自修不定积分：吴传生《经济数学微积分》第四版总习题五（二）

书接上文：自修不定积分：吴传生《经济数学微积分》第四版总习题五（一）
(16)∫ln⁡(1+x2)dx=∫ln⁡[(1+ix)(1−ix)]dx=∫ln⁡(1+ix)+ln⁡(1−ix)]dx=∫ln⁡(1+ix)dx+∫ln⁡(1−ix)dx=−i∫ln⁡(1+ix)d(1+ix)+i∫ln⁡(1−ix)d(1−ix)=−i(1+ix)[ln⁡(1+ix)−1]+i(1−ix)[ln⁡(1−ix)−1]+C=(−i+x)ln⁡(1+ix)+(i−x)+(i+x)ln⁡(1−ix)−(i+x)+C=−2x−iln⁡(1+ix)+xln⁡(1+ix)+iln⁡(1−ix)+xln⁡(1−ix)+C=−2x+xln⁡(x2+1)+i[ln⁡(1−ix)−ln⁡(1+ix)]+C=−2x+xln⁡(x2+1)+i{ln⁡[1+x2(11+x2+−xi1+x2)]−ln⁡[1+x2(11+x2+xi1+x2)]}+C={−2x+xln⁡(x2+1)+i[ln⁡1+x2+ln⁡(cos⁡α+isin⁡α)−ln⁡1+x2−ln⁡(cos⁡β+isin⁡β)]+C}tan⁡α=−x,tan⁡β=x=−2x+xln⁡(x2+1)+i(iα−iβ)+C=−2x+xln⁡(x2+1)+−α+β+C=−2x+xln⁡(x2+1)+2arctan⁡x+C\begin{aligned} (16)\int\ln\left(1+x^2\right)\mathrm dx &=\int\ln\left[\left(1+\mathrm ix\right)\left(1-\mathrm ix\right)\right]\mathrm dx\\ &=\int\ln(1+\mathrm ix)+\ln(1-\mathrm ix)]\mathrm dx\\ &=\int\ln(1+\mathrm ix)\mathrm dx+\int\ln(1-\mathrm ix)\mathrm dx\\ &=-\mathrm i\int\ln(1+\mathrm ix)\mathrm d(1+\mathrm ix)+\mathrm i\int\ln(1-\mathrm ix)\mathrm d(1-\mathrm ix)\\ &=-\mathrm i(1+\mathrm ix)[\ln(1+\mathrm ix)-1]+\mathrm i(1-\mathrm ix)[\ln(1-\mathrm ix)-1]+C\\ &=(-\mathrm i+x)\ln(1+\mathrm ix)+(\mathrm i-x)+(\mathrm i+x)\ln(1-\mathrm ix)-(\mathrm i+x)+C\\ &=-2x-\mathrm i\ln(1+\mathrm ix)+x\ln(1+\mathrm ix)+\mathrm i\ln(1-\mathrm ix)+x\ln(1-\mathrm ix)+C\\ &=-2x+x\ln\left(x^2+1\right)+\mathrm i[\ln(1-\mathrm ix)-\ln(1+\mathrm ix)]+C\\ &=-2x+x\ln\left(x^2+1\right)+\mathrm i\left\{\ln\left[\sqrt{1+x^2}\left(\frac{1}{\sqrt{1+x^2}}+\frac{-x\mathrm i}{\sqrt{1+x^2}}\right)\right]-\ln\left[\sqrt{1+x^2}\left(\frac{1}{\sqrt{1+x^2}}+\frac{x\mathrm i}{\sqrt{1+x^2}}\right)\right]\right\}+C\\ &=\left\{-2x+x\ln\left(x^2+1\right)+\mathrm i\left[\ln\sqrt{1+x^2}+\ln(\cos\alpha+\mathrm i\sin\alpha)-\ln\sqrt{1+x^2}-\ln(\cos\beta+\mathrm i\sin\beta)\right]+C\right\}_{\tan\alpha=-x,\tan\beta=x}\\ &=-2x+x\ln\left(x^2+1\right)+\mathrm i(\mathrm i\alpha-\mathrm i\beta)+C\\ &=-2x+x\ln\left(x^2+1\right)+-\alpha+\beta+C\\ &=-2x+x\ln\left(x^2+1\right)+2\arctan x+C \end{aligned} (16)∫ln(1+x2)dx=∫ln[(1+ix)(1−ix)]dx=∫ln(1+ix)+ln(1−ix)]dx=∫ln(1+ix)dx+∫ln(1−ix)dx=−i∫ln(1+ix)d(1+ix)+i∫ln(1−ix)d(1−ix)=−i(1+ix)[ln(1+ix)−1]+i(1−ix)[ln(1−ix)−1]+C=(−i+x)ln(1+ix)+(i−x)+(i+x)ln(1−ix)−(i+x)+C=−2x−iln(1+ix)+xln(1+ix)+iln(1−ix)+xln(1−ix)+C=−2x+xln(x2+1)+i[ln(1−ix)−ln(1+ix)]+C=−2x+xln(x2+1)+i{ln[1+x2(1+x21+1+x2−xi)]−ln[1+x2(1+x21+1+x2xi)]}+C={−2x+xln(x2+1)+i[ln1+x2+ln(cosα+isinα)−ln1+x2−ln(cosβ+isinβ)]+C}tanα=−x,tanβ=x=−2x+xln(x2+1)+i(iα−iβ)+C=−2x+xln(x2+1)+−α+β+C=−2x+xln(x2+1)+2arctanx+C

