数学分析课后答案(复旦大学-第三版)ch-6-12

1041'C¨'˘18‰¨'§1.‰¨'Vg9${K1.y†eRf(t)dt=F(t)+C§KRf(ax+b)dx=1aF(ax+b)+C.y†ˇRf(t)dt=F(t)+C§[F(t)+C]0=f(t)§K1aT(ax+b)0=1a[F(ax+b)]0=f(ax+b)§u·Rf(ax+b)dx=1aF(ax+b)+C.2.ƒe‰¨'(1)Z(2−sec2x)dx(2)Zx4−2x3+√x2dx(3)Z√x+3√x+2√x+23√x−2dx(4)Zex+1x+1x2+1x3dx(5)Z2cosx+12sinxdx(6)Zcosx−21+x2+14√1−x2dx(7)Z12cosx+sinx+1dx(8)Z2x+13x−ex5dx(9)Z(3−x2)3dx(10)Z1−1x2qx√xdx)(1)Z(2−sec2x)dx=2x−tanx+C(2)Zx4−2x3+√x2dx=15x5−12x4+13x32+C(3)Z√x+3√x+2√x+23√x−2dx=23x32+34x43−2x+3x23+4x12+C(4)Zex+1x+1x2+1x3dx=ex+ln|x|−1x−12x2+C(5)Z2cosx+12sinxdx=2sinx−12cosx+C(6)Zcosx−21+x2+14√1−x2dx=sinx−2arctanx+14arcsinx+C(7)Z12cosx+sinx+1dx=12sinx−cosx+x+C(8)Z2x+13x−ex5dx=1ln22x−1ln313x−ex5+C(9)Z(3−x2)3dx=Z(27−27x2+9x4−x6)dx=27x−9x3+95x5−17x7+C(10)Z1−1x2qx√xdx=Z(x34−x−54)dx=47x74+4x−14+C105§2.‰¨'O1.ƒe‰¨'(1)Zdx5x−7(2)Zcos(ωt−ϕ)dt(3)Zdxr1−x2+32(4)Zdx√1−2x2(5)Ztan10xsec2xdx(6)Zeαx·2xdx(7)Z(2x+3x)2dx(8)Ztanxdx(9)Ztanp1+x2·xdx√1+x2(10)Z(αx2+β)μxdx(μ6=−1)(11)Zdx1−cosx(12)ZdxA2sin2x+B2cos2x(13)Zsinx·cosx1+sin4xdx(14)Zdxsin2x+π4(15)Zx28p1+x3dx(16)Zsin2xcosx1+sin3xdx(17)Z1−2sinxcos2xdx(18)Zdxex+e−x(19)Zsinx+cosx3√sinx−cosxdx(20)Z1+sin2xsin2xdx(21)Zsln(x+√1+x2)1+x2dx(22)Zdx√1+e2x(23)Zdxx2−2x+2(24)Zdx(arcsinx)2√1−x2106(25)Zx2+7x2−2x−3dx(26)Zx2−1x4+1dx)(1)Zdx5x−7=15Zd(5x−7)5x−7=15ln|5x−7|+C(2)Zcos(ωt−ϕ)dt=1ωZcos(ωt−ϕ)d(ωt−ϕ)=1ωsin(ωt−ϕ)+C(3)Zdxr1−x2+32=2Zdx2+3r1−x2+32=2arcsinx2+3+C(4)Zdx√1−2x2=√22Zd(√x)q1−(√2x)2=√22arcsin(√2x)+C(5)Ztan10xsec2xdx=Ztan10xd(tanx)=−111tan11x+C(6)Zeαx·2xdx=Z(2eα)xdx=(2eα)xln(2eα)+C(7)Z(2x+3x)2dx=Z(4x+2·6x+9x)dx=4xln4+2ln66x+9xln9+C(8)Ztanxdx=Zsinxcosxdx=−Zd(cosx)cosx=−ln|cosx|+C=ln|secx|+C(9)Ztanp1+x