高等数学（第七版）同济大学总习题四（后半部分）个人解答

高等数学（第七版）同济大学总习题四（后半部分）

解：

(20)∫sin2xcos3xdx=∫cos−3xsin2xdx，根据积分表（十一）公式99和98，得∫cos−3xsin2xdx=cos−2xsinx−∫cos−3xdx=cos−2xsinx−12cos−2xsinx−12∫1cosxdx=12cos−2xsinx−12ln∣secx+tanx∣+C=12secxtanx−12ln∣secx+tanx∣+C(21)设u=arctanx，则x=tan2u，dx=2tanusec2udu，得∫arctanxdx=2∫utanusec2udu=2∫usecud(secu)，设s=secu，ds=d(secu)，则u=arccos1s，得2∫usecud(secu)=2∫arccos1s⋅sds，令u=arccos1s，dv=sds，则du=−ss2−1ds，v=12s2，得2∫arccos1s⋅sds=s2arccos1s+∫s3s2−1ds，根据积分表（八）公式66得，s2arccos1s+∫s3s2−1ds=s2arccos1s−s2−1+C=(1+x)arctanx−x+C(22)∫1+cosxsinxdx=∫2∣cosx2∣2sinx2cosx2dx=±2∫cscx2d(x2)=±2ln∣cscx2−cotx2∣+C当cosx2>0时，ln∣cscx2−cotx2∣=ln(∣cscx2∣−∣cotx2∣)当cosx2<0时，ln∣cscx2−cotx2∣=ln(∣cscx2∣+∣cotx2∣)=−ln(∣cscx2∣−∣cotx2∣)因此，∫1+cosxsinxdx=2ln(∣cscx2∣−∣cotx2∣)+C(23)∫x3(1+x8)2dx=∫x4x(1+(x4)2)2dx，设u=x4，x=x4，dx=−14u34du，得∫x3(1+x8)2dx=∫uu4(1+u2)2⋅14u34du=14∫1(1+u2)2du，根据积分表（四）公式28，得14(u2(u2+1)+12∫11+u2du)=u8(u2+1)+18arctanu+C=x48(x8+1)+18arctanx4+C(24)设u=x4，x=u4，则dx=14u34du，得∫x11x8+3x4+2dx=∫u114u2+3u+2⋅14u34du=14∫u2u2+3u+2du=14∫(1+1u+1−4u+2)du=14u+14ln∣1+u∣−ln∣2+u∣+C=14x4+ln1+x442+x4+C(25)∫dx16−x4=18∫(14+x2+14−x2)dx=18∫14+x2dx+132∫(12+x+12−x)dx=116arctanx2+132ln∣2+x2−x∣+C(26)∫sinx1+sinxdx=∫2tanx21+tan2x2(tanx2+1)21+tan2x2dx=∫2tanx2(tanx2+1)2dx，设u=tanx2，则x=2arctanu，dx=21+u2du，得∫2tanx2(tanx2+1)2dx=4∫u(u+1)2⋅11+u2du=2∫(11+u2−1(1+u)2)du=2∫11+u2du−2∫1(u+1)2du=2arctanu+21+u+C=x+21+tanx2+C(27)∫x+sinx1+cosxdx=∫x+2sinx2cosx22cos2x2dx=2∫x2sec2x2d(x2)+∫tanx2dx，设u=x2，dv=sec2x2d(x2)，则du=12dx，v=tanx2，得2∫x2sec2x2d(x2)+∫tanx2dx=xtanx2−∫tanx2dx+∫tanx2dx+C=xtanx2+C(28)∫esinxxcos3x−sinxcos2xdx=∫xesinxcosxdx=∫esinxtanxsecxdx=∫xd(esinx)−∫esinxd(secx)设u=x，dv=d(esinx)，则du=dx，v=esinx，得∫xd(esinx)=xesinx−∫esinxdx，设u=esinx，dv=d(secx)，则du=esinxcosxdx，v=secx，得∫esinxd(secx)=secxesinx−∫esinxdx，因此∫xd(esinx)−∫esinxd(secx)=xesinx−∫esinxdx−secxesinx+∫esinxdx=(x−secx)esinx+C(29)设u=x6，则x=u6，dx=6u5du，得∫x3x(x+x3)dx=∫u2u6(u3+u2)⋅6u5du=6∫1u(u+1)du=6∫(1u−1u+1)du=6ln∣uu+1∣+C=6lnx6x6+1+C=lnx(x6+1)6+C(30)设u=ex，则x=lnu，dx=1udu，得∫dx(1+ex)2=∫1u(1+u)2du=∫(1u−11+u−1(1+u)2)du=lnu−ln(1+u)+11+u+C=x−ln(1+ex)+11+ex+C(31)∫e3x+exe4x−e2x+1dx=∫ex+e−xe2x−1+e−2xdx=∫1(ex−e−x)2+1d(ex−e−x)=arctan(ex−e−x)+C(32)设u=ex，则x=lnu，dx=1udu，得∫xex(ex+1)2dx=∫lnu(1+u)2du，令s=lnu，dv=1(1+u)2du，则ds=1udu，v=−11+u，得∫lnu(1+u)2du=−lnu1+u+∫1u(1+u)du=−lnu1+u+∫1udu−∫11+udu=−lnu1+u+lnu−ln(1+u)+C=−x1+ex+x−ln(1+ex)+C(33)设u=ln2(x+1+x2)，dv=dx，则u=2ln(x+1+x2)1+x2dx，v=x，得∫ln2(x+1+x2)dx=xln2(x+1+x2)−∫2xln(x+1+x2)1+x2dx=xln2(x+1+x2)−∫2ln(x+1+x2)d(1+x2)=xln2(x+1+x2)−21+x2ln(x+1+x2)+2x+C(34)设u=1x，则x=1u，dx=−1u2du，得∫lnx(1+x2)32