(17)∫arctan⁡xdx=(∫udtan⁡2u)tan⁡2u=x=utan⁡2u−∫tan⁡2udu=utan⁡2u−∫tan⁡usec⁡u⋅tan⁡usec⁡udu=utan⁡2u−∫sin⁡udsec⁡u=utan⁡2u−sin⁡usec⁡u+∫sec⁡ucos⁡udu=utan⁡2u−tan⁡u+u+C=xarctan⁡x−x+arctan⁡x+C\begin{aligned} (17)\int\arctan\sqrt x\mathrm dx &=\left(\int u\mathrm d\tan^2u\right)_{\tan^2u=x}\\ &=u\tan^2u-\int\tan^2u\mathrm du\\ &=u\tan^2u-\int\frac{\tan u}{\sec u}\cdot\tan u\sec u\mathrm du\\ &=u\tan^2u-\int\sin u\mathrm d\sec u\\ &=u\tan^2u-\sin u\sec u+\int\sec u\cos u\mathrm du\\ &=u\tan^2u-\tan u+u+C\\ &=x\arctan\sqrt x-\sqrt x+\arctan\sqrt x+C \end{aligned} (17)∫arctanxdx=(∫udtan2u)tan2u=x=utan2u−∫tan2udu=utan2u−∫secutanu⋅tanusecudu=utan2u−∫sinudsecu=utan2u−sinusecu+∫secucosudu=utan2u−tanu+u+C=xarctanx−x+arctanx+C

(18)∫1+cos⁡2xsin⁡2xdx=∫1+cos⁡2x−sin⁡2x2cos⁡xsin⁡xdx=∫cos⁡2x+sin⁡2x+cos⁡2x−sin⁡2x2cos⁡xsin⁡xdx=∫2cos⁡x2cos⁡xsin⁡xdx=22∫csc⁡xdx=22ln⁡∣csc⁡x−cot⁡x∣+C\begin{aligned} (18)\int\frac{\sqrt{1+\cos2x}}{\sin2x}\mathrm dx &=\int\frac{\sqrt{1+\cos^2x-\sin^2x}}{2\cos x\sin x}\mathrm dx\\ &=\int\frac{\sqrt{\cos^2x+\sin^2x+\cos^2x-\sin^2x}}{2\cos x\sin x}\mathrm dx\\ &=\int\frac{\sqrt{2}\cos x}{2\cos x\sin x}\mathrm dx\\ &=\frac{\sqrt2}{2}\int\csc x\mathrm dx\\ &=\frac{\sqrt2}{2}\ln|\csc x-\cot x|+C \end{aligned} (18)∫sin2x1+cos2xdx=∫2cosxsinx1+cos2x−sin2xdx=∫2cosxsinxcos2x+sin2x+cos2x−sin2xdx=∫2cosxsinx2cosxdx=22∫cscxdx=22ln∣cscx−cotx∣+C

(19)∫x+2sin⁡xcos⁡x1+cos⁡2xdx=∫x+2sin⁡xcos⁡x2cos⁡2xdx=12∫xcos⁡2xdx+∫tan⁡xdx=12∫xsec⁡2xdx+∫tan⁡xdx=12∫x(tan⁡2x+1)dx+∫tan⁡xdx=12∫xtan⁡2xdx+12∫xdx+∫tan⁡xdx=12∫x⋅tan⁡xsec⁡x⋅tan⁡xsec⁡xdx+12∫xdx+∫tan⁡xdx=12xtan⁡x−12∫sec⁡xd(xsin⁡x)+12∫xdx+∫tan⁡xdx=12xtan⁡x−12∫sec⁡x(sin⁡x+xcos⁡x)dx+12∫xdx+∫tan⁡xdx=12xtan⁡x−12∫tan⁡xdx−12∫xdx+12∫xdx+∫tan⁡xdx=12xtan⁡x+12∫tan⁡xdx=12xtan⁡x−12ln⁡∣cos⁡x∣+C\begin{aligned} (19)\int\frac{x+2\sin x\cos x}{1+\cos 2x}\mathrm dx &=\int\frac{x+2\sin x\cos x}{2\cos^2x}\mathrm dx\\ &=\frac{1}{2}\int \frac{x}{\cos^2x}\mathrm dx+\int\tan x\mathrm dx\\ &=\frac{1}{2}\int x\sec^2x\mathrm dx+\int\tan x\mathrm dx\\ &=\frac{1}{2}\int x\left(\tan^2x+1\right)\mathrm dx+\int\tan x\mathrm dx\\ &=\frac{1}{2}\int