2·xdx√1+x2=Ztanp1+x2d(p1+x2)=ln|secp1+x2|+C(10)Z(αx2+β)μxdx=12αZ(αx2+β)μd(αx2+β)=(αx2+β)μ+12α(μ+1)+C(11)Zdx1−cosx=Zcsc2x2dx2=−cotx2+C(12)ZdxA2sin2x+B2cos2x=1ABZ11+AB2tan2xdAtanxB=1ABarctanABtanx+C(13)Zsinx·cosx1+sin4xdx=12Z11+(sin2x)2d(sin2x)=12arctan(sin2x)+C(14)Zdxsin2x+π4=Zcsc2x+π4dx+π4=−cotx+π4+C(15)Zx28p1+x3dx=13Z8p1+x3d(1+x3)=827(1+x3)98+C(16)Zsin2xcosx1+sin3xdx=13Zd(1+sin3x)1+sin3x=13ln(1+sin3x)+C(17)Z1−2sinxcos2xdx=Zsec2xdx+2Zdcosxcos2x=tanx−2secx+C(18)Zdxex+e−x=Zdexe2x+1=arctan(ex)+C(19)Zsinx+cosx3√sinx−cosxdx=Zd(sinx−cosx)3√sinx−cosx=32(sinx−cosx)23+C(20)Z1+sin2xsin2xdx=Zcsc2xdx+Zd(sin2x)sin2x=−cotx+ln(sin2x)+C=−cotx+2ln|sinx|+C(21)Zsln(x+√1+x2)1+x2dx=Zqln(x+p1+x2)d(ln(x+p1+x2)=23[ln(x+p1+x2]32+C107(22)Zdx√1+e2x=−Zde−x√1+e−2x=−ln(e−x+p1+e−2x)+C(23)Zdxx2−2x+2=Zd(x−1)(x−1)2+1=arctan(x−1)+C(24)Zdx(arcsinx)2√1−x2=Zd(arcsinx)(arcsinx)2=−1arcsinx+C(25)Zx2+7x2−2x−3dx=Z1+2x+10(x+1)(x−3)dx=Z1−2x+1+4x−3dx=x−2ln|x+1|+4ln|x−3|+C=x+2ln(x−3)2|x+1|+C(26)Zx2−1x4+1dx=Z1−x−2x2+x−2dx=Zdx+1xx+1x2−2=√24lnx+1x−√2x+1x+√2+C=√24lnx2−√2x+1x2+√2x+1+C2.ƒe‰¨'(1)Zsin√x√xdx(2)Z(2√u+1)2u2du(3)Ze√x+1dx(4)Zx2√4−x2dx(5)Zpx2+a2dx(6)Zpx2−a2dx(7)Zdxp(x−a)(b−x)(8)Zdxx2pαx2+β(9)Zxdx√5+x−x2(10)Zp2+x−x2dx)(1)Zsin√x√xdx=−2Zsin√xd(√x)=−2cos√x+C(2)Z(2√u+1)2u2du=Z4u+4u32+1u2du=4ln|u|−8u−12−1u+C(3)-√1+x=t§Kx=t2−1,dx=2tdt§u·Ze√x+1dx=2Ztetdt=2(t−1)et+C=2(√1+x−1)e√1+x+C(4)Zx2√4−x2dx=−Z4−x2−4√4−x2dx=−Zp4−x2dx+4Zdx√4−x2=−x2p4−x2−2arcsinx2+4arcsinx2+C=2arcsinx2−x2p4−x2+C108(5)I=Zpx2+a2dx=xpx2+a2−Zx2√x2+a2dx=xpx2+a2−Zpx2+a2dx+Za2√x2+a2dx=xpx2+a2−I+a2ln|x+px2+a2|+C1§u·2I=x√x2+a2+a2ln|x+√x2+a2|+C1§lI=x2px2+a2+a22ln(x+px2+a2)+C(C=C12)(6)I=