数学分析课后习题答案(华东师范大学版)

（2）因为duuC,所以df(x)f(x)C.证明当x0时,y1)dx1dxxdxx3dxx3dxxxx3x3C3x2---------------------------------------------------------------------12x12C4x26x9xCln4ln6ln921x2dxx11dx1(11)dx1(1arctanx)Ccos2xdxcosxsinxdx(cosxsinx)dxsinxcosxCcos2xdxcosxsinxdx(11)dxcotxtanxCsin2xcos2x32tdt(109)tdt(109)C90Cxxxdxx8dx8x8C1x1x)dx(1x1x)dx2dx2arcsinxC---------------------------------------------------------------------2x12(1x)ndx(1x)nd(1x)1(1x)n1Cdx1)d3x313x2arcsinx1arcsin3xC41cot(2x)Csin2(2x)2sin2(2x)---------------------------------------------------------------------dx(1cosx)dxdxcosxdxcotxdsinxcotx1Cdxdx(1sinx)dxdxdcosxtanx1Cdxdxdtanx2(tanx21)2tanx21⒀解法一：cscxdxsec(2x)dxsec(2x)d(2x)1dxsinxdxdcosx1lncosx1Csinxcosx12cosx1---------------------------------------------------------------------1xdcotxln|cotx|Cln|tanx|Cdx21arctanxC24(x2)2dxdlnxln|lnx|C5(1x5)3xdx1x24(x11ln|x2|C1ln|x2|Cx42x42dx(11)dxln|x|ln|1x|Cln|x|Cx1x(21)dxcosxdxdtanxln|tanx|Cdx(sinxcosx)dxcosxsinx---------------------------------------------------------------------dxexdxdexarctanexCx2dx(x1)2(x1)3dx(13)dxln|x1|2x1(x1)2(x1)3x12(x1)2asectdtln|secttant|C1ln|xx2a2|C1dasectdt1costdt1sintCa3sec3ta2a2x2a2xdx1x解令x6t,则xt6,dx6t5---------------------------------------------------------------------xdx6ttdt6(t1xtt1)1)dt6(t7t5t3t)6ln|t1|C(30)解令x1t,则x1t2,dx2tdt,x11dxt12tdt(12)2tdt(2t4t)dt(2t44)dt1lnx11dxlnx1C2x22x32x24x21x2arctanx1xdx21x21x2arctanx1(11)dx1x2arctanx1(xarctanx)C---------------------------------------------------------------------[ln(lnx)1]dxln(lnx)dx1dx1dx1dxxln(lnx)Cx(arcsinx)221x2arcsinx21x2xx2a2(x2a2xx2a2x2a2dxxx2a2x2a2dxa2ln(xx2a2)x2a2dx1(xx2a2a2ln(xx2a2))Cx2a2dx1(xx2a2a2ln(xx2a2))C---------------------------------------------------------------------f(x)dx[f(x)]adf(x)1[f(x)]a1Cf(x)1tann1xI所以tann2xdtanx1tann1x.1tann1xI.I(m,n)cosxsinxm1I(m2,n)I(m,n)cosmxsinnxdx1cosm1xdsinn1x1[cosm1xsinn1x(m1)cosm2xsinn2xdx]1[cosm1xsinn1x(m1)cosm2xsinnx(1cos2x)dx]1[cosm1xsinn1x(m1)(I(m2,n)I(m,n))]---------------------------------------------------------------------xdxx11dx(xx11)dxxxxln|x1|Cdx(21)dxln(x4)Cx4x3dx12x7dx12x27x122x27x122(x7)212x3ABxC1x3(1x)(1xx2)1x1xx2dx1dx1x2dx31x31xx2---------------------------------------------------------------------61xx221xx21ln(1x)21arctan2x1C61xx2dx21x421x4d(x1)1d(x1)x2dx1x2x212x212x212x21x12(x1)222(x1)22x121arctanx1ln42x12C2arctanx212lnx22x1Cx22x1ABxCDxE,解得(x1)(x21)2x1x21(x21)2A1,BC1,DE1,于是1dx1x1dx1x1dx1)24x14x212(x21)2---------------------------------------------------------------------1ln|x1|1ln(x21)1arctanx111(arctanxx)C4x2144(2x2(2x4x22x1arctan(2x1)2x15arctan(2x1)C42x22x12(2x1)2122(2x22x1)2dx2dtdt1d(2t)1arctan2tC14t221(2t)22解令ttandxdx---------------------------------------------------------------------d(3tanx)1arctan(3tanx)C6(13tan2x)dxcosxdx1cosxsinxsinxcosxdxcosxsinx2cosxdx，Isinxdx，dxI1(xln|cosxsinx|)C1xx221xx2)21xx2---------------------------------------------------------------------ln|x1x2x|C解令t1t2dx4tdt2t4tdt4t2dt41t21dt1dt4dt41dt[112]dt(1t)2(1t)21t21dt[11]dtln|1t|11C(1t)2(1t)21t1t1tln|11x|1xC---------------------------------------------------------------------dx1x解令xu，则dx2ududx2udu2(11)du2(uln|1u|)C1x2(xln|1x|)C(x1)Cxx21x211xx21---------------------------------------------------------------------xxdx(x2)3(x2)2dxln|x2|3x2(x2)2dxseccosxxdx(1tan2x)dtanxtanx1tan3xC⑾3x3x(x1)(x2)2x1x2(x2)22B2C1dx21dx21dx3x13x22ln|x2|1C解令1---------------------------------------------------------------------xarctan(1x)xln(2x