x\tan^2x\mathrm dx+\frac{1}{2}\int x\mathrm dx+\int\tan x\mathrm dx\\ &=\frac{1}{2}\int x\cdot\frac{\tan x}{\sec x}\cdot\tan x\sec x\mathrm dx+\frac{1}{2}\int x\mathrm dx+\int\tan x\mathrm{d}x\\ &=\frac{1}{2}x\tan x-\frac{1}{2}\int\sec x\mathrm d(x\sin x)+\frac{1}{2}\int x\mathrm dx+\int\tan x\mathrm dx\\ &=\frac{1}{2}x\tan x-\frac{1}{2}\int\sec x(\sin x+x\cos x)\mathrm dx+\frac{1}{2}\int x\mathrm dx+\int\tan x\mathrm dx\\ &=\frac{1}{2}x\tan x-\frac{1}{2}\int\tan x\mathrm dx-\frac{1}{2}\int x\mathrm dx+\frac{1}{2}\int x\mathrm dx+\int\tan x\mathrm dx\\ &=\frac{1}{2}x\tan x+\frac{1}{2}\int\tan x\mathrm dx\\ &=\frac{1}{2}x\tan x-\frac{1}{2}\ln|\cos x|+C \end{aligned} (19)∫1+cos2xx+2sinxcosxdx=∫2cos2xx+2sinxcosxdx=21∫cos2xxdx+∫tanxdx=21∫xsec2xdx+∫tanxdx=21∫x(tan2x+1)dx+∫tanxdx=21∫xtan2xdx+21∫xdx+∫tanxdx=21∫x⋅secxtanx⋅tanxsecxdx+21∫xdx+∫tanxdx=21xtanx−21∫secxd(xsinx)+21∫xdx+∫tanxdx=21xtanx−21∫secx(sinx+xcosx)dx+21∫xdx+∫tanxdx=21xtanx−21∫tanxdx−21∫xdx+21∫xdx+∫tanxdx=21xtanx+21∫tanxdx=21xtanx−21ln∣cosx∣+C

(20)∫sin⁡2xcos⁡3xdx=∫tan⁡2xsec⁡xdx=∫tan⁡xdsec⁡x=tan⁡xsec⁡x−∫sec⁡3x=tan⁡xsec⁡x−∫sec⁡xtan⁡2xdx−∫sec⁡xdx=tan⁡xsec⁡x−∫tan⁡xdsec⁡x−ln⁡∣sec⁡x+tan⁡x∣∴∫sin⁡2xcos⁡3xdx=∫tan⁡xdsec⁡x=12tan⁡xsec⁡x−12ln⁡∣sec⁡x+tan⁡x∣+C\begin{aligned} (20)\int\frac{\sin^2x}{\cos^3x}\mathrm dx &=\int\tan^2x\sec x\mathrm dx\\ &=\int\tan x\mathrm d\sec x\\ &=\tan x\sec x-\int\sec^3x\\ &=\tan x\sec x-\int\sec x\tan^2x\mathrm dx-\int\sec x\mathrm dx\\ &=\tan x\sec x-\int\tan x\mathrm d\sec x-\ln|\sec x+\tan x|\\ \end{aligned}\\ \therefore \int\frac{\sin^2x}{\cos^3x}\mathrm dx=\int\tan x\mathrm d\sec x=\frac{1}{2}\tan x\sec x-\frac{1}{2}\ln|\sec x+\tan x|+C (20)∫cos3xsin2xdx=∫tan2xsecxdx=∫tanxdsecx=tanxsecx−∫sec3x=tanxsecx−∫secxtan2xdx−∫secxdx=tanxsecx−∫tanxdsecx−ln∣secx+tanx∣∴∫cos3xsin2xdx=∫tanxdsecx=21tanxsecx−21ln∣secx+tanx∣+C

(21)∫x3x(x+x3)dx=∫x13x(x12+x13)dx=∫x13x32+x43dx=∫1x76+xdx=(∫da6a7+a6)x=a6=∫6a5daa7+a6=6∫daa2+a=6∫1a−1a+1da=6∫1ada−6∫1a+1da=6ln⁡∣a∣−6ln⁡∣a+1∣+C=ln⁡∣a6∣−6ln⁡∣a+1∣+C=ln⁡x−6ln⁡(x6+1)+C=ln⁡x(x6+1)6+C\begin{aligned} (21)\int\frac{\sqrt[3]{x}}{x\left(\sqrt{x}+\sqrt[3]{x}\right)}\mathrm dx &=\int\frac{x^\frac{1}{3}}{x\left(x^\frac{1}{2}+x^\frac{1}{3}\right)}\mathrm dx\\ &=\int\frac{x^\frac{1}{3}}{x^\frac{3}{2}+x^\frac{4}{3}}\mathrm dx\\ &=\int\frac{1}{x^\frac{7}{6}+x}\mathrm