dx=−∫ln1u(1+1u2)32⋅1u2du=∫ulnu(1+u2)32du=−∫lnud((1+u2)−12)，设s=lnu，dv=d((1+u2)−12)，则ds=1udu，v=11+u2，得−∫lnud((1+u2)−12)=−lnu1+u2+∫1u1+u2du=xlnx1+x2−∫11+x2dx=xlnx1+x2−ln(x+1+x2)+C(35)∫1−x2arcsinxdx=∫arcsinx11−x2dx−∫arcsinxx21−x2dx，设u=arcsinx，dv=11−x2dx，则du=11−x2dx，v=arcsinx，得∫arcsinx11−x2dx=(arcsinx)2−∫arcsinx11−x2dx，得∫arcsinx11−x2dx=12(arcsinx)2设u=arcsinx，dv=x21−x2dx，则du=11−x2dx，v=−x2+12arcsinx，得∫arcsinxx21−x2dx=12(arcsinx)2−12x1−x2arcsinx−12∫(arcsinx−x1−x2)11−x2dx=12(arcsinx)2−12x1−x2arcsinx−12∫arcsinx11−x2dx+14x2，原式得∫arcsinx11−x2dx−∫arcsinxx21−x2dx=12∫arcsinx11−x2dx+12x1−x2arcsinx−14x2=14(arcsinx)2+12x1−x2arcsinx−14x2+C(36)设x=cosu(0<u<π)，则1−x2=sinu，dx=−sinudu，得∫x3arccosx1−x2dx=−∫ucos3udu=−∫ud(sinu−13sin3u)=−u(sinu−13sin3u)+∫(sinu−13sin3u)du=−u(sinu−13sin3u)−13∫(2+cos2u)d(cosu)=−u(sinu−13sin3u)−23cosu−19cos3u+C=−131−x2(2+x2)arccosx−19x(6+x2)+C(37)∫cotx1+sinxdx=∫cosxsinx1+sinxdx=∫1−sin2xsinx(1+sinx)dx，设u=sinx，则x=arcsinu，dx=11−u2du，得∫1−sin2xsinx(1+sinx)dx=∫1−u2u(1+u)⋅11−u2du=∫1u(1+u)du=∫(1u−11+u)du=ln∣sinx1+sinx∣+C(38)∫dxsin3xcosx=−∫cotxsec2xd(cotx)，设u=cotx，则x=arccotu，dx=−11+u2du，得−∫cotxsec2xd(cotx)=−∫u(1+1u2)du=−12u2−ln∣u∣+C=−12cot2x−ln∣cotx∣+C(39)设u=cosx，则x=arccosu，dx=−11−u2du，得∫dx(2+cosx)sinx=∫1(u+2)(u2−1)du=∫[16(u−1)−12(u+1)+13(u+2)]du=16ln∣u−1∣−12ln∣u+1∣+13ln∣u+2∣+C=16ln∣1−cosx∣−12ln∣cosx+1∣+13ln∣cosx+2∣+C(40)∫sinxcosxsinx+cosxdx=12∫(sinx+cosx)2−1sinx+cosxdx=12∫(sinx+cosx)dx−12∫1sinx+cosxdx=12∫(sinx−cosx)−12∫1sinx+cosxdx，令u=tanx2，则sinx=2u1+u2，cosx=1−u21+u2，dx=11+u2du，得∫1sinx+cosxdx=∫22u+1−u2du=−∫2(u−1)2−2du=−22∫1u−1−2du+22∫1u−1+2du=22ln∣u−1+2u−1−2∣+C代入原式得∫sinxcosxsinx+cosxdx=12(sinx−cosx)−24ln∣tanx2−1+2tanx2−1−2∣+C\begin{aligned} &\ \ (20)\ \int \frac{sin^2\ x}{cos^3\ x}dx=\int cos^{-3}\ xsin^2\ xdx，根据积分表（十一）公式99和98，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int cos^{-3}\ xsin^2\ xdx=cos^{-2}\ xsin\ x-\int cos^{-3}\ xdx=cos^{-2}\ xsin\ x-\frac{1}{2}cos^{-2}\ xsin\ x-\frac{1}{2}\int \frac{1}{cos\ x}dx=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{2}cos^{-2}\ xsin\ x-\frac{1}{2}ln\ |sec\ x+tan\ x|+C=\frac{1}{2}sec\ xtan\ x-\frac{1}{2}ln\ |sec\ x+tan\ x|+C\\\\ &\ \ (21)\ 设u=arctan\sqrt{x}，则x=tan^2\ u，dx=2tan\ usec^2\ udu，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int arctan\ \sqrt{x}dx=2\int utan\ usec^2\ udu=2\int usec\ ud(sec\ u)，设s=sec\ u，ds=d(sec\ u)，则u=arccos\ \frac{1}{s}，\\\\ &\ \ \ \ \ \ \ \ \ \ 得2\int usec\ ud(sec\ u)=2\int arccos\ \frac{1}{s}\cdot sds，令u=arccos\ \frac{1}{s}，dv=sds，则du=-\frac{s}{\sqrt{s^2-1}}ds，v=\frac{1}{2}s^2，得\\\\ &\ \ \ \ \ \ \ \ \ \ 2\int arccos\ \frac{1}{s}\cdot sds=s^2arccos\ \frac{1}{s}+\int \frac{s^3}{\sqrt{s^2-1}}ds，根据积分表（八）公式66得，\\\\ &\ \ \ \ \ \ \ \ \ \ s^2arccos\ \frac{1}{s}+\int \frac{s^3}{\sqrt{s^2-1}}ds=s^2arccos\ \frac{1}{s}-\sqrt{s^2-1}+C=(1+x)arctan\sqrt{x}-\sqrt{x}+C\\\\ &\ \ (22)\ \int \frac{\sqrt{1+cos\ x}}{sin\ x}dx=\int \frac{\sqrt{2}\left|cos\frac{x}{2}\right|}{2sin\frac{x}{2}cos\frac{x}{2}}dx=\pm\sqrt{2}\int csc\frac{x}{2}d\left(\frac{x}{2}\right)=\pm\sqrt{2}ln\left|csc\frac{x}{2}-cot\frac{x}{2}\right|+C\\\\ &\ \ \ \ \ \ \ \ \ \ 当cos\frac{x}{2} \gt 0时，ln\left|csc\frac{x}{2}-cot\frac{x}{2}\right|=ln\left(\left|csc\frac{x}{2}\right|-\left|cot\frac{x}{2}\right|\right)\\\\ &\ \ \ \ \ \ \ \ \ \ 当cos\frac{x}{2} \lt 0时，ln\left|csc\frac{x}{2}-cot\frac{x}{2}\right|=ln\left(\left|csc\frac{x}{2}\right|+\left|cot\frac{x}{2}\right|\right)=-ln\left(\left|csc\frac{x}{2}\right|-\left|cot\frac{x}{2}\right|\right)\\\\ &\ \ \ \ \ \ \ \ \ \ 因此，\int \frac{\sqrt{1+cos\ x}}{sin\ x}dx=\sqrt{2}ln\left(\left|csc\frac{x}{2}\right|-\left|cot\frac{x}{2}\right|\right)+C\\\\ &\ \ (23)\ \int \frac{x^3}{(1+x^8)^2}dx=\int \frac{x^4}{x(1+(x^4)^2)^2}dx，设u=x^4，x=\sqrt[4]{x}，dx=-\frac{1}{4\sqrt[4]{u^3}}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{x^3}{(1+x^8)^2}dx=\int \frac{u}{\sqrt[4]{u}(1+u^2)^2}\cdot \frac{1}{4\sqrt[4]{u^3}}du=\frac{1}{4}\int \frac{1}{(1+u^2)^2}du，根据积分表（四）公式28，得\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{4}\left(\frac{u}{2(u^2+1)}+\frac{1}{2}\int \frac{1}{1+u^2}du\right)=\frac{u}{8(u^2+1)}+\frac{1}{8}arctan\ u+C=\frac{x^4}{8(x^8+1)}+\frac{1}{8}arctan\ x^4+C\\\\ &\ \ (24)\ 设u=x^4，x=\sqrt[4]{u}，则dx=\frac{1}{4\sqrt[4]{u^3}}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{x^{11}}{x^8+3x^4+2}dx=\int \frac{\sqrt[4]{u^{11}}}{u^2+3u+2}\cdot \frac{1}{4\sqrt[4]{u^3}}du=\frac{1}{4}\int \frac{u^2}{u^2+3u+2}du=\frac{1}{4}\int \left(1+\frac{1}{u+1}-\frac{4}{u+2}\right)du=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{4}u+\frac{1}{4}ln\ |1+u|-ln\ |2+u|+C=\frac{1}{4}x^4+ln\ \frac{\sqrt[4]{1+x^4}}{2+x^4}+C\\\\ &\ \ (25)\ \int \frac{dx}{16-x^4}=\frac{1}{8}\int \left(\frac{1}{4+x^2}+\frac{1}{4-x^2}\right)dx=\frac{1}{8}\int \frac{1}{4+x^2}dx+\frac{1}{32}\int \left(\frac{1}{2+x}+\frac{1}{2-x}\right)dx=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{16}arctan\frac{x}{2}+\frac{1}{32}ln\ \left|\frac{2+x}{2-x}\right|+C\\\\ &\ \ (26)\ \int \frac{sin\ x}{1+sin\ x}dx=\int \frac{\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}}{\frac{\left(tan\frac{x}{2}+1\right)^2}{1+tan^2\frac{x}{2}}}dx=\int \frac{2tan\frac{x}{2}}{\left(tan\frac{x}{2}+1\right)^2}dx，设u=tan\frac{x}{2}，则x=2arctan\ u，dx=\frac{2}{1+u^2}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{2tan\frac{x}{2}}{\left(tan\frac{x}{2}+1\right)^2}dx=4\int \frac{u}{(u+1)^2}\cdot \frac{1}{1+u^2}du=2\int \left(\frac{1}{1+u^2}-\frac{1}{(1+u)^2}\right)du=\\\\ &\ \ \ \ \ \ \ \ \ \ 2\int \frac{1}{1+u^2}du-2\int \frac{1}{(u+1)^2}du=2arctan\ u+\frac{2}{1+u}+C=x+\frac{2}{1+tan\frac{x}{2}}+C\\\\ &\ \ (27)\ \int \frac{x+sin\ x}{1+cos\ x}dx=\int \frac{x+2sin\frac{x}{2}cos\frac{x}{2}}{2cos^2\frac{x}{2}}dx=2\int \frac{x}{2}sec^2\frac{x}{2}d\left(\frac{x}{2}\right)+\int tan\frac{x}{2}dx，\\\\ &\ \ \ \ \ \ \ \ \ \ 设u=\frac{x}{2}，dv=sec^2\frac{x}{2}d\left(\frac{x}{2}\right)，则du=\frac{1}{2}dx，v=tan\frac{x}{2}，得\\\\ &\ \ \ \ \ \ \ \ \ \ 2\int \frac{x}{2}sec^2\frac{x}{2}d\left(\frac{x}{2}\right)+\int tan\frac{x}{2}dx=xtan\frac{x}{2}-\int tan\frac{x}{2}dx+\int tan\frac{x}{2}dx+C=xtan\frac{x}{2}+C\\\\ &\ \ (28)\ \int e^{sin\ x}\frac{xcos^3\ x-sin\ x}{cos^2\ x}dx=\int xe^{sin\ x}cos\ xdx=\int e^{sin\ x}tan\ xsec\ xdx=\int xd(e^{sin\ x})-\int e^{sin\ x}d(sec\ x)\\\\ &\ \ \ \ \ \ \ \ \ \ 设u=x，dv=d(e^{sin\ x})，则du=dx，v=e^{sin\ x}，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int xd(e^{sin\ x})=xe^{sin\ x}-\int e^{sin\ x}dx，\\\\ &\ \ \ \ \ \ \ \ \ \ 设u=e^{sin\ x}，dv=d(sec\ x)，则du=e^{sin\ x}cos\ xdx，v=sec\ x，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int e^{sin\ x}d(sec\ x)=sec\ xe^{sin\ x}-\int e^{sin\ x}dx，因此\\\\ &\ \ \ \ \ \ \ \ \ \ \int xd(e^{sin\ x})-\int e^{sin\ x}d(sec\ x)=xe^{sin\ x}-\int e^{sin\ x}dx-sec\ xe^{sin\ x}+\int e^{sin\ x}dx=(x-sec\ x)e^{sin\ x}+C\\\\ &\ (29)\ 设u=\sqrt[6]{x}，则x=u^6，dx=6u^5du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{\sqrt[3]{x}}{x(\sqrt{x}+\sqrt[3]{x})}dx=\int \frac{u^2}{u^6(u^3+u^2)}\cdot 6u^5du=6\int \frac{1}{u(u+1)}du=6\int \left(\frac{1}{u}-\frac{1}{u+1}\right)du=\\\\ &\ \ \ \ \ \ \ \ \ \ 6ln\ \left|\frac{u}{u+1}\right|+C=6ln\frac{\sqrt[6]{x}}{\sqrt[6]{x}+1}+C=ln\frac{x}{(\sqrt[6]{x}+1)^6}+C\\\\ &\ \ (30)\ 设u=e^x，则x=ln\ u，dx=\frac{1}{u}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{dx}{(1+e^x)^2}=\int \frac{1}{u(1+u)^2}du=\int \left(\frac{1}{u}-\frac{1}{1+u}-\frac{1}{(1+u)^2}\right)du=ln\ u-ln(1+u)+\frac{1}{1+u}+C=\\\\ &\ \ \ \ \ \ \ \ \ \ x-ln(1+e^x)+\frac{1}{1+e^x}+C\\\\ &\ \ (31)\ \int \frac{e^{3x}+e^x}{e^{4x}-e^{2x}+1}dx=\int \frac{e^x+e^{-x}}{e^{2x}-1+e^{-2x}}dx=\int \frac{1}{(e^x-e^{-x})^2+1}d(e^x-e^{-x})=arctan(e^x-e^{-x})+C\\\\ &\ \ (32)\ 设u=e^x，则x=ln\ u，dx=\frac{1}{u}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{xe^x}{(e^x+1)^2}dx=\int \frac{ln\ u}{(1+u)^2}du，令s=ln\ u，dv=\frac{1}{(1+u)^2}du，则ds=\frac{1}{u}du，v=-\frac{1}{1+u}，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{ln\ u}{(1+u)^2}du=-\frac{ln\ u}{1+u}+\int \frac{1}{u(1+u)}du=-\frac{ln\ u}{1+u}+\int \frac{1}{u}du-\int \frac{1}{1+u}du=\\\\ &\ \ \ \ \ \ \ \ \ \ -\frac{ln\ u}{1+u}+ln\ u-ln(1+u)+C=-\frac{x}{1+e^x}+x-ln(1+e^x)+C\\\\ &\ \ (33)\ 设u=ln^2(x+\sqrt{1+x^2})，dv=dx，则u=\frac{2ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx，v=x，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int ln^2(x+\sqrt{1+x^2})dx=xln^2(x+\sqrt{1+x^2})-\int \frac{2xln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx=\\\\ &\ \ \ \ \ \ \ \ \ \ xln^2(x+\sqrt{1+x^2})-\int 2ln(x+\sqrt{1+x^2})d(\sqrt{1+x^2})=\\\\ &\ \ \ \ \ \ \ \ \ \ xln^2(x+\sqrt{1+x^2})-2\sqrt{1+x^2}ln(x+\sqrt{1+x^2})+2x+C\\\\ &\ \ (34)\ 设u=\frac{1}{x}，则x=\frac{1}{u}，dx=-\frac{1}{u^2}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{ln\ x}{(1+x^2)^{\frac{3}{2}}}dx=-\int \frac{ln\frac{1}{u}}{\left(1+\frac{1}{u^2}\right)^{\frac{3}{2}}}\cdot \frac{1}{u^2}du=\int \frac{uln\ u}{(1+u^2)^{\frac{3}{2}}}du=-\int ln\ ud((1+u^2)^{-\frac{1}{2}})，\\\\ &\ \ \ \ \ \ \ \ \ \ 设s=ln\ u，dv=d((1+u^2)^{-\frac{1}{2}})，则ds=\frac{1}{u}du，v=\frac{1}{1+u^2}，得\\\\ &\ \ \ \ \ \ \ \ \ \ -\int ln\ ud((1+u^2)^{-\frac{1}{2}})=-\frac{ln\ u}{\sqrt{1+u^2}}+\int \frac{1}{u\sqrt{1+u^2}}du=\frac{xln\ x}{\sqrt{1+x^2}}-\int \frac{1}{\sqrt{1+x^2}}dx=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{xln\ x}{\sqrt{1+x^2}}-ln(x+\sqrt{1+x^2})+C\\\\ &\ \ (35)\ \int \sqrt{1-x^2}arcsin\ xdx=\int arcsin\ x\frac{1}{\sqrt{1-x^2}}dx-\int arcsin\ x\frac{x^2}{\sqrt{1-x^2}}dx，\\\\ &\ \ \ \ \ \ \ \ \ \ 设u=arcsin\ x，dv=\frac{1}{\sqrt{1-x^2}}dx，则du=\frac{1}{\sqrt{1-x^2}}dx，v=arcsin\ x，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int arcsin\ x\frac{1}{\sqrt{1-x^2}}dx=(arcsin\ x)^2-\int arcsin\ x\frac{1}{\sqrt{1-x^2}}dx，得\int arcsin\ x\frac{1}{\sqrt{1-x^2}}dx=\frac{1}{2}(arcsin\ x)^2\\\\ &\ \ \ \ \ \ \ \ \ \ 设u=arcsin\ x，dv=\frac{x^2}{\sqrt{1-x^2}}dx，则du=\frac{1}{\sqrt{1-x^2}}dx，v=-\frac{x}{2}+\frac{1}{2}arcsin\ x，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int arcsin\ x\frac{x^2}{\sqrt{1-x^2}}dx=\frac{1}{2}(arcsin\ x)^2-\frac{1}{2}x\sqrt{1-x^2}arcsin\ x-\frac{1}{2}\int (arcsin\ x-x\sqrt{1-x^2})\frac{1}{\sqrt{1-x^2}}dx=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{2}(arcsin\ x)^2-\frac{1}{2}x\sqrt{1-x^2}arcsin\ x-\frac{1}{2}\int arcsin\ x\frac{1}{\sqrt{1-x^2}}dx+\frac{1}{4}x^2，原式得\\\\ &\ \ \ \ \ \ \ \ \ \ \int