sinx)sinx1+sinxcotx111cotx =ln|sin x|-ln|1+sin x|+C=-ln|csc x+1|+C . 38. 解 dxsin3xcosxxcosx1; =-1sinxcosxdcotx=-cosxsinxcos12dxsin3xdcotx=-cotx11cos22xdcotx 2 =-(+cotx)dc
17、otx=-ln|cotx|-cotx+C=ln|tanx|-cotx22sinx+C1. 39. ; (2+cosx)sinx 解 令u=tanx2dx, 则 1(2+1-u1+u22= (2+cosx)sinxdx)2u1+u221+u2du=1+u22(u+3)udu 12uu+33 =3132du+x2113ux2du=13ln(u+3)+213ln|u|+C =ln|tan+3tan|+C. 40. dx; sinx+cosx 解 dx=sinx+cosx =1sin2sin22sinxcosxsinxcosx(sinx-cosx)sinxcosxsinx-cos22xdu=sin2xcosx22sinx-1dx+cos2xsinx22cosx-1dx xx-1dsinx-cos2cos22xx-1dcosx)dcosx =(1-2112sin122x-1)dsinx-12(1-1212cos2x-1ln| =sinx-2ln|2sinx-12sinx+1|-cosx+122cosx-12cosx+1|+C =(sinx-cosx)+2ln|212sinx+12cosx+1|+C.