dx\\ &=\left(\int\frac{\mathrm da^6}{a^7+a^6}\right)_{x=a^6}\\ &=\int\frac{6a^5\mathrm da}{a^7+a^6}\\ &=6\int\frac{\mathrm da}{a^2+a}\\ &=6\int\frac{1}{a}-\frac{1}{a+1}\mathrm da\\ &=6\int\frac{1}{a}\mathrm da-6\int\frac{1}{a+1}\mathrm da\\ &=6\ln|a|-6\ln|a+1|+C\\ &=\ln\left|a^6\right|-6\ln|a+1|+C\\ &=\ln x-6\ln\left(\sqrt[6]{x}+1\right)+C\\ &=\ln\frac{x}{\left(\sqrt[6]{x}+1\right)^6}+C \end{aligned} (21)∫x(x+3x)3xdx=∫x(x21+x31)x31dx=∫x23+x34x31dx=∫x67+x1dx=(∫a7+a6da6)x=a6=∫a7+a66a5da=6∫a2+ada=6∫a1−a+11da=6∫a1da−6∫a+11da=6ln∣a∣−6ln∣a+1∣+C=ln∣∣a6∣∣−6ln∣a+1∣+C=lnx−6ln(6x+1)+C=ln(6x+1)6x+C

(22)∫1(1+ex)2dx=∫1+ex−ex(1+ex)2dx=∫11+ex−ex(1+ex)2dx=∫1−ex1+exdx−∫d(ex+1)(1+ex)2dx=x−∫dex1+ex+11+ex=x−ln⁡(ex+1)+11+ex+C\begin{aligned} (22)\int\frac{1}{\left(1+\mathrm{e}^x\right)^2}\mathrm dx &=\int\frac{1+\mathrm e^x-\mathrm e^x}{\left(1+\mathrm{e}^x\right)^2}\mathrm dx\\ &=\int\frac{1}{1+\mathrm e^x}-\frac{\mathrm e^x}{\left(1+\mathrm{e}^x\right)^2}\mathrm dx\\ &=\int1-\frac{\mathrm e^x}{1+\mathrm e^x}\mathrm dx-\int\frac{\mathrm d\left(\mathrm e^x+1\right)}{\left(1+\mathrm{e}^x\right)^2}\mathrm dx\\ &=x-\int\frac{\mathrm{de}^x}{1+\mathrm e^x}+\frac{1}{1+\mathrm e^x}\\ &=x-\ln\left(\mathrm e^x+1\right)+\frac{1}{1+\mathrm e^x}+C \end{aligned} (22)∫(1+ex)21dx=∫(1+ex)21+ex−exdx=∫1+ex1−(1+ex)2exdx=∫1−1+exexdx−∫(1+ex)2d(ex+1)dx=x−∫1+exdex+1+ex1=x−ln(ex+1)+1+ex1+C

(23)∫cos⁡x1+sin⁡xdx=∫1−sin⁡xcos⁡xdx=∫sec⁡x−tan⁡xdx=ln⁡∣sec⁡x+tan⁡x∣−ln⁡∣sec⁡x∣+C=ln⁡∣1+sin⁡x∣+C\begin{aligned} (23)\int\frac{\cos x}{1+\sin x}\mathrm dx &=\int\frac{1-\sin x}{\cos x}\mathrm dx\\ &=\int\sec x-\tan x\mathrm dx\\ &=\ln|\sec x+\tan x|-\ln|\sec x|+C\\ &=\ln|1+\sin x|+C \end{aligned} (23)∫1+sinxcosxdx=∫cosx1−sinxdx=∫secx−tanxdx=ln∣secx+tanx∣−ln∣secx∣+C=ln∣1+sinx∣+C

(24)∫cos⁡x1+cos⁡xdx=∫cos⁡x−cos⁡2xsin⁡2xdx=∫cot⁡xcsc⁡xdx−∫cot⁡2xdx=−csc⁡x−∫cos⁡xcot⁡xcsc⁡xdx=−csc⁡x+∫cos⁡xdcsc⁡x=−csc⁡x+cot⁡x−∫csc⁡xdcos⁡x=−csc⁡x+cot⁡x+∫csc⁡xsin⁡xdx=−csc⁡x+cot⁡x+x+C\begin{aligned} (24)\int\frac{\cos x}{1+\cos x}\mathrm dx &=\int\frac{\cos x-\cos^2x}{\sin^2x}\mathrm dx\\ &=\int\cot x\csc x\mathrm dx-\int\cot^2x\mathrm dx\\ &=-\csc x-\int\cos x\cot x\csc x\mathrm dx\\ &=-\csc x+\int\cos x\mathrm d\csc x\\ &=-\csc x+\cot x-\int\csc x\mathrm d\cos x\\ &=-\csc x+\cot x+\int\csc x\sin x\mathrm dx\\ &=-\csc x+\cot x+x+C \end{aligned} (24)∫1+cosxcosxdx=∫sin2xcosx−cos2xdx=∫cotxcscxdx−∫cot2xdx=−cscx−∫cosxcotxcscxdx=−cscx+∫cosxdcscx=−cscx+cotx−∫cscxdcosx=−cscx+cotx+∫cscxsinxdx=−cscx+cotx+x+C