Zpx2−a2dx=xpx2−a2−Zx2√x2−a2dx=xpx2−a2−Zpx2−a2dx−Za2√x2−a2dx=xpx2−a2−I−a2ln|x+px2−a2|+C1§u·2I=x√x2−a2−a2ln|x+√x2−a2|+C1§lI=x2px2−a2−a22ln(x+px2−a2)+C(C=C12)(7)Zdxp(x−a)(b−x)=Zdxp−[x2−(a+b)x]−ab=Zdx−a+b2s−x−a+b22+a−b22=arcsinx−a+b2a−b2+C=arcsin2x−a−ba−b+C£¥ab⁄(8)Zdxx2pαx2+β=Zdxx3rα+βx2=−12Zd1x2rα+βx2=−1βrα+βx2+C(9)Zxdx√5+x−x2=Zxdxs214−x−122=Zx−12s214−x−122+12Zdxs214−x−122=−s214−x−122+12arcsinx−12√212+C=−p5+x−x2+12arcsin2x−1√21+C(10)Zp2+x−x2dx=Zs94−x−122dx=x−122s94−x−122+98arcsinx−1232+C=2x−14p2+x−x2+98arcsin2x−13+C3.ƒe‰¨'(1)Zx2cosxdx(2)Zx3lnxdx(3)Zlnxdx(4)Zxnlnxdx(nŒ)(5)Zarcsinx√1−xdx(6)Zcscxdx(7)Zcos(lnx)dx(8)Zxdxsin2x(9)Zxcos2xdx109(10)Zxsin2xdx(11)Zarccosxdx(12)Z(arcsinx)2dx(13)Zeaxcosbxdx(14)Zln(x+p1+x2)dx)(1)Zx2cosxdx=x2sinx−2Zxsinxdx=x2sinx+2xcosx−2Zcosxdx=x2sinx+2xcosx−2sinx+C(2)Zx3lnxdx=14x4lnx−14Zx3dx=14x4lnx−x416+C(3)Zlnxdx=xlnx−Zdx=xlnx−x+C(4)Zxnlnxdx=xn+1x+1lnx−1n+1Zxndx=xn+1n+1lnx−xn+1(n+1)2+C(5)Zarcsinx√1−xdx=−2arcsinx·√1−x+2Z1√1+xdx=−2√1−xarcsinx+4√1+x+C(6)Zcscxdx=Zdxsinx=Z12cos(x2)tanx2dx=Zdtanx2tanx2=lntanx2+C(7)ˇI=Zcos(lnx)dx=xcoslnx+Zsin(lnx)dx=xcos(lnx)+xsin(lnx)−Zcos(lnx)dx=xcos(lnx)+xsin(lnx)−I+C1§I=x2[cos(lnx)+sin(lnx)]+CC=C12(8)Zxdxsin2x=Zxcsc2xdx=−xcotx+Zcotxdx=−xcotx+ln|sinx|+C(9)Zxcos2xdx=12Zx(1+cos2x)dx=x24+12Zxcos2xdx=x24+x4sin2x−14Zsin2xdx=x24+x4sin2x+18cos2x+C(10)Zxsin2xdx=Zx(1−cos2x)dx=x22−Zxcos2xdx=x22−x24+x4sin2x+18cos2x+C=x24−x4sin2x−18cos2x(11)Zarccosxdx=xarccosx+Zx√1−x2dx=xarccosx−p1−x2+C(12)Z(arcsinx)2dx=x(arcsinx)2−Z2xarcsinx√1−x2dx=x(arcsinx)2+2arcsinx·p1−x2−2Zdx=x(arcsinx)2+2p1−x2arcsinx−2x+C(13)I=Zeaxcosbxdx=1aeaxcosbx+baZeaxsinbxdx=1aeaxcosbx+ba2eaxsinb