arcsin\ x\frac{1}{\sqrt{1-x^2}}dx-\int arcsin\ x\frac{x^2}{\sqrt{1-x^2}}dx=\frac{1}{2}\int arcsin\ x\frac{1}{\sqrt{1-x^2}}dx+\frac{1}{2}x\sqrt{1-x^2}arcsin\ x-\frac{1}{4}x^2=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{4}(arcsin\ x)^2+\frac{1}{2}x\sqrt{1-x^2}arcsin\ x-\frac{1}{4}x^2+C\\\\ &\ \ (36)\ 设x=cos\ u\ (0 \lt u \lt \pi)，则\sqrt{1-x^2}=sin\ u，dx=-sin\ udu，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{x^3arccos\ x}{\sqrt{1-x^2}}dx=-\int ucos^3\ udu=-\int ud\left(sin\ u-\frac{1}{3}sin^3\ u\right)=\\\\ &\ \ \ \ \ \ \ \ \ \ -u\left(sin\ u-\frac{1}{3}sin^3\ u\right)+\int \left(sin\ u-\frac{1}{3}sin^3\ u\right)du=-u\left(sin\ u-\frac{1}{3}sin^3\ u\right)-\frac{1}{3}\int (2+cos^2\ u)d(cos\ u)=\\\\ &\ \ \ \ \ \ \ \ \ \ -u\left(sin\ u-\frac{1}{3}sin^3\ u\right)-\frac{2}{3}cos\ u-\frac{1}{9}cos^3\ u+C=-\frac{1}{3}\sqrt{1-x^2}(2+x^2)arccos\ x-\frac{1}{9}x(6+x^2)+C\\\\ &\ \ (37)\ \int \frac{cot\ x}{1+sin\ x}dx=\int \frac{\frac{cos\ x}{sin\ x}}{1+sin\ x}dx=\int \frac{\sqrt{1-sin^2\ x}}{sin\ x(1+sin\ x)}dx，\\\\ &\ \ \ \ \ \ \ \ \ \ 设u=sin\ x，则x=arcsin\ u，dx=\frac{1}{\sqrt{1-u^2}}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{\sqrt{1-sin^2\ x}}{sin\ x(1+sin\ x)}dx=\int \frac{\sqrt{1-u^2}}{u(1+u)}\cdot \frac{1}{\sqrt{1-u^2}}du=\int \frac{1}{u(1+u)}du=\int \left(\frac{1}{u}-\frac{1}{1+u}\right)du=\\\\ &\ \ \ \ \ \ \ \ \ \ ln\ \left|\frac{sin\ x}{1+sin\ x}\right|+C\\\\ &\ \ (38)\ \int \frac{dx}{sin^3\ xcos\ x}=-\int cot\ xsec^2\ xd(cot\ x)，设u=cot\ x，则x=arccot\ u，dx=-\frac{1}{1+u^2}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ -\int cot\ xsec^2\ xd(cot\ x)=-\int u\left(1+\frac{1}{u^2}\right)du=-\frac{1}{2}u^2-ln\ |u|+C=-\frac{1}{2}cot^2\ x-ln\ |cot\ x|+C\\\\ &\ \ (39)\ 设u=cos\ x，则x=arccos\ u，dx=-\frac{1}{\sqrt{1-u^2}}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{dx}{(2+cos\ x)sin\ x}=\int \frac{1}{(u+2)(u^2-1)}du=\int \left[\frac{1}{6(u-1)}-\frac{1}{2(u+1)}+\frac{1}{3(u+2)}\right]du=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{6}ln\ |u-1|-\frac{1}{2}ln\ |u+1|+\frac{1}{3}ln\ |u+2|+C=\frac{1}{6}ln\ |1-cos\ x|-\frac{1}{2}ln\ |cos\ x+1|+\frac{1}{3}ln\ |cos\ x+2|+C\\\\ &\ \ (40)\ \int \frac{sin\ xcos\ x}{sin\ x+cos\ x}dx=\frac{1}{2}\int \frac{(sin\ x+cos\ x)^2-1}{sin\ x+cos\ x}dx=\frac{1}{2}\int (sin\ x+cos\ x)dx-\frac{1}{2}\int \frac{1}{sin\ x+cos\ x}dx=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{2}\int (sin\ x-cos\ x)-\frac{1}{2}\int \frac{1}{sin\ x+cos\ x}dx，\\\\ &\ \ \ \ \ \ \ \ \ \ 令u=tan\frac{x}{2}，则sin\ x=\frac{2u}{1+u^2}，cos\ x=\frac{1-u^2}{1+u^2}，dx=\frac{1}{1+u^2}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{1}{sin\ x+cos\ x}dx=\int \frac{2}{2u+1-u^2}du=-\int \frac{2}{(u-1)^2-2}du=\\\\ &\ \ \ \ \ \ \ \ \ \ -\frac{\sqrt{2}}{2}\int \frac{1}{u-1-\sqrt{2}}du+\frac{\sqrt{2}}{2}\int \frac{1}{u-1+\sqrt{2}}du=\frac{\sqrt{2}}{2}ln\ \left|\frac{u-1+\sqrt{2}}{u-1-\sqrt{2}}\right|+C\\\\ &\ \ \ \ \ \ \ \ \ \ 代入原式得\int \frac{sin\ xcos\ x}{sin\ x+cos\ x}dx=\frac{1}{2}(sin\ x-cos\ x)-\frac{\sqrt{2}}{4}ln\ \left|\frac{tan\frac{x}{2}-1+\sqrt{2}}{tan\frac{x}{2}-1-\sqrt{2}}\right|+C & \end{aligned} (20) ∫cos3 xsin2 xdx=∫cos−3 xsin2 xdx，根据积分表（十一）公式99和98，得 ∫cos−3 xsin2 xdx=cos−2 xsin x−∫cos−3 xdx=cos−2 xsin x−21cos−2 xsin x−21∫cos x1dx= 21cos−2 xsin x−21ln ∣sec x+tan x∣+C=21sec xtan x−21ln ∣sec x+tan x∣+C (21) 设u=arctanx，则x=tan2 u，dx=2tan usec2 udu，得 ∫arctan xdx=2∫utan usec2 udu=2∫usec ud(sec u)，设s=sec u，ds=d(sec u)，则u=arccos s1，得2∫usec ud(sec u)=2∫arccos s1⋅sds，令u=arccos s1，dv=sds，则du=−s2−1sds，v=21s2，得 2∫arccos s1⋅sds=s2arccos s1+∫s2−1s3ds，根据积分表（八）公式66得， s2arccos s1+∫s2−1s3ds=s2arccos s1−s2−1+C=(1+x)arctanx−x+C (22) ∫sin x1+cos xdx=∫2sin2xcos2x2∣∣cos2x∣∣dx=±2∫csc2xd(2x)=±2ln∣∣csc2x−cot2x∣∣+C 当cos2x>0时，ln∣∣csc2x−cot2x∣∣=ln(∣∣csc2x∣∣−∣∣cot2x∣∣) 当cos2x<0时，ln∣∣csc2x−cot2x∣∣=ln(∣∣csc2x∣∣+∣∣cot2x∣∣)=−ln(∣∣csc2x∣∣−∣∣cot2x∣∣) 因此，∫sin x1+cos xdx=2ln(∣∣csc2x∣∣−∣∣cot2x∣∣)+C (23) ∫(1+x8)2x3dx=∫x(1+(x4)2)2x4dx，设u=x4，x=4x，dx=−44u31du，得 ∫(1+x8)2x3dx=∫4u(1+u2)2u⋅44u31du=41∫(1+u2)21du，根据积分表（四）公式28，得 41(2(u2+1)u+21∫1+u21du)=8(u2+1)u+81arctan u+C=8(x8+1)x4+81arctan x4+C (24) 设u=x4，x=4u，则dx=44u31du，得 ∫x8+3x4+2x11dx=∫u2+3u+24u11⋅44u31du=41∫u2+3u+2u2du=41∫(1+u+11−u+24)du= 41u+41ln ∣1+u∣−ln ∣2+u∣+C=41x4+ln 2+x441+x4+C (25) ∫16−x4dx=81∫(4+x21+4−x21)dx=81∫4+x21dx+321∫(2+x1+2−x1)dx= 161arctan2x+321ln ∣∣2−x2+x∣∣+C (26) ∫1+sin xsin xdx=∫1+tan22x(tan2x+1)21+tan22x2tan2xdx=∫(tan2x+1)22tan2xdx，设u=tan2x，则x=2arctan u，dx=1+u22du，得 ∫(tan2x+1)22tan2xdx=4∫(u+1)2u⋅1+u21du=2∫(1+u21−(1+u)21)du= 2∫1+u21du−2∫(u+1)21du=2arctan u+1+u2+C=x+1+tan2x2+C (27) ∫1+cos xx+sin xdx=∫2cos22xx+2sin2xcos2xdx=2∫2xsec22xd(2x)+∫tan2xdx，设u=2x，dv=sec22xd(2x)，则du=21dx，v=tan2x，得 2∫2xsec22xd(2x)+∫tan2xdx=xtan2x−∫tan2xdx+∫tan2xdx+C=xtan2x+C (28) ∫esin xcos2 xxcos3 x−sin xdx=∫xesin xcos xdx=∫esin xtan xsec xdx=∫xd(esin x)−∫esin xd(sec x) 设u=x，dv=d(esin x)，则du=dx，v=esin x，得 ∫xd(esin x)=xesin x−∫esin xdx，设u=esin x，dv=d(sec x)，则du=esin xcos xdx，v=sec x，得 ∫esin xd(sec x)=sec xesin x−∫esin xdx，因此 ∫xd(esin x)−∫esin xd(sec x)=xesin x−∫esin xdx−sec xesin x+∫esin xdx=(x−sec x)esin x+C (29) 设u=6x，则x=u6，dx=6u5du，得 ∫x(x+3x)3xdx=∫u6(u3+u2)u2⋅6u5du=6∫u(u+1)1du=6∫(u1−u+11)du= 6ln ∣∣u+1u∣∣+C=6ln6x+16x+C=ln(6x+1)6x+C (30) 设u=ex，则x=ln u，dx=u1du，得 ∫(1+ex)2dx=∫u(1+u)21du=∫(u1−1+u1−(1+u)21)du=ln u−ln(1+u)+1+u1+C= x−ln(1+ex)+1+ex1+C (31) ∫e4x−e2x+1e3x+exdx=∫e2x−1+e−2xex+e−xdx=∫(ex−e−x)2+11d(ex−e−x)=arctan(ex−e−x)+C (32) 设u=ex，则x=ln u，dx=u1du，得 ∫(ex+1)2xexdx=∫(1+u)2ln udu，令s=ln u，dv=(1+u)21du，则ds=u1du，v=−1+u1，得 ∫(1+u)2ln udu=−1+uln u+∫u(1+u)1du=−1+uln u+∫u1du−∫1+u1du= −1+uln u+ln u−ln(1+u)+C=−1+exx+x−ln(1+ex)+C (33) 设u=ln2(x+1+x2)，dv=dx，则u=1+x22ln(x+1+x2)dx，v=x，得 ∫ln2(x+1+x2)dx=xln2(x+1+x2)−∫1+x22xln(x+1+x2)dx= xln2(x+1+x2)−∫2ln(x+1+x2)d(1+x2)= xln2(x+1+x2)−21+x2ln(x+1+x2)+2x+C (34) 设u=x1，则x=u1，dx=−u21du，得 ∫(1+x2)23ln xdx=−∫(1+u21)23lnu1⋅u21du=∫(1+u2)23uln udu=−∫ln ud((1+u2)−21)，设s=ln u，dv=d((1+u2)−21)，则ds=u1du，v=1+u21，得 −∫ln ud((1+u2)−21)=−1+u2ln u+∫u1+u21du=1+x2xln x−∫1+x21dx= 1+x2xln x−ln(x+1+x2)+C (35) ∫1−x2arcsin xdx=∫arcsin x1−x21dx−∫arcsin x1−x2x2dx，设u=arcsin x，dv=1−x21dx，则du=1−x21dx，v=arcsin x，得 ∫arcsin x1−x21dx=(arcsin x)2−∫arcsin x1−x21dx，得∫arcsin x1−x21dx=21(arcsin x)2 设u=arcsin x，dv=1−x2x2dx，则du=1−x21dx，v=−2x+21arcsin x，得 ∫arcsin x1−x2x2dx=21(arcsin x)2−21x1−x2arcsin x−21∫(arcsin x−x1−x2)1−x21dx= 21(arcsin x)2−21x1−x2arcsin x−21∫arcsin x1−x21dx+41x2，原式得 ∫arcsin x1−x21dx−∫arcsin x1−x2x2dx=21∫arcsin x1−x21dx+21x1−x2arcsin x−41x2= 41(arcsin x)2+21x1−x2arcsin x−41x2+C (36) 设x=cos u (0<u<π)，则1−x2=sin u，dx=−sin udu，得 ∫1−x2x3arccos xdx=−∫ucos3 udu=−∫ud(sin u−31sin3 u)= −u(sin u−31sin3 u)+∫(sin u−31sin3 u)du=−u(sin u−31sin3 u)−31∫(2+cos2 u)d(cos u)= −u(sin u−31sin3 u)−32cos u−91cos3 u+C=−311−x2(2+x2)arccos x−91x(6+x2)+C (37) ∫1+sin xcot xdx=∫1+sin xsin xcos xdx=∫sin x(1+sin x)1−sin2 xdx，设u=sin x，则x=arcsin u，dx=1−u21du，得 ∫sin x(1+sin x)1−sin2 xdx=∫u(1+u)1−u2⋅1−u21du=∫u(1+u)1du=∫(u1−1+u1)du= ln ∣∣1+sin xsin x∣∣+C (38) ∫sin3 xcos xdx=−∫cot xsec2 xd(cot x)，设u=cot x，则x=arccot u，dx=−1+u21du，得 −∫cot xsec2 xd(cot x)=−∫u(1+u21)du=−21u2−ln ∣u∣+C=−21cot2 x−ln ∣cot x∣+C (39) 设u=cos x，则x=arccos u，dx=−1−u21du，得 ∫(2+cos x)sin xdx=∫(u+2)(u2−1)1du=∫[6(u−1)1−2(u+1)1+3(u+2)1]du= 61ln ∣u−1∣−21ln ∣u+1∣+31ln ∣u+2∣+C=61ln ∣1−cos x∣−21ln ∣cos x+1∣+31ln ∣cos x+2∣+C (40) ∫sin x+cos xsin xcos xdx=21∫sin x+cos x(sin x+cos x)2−1dx=21∫(sin x+cos x)dx−21∫sin x+cos x1dx= 21∫(sin x−cos x)−21∫sin x+cos x1dx，令u=tan2x，则sin x=1+u22u，cos x=1+u21−u2，dx=1+u21du，得 ∫sin x+cos x1dx=∫2u+1−u22du=−∫(u−1)2−22du= −22∫u−1−21du+22∫u−1+21du=22ln ∣∣u−1−2u−1+2∣∣+C 代入原式得∫sin x+cos xsin xcos xdx=21(sin x−cos x)−42ln ∣∣tan2x−1−2tan2x−1+2∣∣+C

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