(25)∫x1−x2arcsin⁡xdx=(∫sin⁡ucos⁡u⋅udsin⁡u)u=arcsin⁡x=∫usin⁡ucos⁡2udu=∫usin⁡u−usin⁡3udu=∫usin⁡udu−∫usin⁡3udu=−∫udcos⁡u−∫ud(−cos⁡u+13cos⁡3u)=−ucos⁡u+sin⁡u+∫ud(cos⁡u−13cos⁡3u)=−ucos⁡u+sin⁡u+u(cos⁡u−13cos⁡3u)−∫cos⁡u−13cos⁡3udu=sin⁡u−13ucos⁡3u−∫cos⁡udu+13∫cos⁡3udu=−13ucos⁡3u+13∫cos⁡3udu=−13ucos⁡3u+13sin⁡u−19sin⁡3u+C=−13(arcsin⁡x)⋅(1−x2)32+13x−19x3+C\begin{aligned} (25)\int x\sqrt{1-x^2}\arcsin x\mathrm dx &=\left(\int\sin u\cos u\cdot u\mathrm d\sin u\right)_{u=\arcsin x}\\ &=\int u\sin u\cos^2u\mathrm du\\ &=\int u\sin u-u\sin^3u\mathrm du\\ &=\int u\sin u\mathrm du-\int u\sin^3u\mathrm du\\ &=-\int u\mathrm d\cos u-\int u\mathrm d\left(-\cos u+\frac{1}{3}\cos^3u\right)\\ &=-u\cos u+\sin u+\int u\mathrm d\left(\cos u-\frac{1}{3}\cos^3u\right)\\ &=-u\cos u+\sin u+u\left(\cos u-\frac{1}{3}\cos^3u\right)-\int\cos u-\frac{1}{3}\cos^3u\mathrm du\\ &=\sin u-\frac{1}{3}u\cos^3u-\int\cos u\mathrm du+\frac{1}{3}\int\cos^3u\mathrm du\\ &=-\frac{1}{3}u\cos^3u+\frac{1}{3}\int\cos^3u\mathrm du\\ &=-\frac{1}{3}u\cos^3u+\frac{1}{3}\sin u-\frac{1}{9}\sin^3u+C\\ &=-\frac{1}{3}(\arcsin x)\cdot\left(1-x^2\right)^\frac{3}{2}+\frac{1}{3}x-\frac{1}{9}x^3+C \end{aligned} (25)∫x1−x2arcsinxdx=(∫sinucosu⋅udsinu)u=arcsinx=∫usinucos2udu=∫usinu−usin3udu=∫usinudu−∫usin3udu=−∫udcosu−∫ud(−cosu+31cos3u)=−ucosu+sinu+∫ud(cosu−31cos3u)=−ucosu+sinu+u(cosu−31cos3u)−∫cosu−31cos3udu=sinu−31ucos3u−∫cosudu+31∫cos3udu=−31ucos3u+31∫cos3udu=−31ucos3u+31sinu−91sin3u+C=−31(arcsinx)⋅(1−x2)23+31x−91x3+C

(26)∫xarccos⁡x1−x2dx=(∫ucos⁡usin⁡udcos⁡u)u=arccos⁡x=−∫ucos⁡udu=−∫ucos⁡udu=−usin⁡u+∫sin⁡udu=−usin⁡u−cos⁡u+C=−1−x2arccos⁡x−x+C\begin{aligned} (26)\int\frac{x\arccos x}{\sqrt{1-x^2}}\mathrm dx &=\left(\int\frac{u\cos u}{\sin u}\mathrm d\cos u\right)_{u=\arccos x}\\ &=-\int u\cos u\mathrm du\\ &=-\int u\cos u\mathrm du\\ &=-u\sin u+\int\sin u\mathrm du\\ &=-u\sin u-\cos u+C\\ &=-\sqrt{1-x^2}\arccos x-x+C \end{aligned} (26)∫1−x2xarccosxdx=(∫sinuucosudcosu)u=arccosx=−∫ucosudu=−∫ucosudu=−usinu+∫sinudu=−usinu−cosu+C=−1−x2arccosx−x+C

(27)∫dxsin⁡2xcos⁡4x=∫dx14sin⁡22x(cos⁡2x+12)=8∫dxsin⁡2x(cos⁡2x+1)=8∫dx(cos⁡2x+1)2(1−cos⁡2x)设A=∫dx(cos⁡2x+1)2(1−cos⁡2x),B=∫cos⁡2x(cos⁡2x+1)2(1−cos⁡2x)dx∴A+B=∫dxsin⁡2x=12∫d2xsin⁡22x=−12cot⁡2xA−B=∫dx(cos⁡2x+1)2=12∫d2x(cos⁡2x+1)2=[12∫du(cos⁡u+1)2]u=2x=12∫1sin⁡ud11+cos⁡u=12sin⁡u(1+cos⁡u)+B=12sin⁡2x(1+cos⁡2x)+B∴{A+B=−12cot⁡2x+CA−B=12sin⁡2x(1+cos⁡2x)+C解得A=13[12sin⁡2x(1+cos⁡2x)−cot⁡2x]+C原式=8A=83[12sin⁡2x(1+cos⁡2x)−cot⁡2x]+C\begin{aligned} (27)\int\frac{\mathrm dx}{\sin^2x\cos^4x} &=\int\frac{\mathrm dx}{\frac{1}{4}\sin^22x\left(\frac{\cos2x+1}{2}\right)}\\ &=8\int\frac{\mathrm dx}{\sin^2x(\cos2x+1)}\\ &=8\int\frac{\mathrm dx}{(\cos2x+1)^2(1-\cos2x)}\\ \end{aligned}\\ \begin{aligned} 设A&=\int\frac{\mathrm dx}{(\cos2x+1)^2(1-\cos2x)},\\ B&=\int\frac{\cos2x}{(\cos2x+1)^2(1-\cos2x)}\mathrm dx\\ \end{aligned}\\ \begin{aligned} \therefore A+B&=\int\frac{\mathrm dx}{\sin^2x}=\frac{1}{2}\int\frac{\mathrm d2x}{\sin^22x}=-\frac{1}{2}\cot2x\\ A-B&=\int\frac{\mathrm dx}{(\cos2x+1)^2}\\ &=\frac{1}{2}\int\frac{\mathrm d2x}{(\cos2x+1)^2}\\ &=\left[\frac{1}{2}\int\frac{\mathrm du}{(\cos u+1)^2}\right]_{u=2x}\\ &=\frac{1}{2}\int\frac{1}{\sin u}\mathrm d\frac{1}{1+\cos u}\\ &=\frac{1}{2\sin u(1+\cos u)}+B\\ &=\frac{1}{2\sin2x(1+\cos2x)}+B \end{aligned}\\ \therefore \begin{cases} A+B=-\frac{1}{2}\cot2x+C\\ A-B=\frac{1}{2\sin2x(1+\cos2x)}+C\\ \end{cases}\\ \begin{aligned} &解得A=\frac{1}{3}\left[\frac{1}{2\sin2x(1+\cos2x)}-\cot2x\right]+C\\ &原式=8A=\frac{8}{3}\left[\frac{1}{2\sin2x(1+\cos2x)}-\cot2x\right]+C \end{aligned} (27)∫sin2xcos4xdx=∫41sin22x(2cos2x+1)dx=8∫sin2x(cos2x+1)dx=8∫(cos2x+1)2(1−cos2x)dx设AB=∫(cos2x+1)2(1−cos2x)dx,=∫(cos2x+1)2(1−cos2x)cos2xdx∴A+BA−B=∫sin2xdx=21∫sin22xd2x=−21cot2x=∫(cos2x+1)2dx=21∫(cos2x+1)2d2x=[21∫(cosu+1)2du]u=2x=21∫sinu1d1+cosu1=2sinu(1+cosu)1+B=2sin2x(1+cos2x)1+B∴{A+B=−21cot2x+CA−B=2sin2x(1+cos2x)1+C解得A=31[2sin2x(1+cos2x)1−cot2x]+C原式=8A=38[2sin2x(1+cos2x)1−cot2x]+C

(28)∫dxsin⁡3xcos⁡5x=14∫1sin⁡4x⋅4sin⁡xcos⁡5xdx=14∫1sin⁡4xd1cos⁡4x=[14∫1(1−u)2d1u2]u=cos⁡2x=−12∫duu3(1−u)2=12∫1u3d1u−1=12u3(u−1)−12∫1u−1d1u3=12u3(u−1)+12∫11−ud1u3=12u3(u−1)+12∫1+u1−ud1u3=12u3(u−1)+12u3+12∫u1−ud1u3=12u3(u−1)+12u3−32∫u1−u⋅u−4du=12u2(u−1)−32∫duu3(1−u)=12u2(u−1)+32∫duu3(u−1)=12u2(u−1)+32∫du−u2⋅u(1−u)=12u2(u−1)+32∫1u(1−u)d1u=12u2(u−1)−32∫1u−1−1ud1u=12u2(u−1)−32∫d1uu−1+32∫1ud1u=12u2(u−1)+34u−2−32∫1u−1d1u=12u2(u−1)+34u2+32∫du(u−1)u2=12u2(u−1)+34u2+32∫dln⁡u(u−1)u=12u2(u−1)+34u2+32∫1u−1−1udln⁡u=12u2(u−1)+34u2+32∫dln⁡uu−1−32∫dln⁡uu=12u2(u−1)+34u2+32∫duu(u−1)−32∫duu2=12u2(u−1)+34u2+32u+32∫duu(u−1)=12u2(u−1)+34u2+32u+32∫duu−1−32∫duu=12u2(u−1)+34u2+32u+32ln⁡∣u−1∣−32ln⁡∣u∣+C=12u2(u−1)+34u2+32u+32ln⁡∣u−1u∣+C=−12cos⁡4xsin⁡2x+34cos⁡4x+32cos⁡2x+32ln⁡∣tan⁡2x∣+C=−12cos⁡4xsin⁡2x+34cos⁡4x+32cos⁡2x+3ln⁡∣tan⁡x∣+C\begin{aligned} (28)\int\frac{\mathrm dx}{\sin^3x\cos^5x} &=\frac{1}{4}\int\frac{1}{\sin^4x}\cdot\frac{4\sin x}{\cos^5x}\mathrm dx\\ &=\frac{1}{4}\int\frac{1}{\sin^4x}\mathrm d\frac{1}{\cos^4x}\\ &=\left[\frac{1}{4}\int\frac{1}{(1-u)^2}\mathrm d\frac{1}{u^2}\right]_{u=\cos^2x}\\ &=-\frac{1}{2}\int\frac{\mathrm du}{u^3(1-u)^2}\\ &=\frac{1}{2}\int\frac{1}{u^3}\mathrm d\frac{1}{u-1}\\ &=\frac{1}{2u^3(u-1)}-\frac{1}{2}\int\frac{1}{u-1}\mathrm d\frac{1}{u^3}\\ &=\frac{1}{2u^3(u-1)}+\frac{1}{2}\int\frac{1}{1-u}\mathrm d\frac{1}{u^3}\\ &=\frac{1}{2u^3(u-1)}+\frac{1}{2}\int1+\frac{u}{1-u}\mathrm d\frac{1}{u^3}\\ &=\frac{1}{2u^3(u-1)}+\frac{1}{2u^3}+\frac{1}{2}\int\frac{u}{1-u}\mathrm d\frac{1}{u^3}\\ &=\frac{1}{2u^3(u-1)}+\frac{1}{2u^3}-\frac{3}{2}\int\frac{u}{1-u}\cdot u^{-4}\mathrm du\\ &=\frac{1}{2u^2(u-1)}-\frac{3}{2}\int\frac{\mathrm du}{u^3(1-u)}\\ &=\frac{1}{2u^2(u-1)}+\frac{3}{2}\int\frac{\mathrm du}{u^3(u-1)}\\ &=\frac{1}{2u^2(u-1)}+\frac{3}{2}\int\frac{\mathrm du}{-u^2\cdot u(1-u)}\\ &=\frac{1}{2u^2(u-1)}+\frac{3}{2}\int\frac{1}{u(1-u)}\mathrm d\frac{1}{u}\\ &=\frac{1}{2u^2(u-1)}-\frac{3}{2}\int\frac{1}{u-1}-\frac{1}{u}\mathrm d\frac{1}{u}\\ &=\frac{1}{2u^2(u-1)}-\frac{3}{2}\int\frac{\mathrm d\frac{1}{u}}{u-1}+\frac{3}{2}\int\frac{1}{u}\mathrm d\frac{1}{u}\\ &=\frac{1}{2u^2(u-1)}+\frac{3}{4}u^{-2}-\frac{3}{2}\int\frac{1}{u-1}\mathrm d\frac{1}{u}\\ &=\frac{1}{2u^2(u-1)}+\frac{3}{4u^2}+\frac{3}{2}\int\frac{\mathrm du}{(u-1)u^2}\\ &=\frac{1}{2u^2(u-1)}+\frac{3}{4u^2}+\frac{3}{2}\int\frac{\mathrm d\ln u}{(u-1)u}\\ &=\frac{1}{2u^2(u-1)}+\frac{3}{4u^2}+\frac{3}{2}\int\frac{1}{u-1}-\frac{1}{u}\mathrm d\ln u\\ &=\frac{1}{2u^2(u-1)}+\frac{3}{4u^2}+\frac{3}{2}\int\frac{\mathrm d\ln u}{u-1}-\frac{3}{2}\int\frac{\mathrm d\ln u}{u}\\ &=\frac{1}{2u^2(u-1)}+\frac{3}{4u^2}+\frac{3}{2}\int\frac{\mathrm du}{u(u-1)}-\frac{3}{2}\int\frac{\mathrm du}{u^2}\\ &=\frac{1}{2u^2(u-1)}+\frac{3}{4u^2}+\frac{3}{2u}+\frac{3}{2}\int\frac{\mathrm du}{u(u-1)}\\ &=\frac{1}{2u^2(u-1)}+\frac{3}{4u^2}+\frac{3}{2u}+\frac{3}{2}\int\frac{\mathrm du}{u-1}-\frac{3}{2}\int\frac{\mathrm du}{u}\\ &=\frac{1}{2u^2(u-1)}+\frac{3}{4u^2}+\frac{3}{2u}+\frac{3}{2}\ln|u-1|-\frac{3}{2}\ln|u|+C\\ &=\frac{1}{2u^2(u-1)}+\frac{3}{4u^2}+\frac{3}{2u}+\frac{3}{2}\ln\left|\frac{u-1}{u}\right|+C\\ &=-\frac{1}{2\cos^4x\sin^2x}+\frac{3}{4\cos^4x}+\frac{3}{2\cos^2x}+\frac{3}{2}\ln\left|\tan^2x\right|+C\\ &=-\frac{1}{2\cos^4x\sin^2x}+\frac{3}{4\cos^4x}+\frac{3}{2\cos^2x}+3\ln|\tan x|+C \end{aligned} (28)∫sin3xcos5xdx=41∫sin4x1⋅cos5x4sinxdx=41∫sin4x1dcos4x1=[41∫(1−u)21du21]u=cos2x=−21∫u3(1−u)2du=21∫u31du−11=2u3(u−1)1−21∫u−11du31=2u3(u−1)1+21∫1−u1du31=2u3(u−1)1+21∫1+1−uudu31=2u3(u−1)1+2u31+21∫1−uudu31=2u3(u−1)1+2u31−23∫1−uu⋅u−4du=2u2(u−1)1−23∫u3(1−u)du=2u2(u−1)1+23∫u3(u−1)du=2u2(u−1)1+23∫−u2⋅u(1−u)du=2u2(u−1)1+23∫u(1−u)1du1=2u2(u−1)1−23∫u−11−u1du1=2u2(u−1)1−23∫u−1du1+23∫u1du1=2u2(u−1)1+43u−2−23∫u−11du1=2u2(u−1)1+4u23+23∫(u−1)u2du=2u2(u−1)1+4u23+23∫(u−1)udlnu=2u2(u−1)1+4u23+23∫u−11−u1dlnu=2u2(u−1)1+4u23+23∫u−1dlnu−23∫udlnu=2u2(u−1)1+4u23+23∫u(u−1)du−23∫u2du=2u2(u−1)1+4u23+2u3+23∫u(u−1)du=2u2(u−1)1+4u23+2u3+23∫u−1du−23∫udu=2u2(u−1)1+4u23+2u3+23ln∣u−1∣−23ln∣u∣+C=2u2(u−1)1+4u23+2u3+23ln∣∣∣∣uu−1∣∣∣∣+C=−2cos4xsin2x1+4cos4x3+2cos2x3+23ln∣∣tan2x∣∣+C=−2cos4xsin2x1+4cos4x3+2cos2x3+3ln∣tanx∣+C

(29)∫dx1+tan⁡x=(∫darctan⁡u1+u)x=arctan⁡u=∫du(1+u)(1+u2)令1(1+u)(1+u2)=Au+B1+u2+C1+u=(A+C)u2+(A+B)u2+(B+C)(1+u)(1+u2)∴{A+C=0A+B=0B+C=1解得{A=−0.5B=0.5C=0.5∴∫dx1+tan⁡x=∫du(1+u)(1+u2)=∫−0.5u+0.51+u2+0.51+udu=12∫−u+11+u2du+12∫du1+u=12ln⁡∣u+1∣+12∫−u1+u2du+12∫du1+u2=12ln⁡∣u+1∣−12∫u1+u2du+12∫du1+u2=12ln⁡∣u+1∣+12arctan⁡u−12∫11+u2d(12u2)=12ln⁡∣u+1∣+12arctan⁡u−14∫du21+u2=12ln⁡∣u+1∣+12arctan⁡u−14ln⁡∣1+u2∣+C=12ln⁡∣tan⁡x+1∣+12x−14ln⁡∣sec⁡2x∣+C=12ln⁡∣tan⁡x+1∣−12ln⁡∣sec⁡x∣+12x+C=12ln⁡∣tan⁡x+1sec⁡x∣+12x+C=12ln⁡∣sin⁡x+cos⁡x∣+12x+C\begin{aligned} (29)\int\frac{\mathrm dx}{1+\tan x} &=\left(\int\frac{\mathrm d\arctan u}{1+u}\right)_{x=\arctan u}\\ &=\int\frac{\mathrm du}{(1+u)\left(1+u^2\right)} \end{aligned}\\ \begin{aligned} 令\frac{1}{(1+u)\left(1+u^2\right)}&=\frac{Au+B}{1+u^2}+\frac{C}{1+u}\\ &=\frac{(A+C)u^2+(A+B)u^2+(B+C)}{(1+u)\left(1+u^2\right)} \end{aligned}\\ \therefore \begin{cases} A+C=0\\ A+B=0\\ B+C=1 \end{cases} 解得\begin{cases} A=-0.5\\ B=0.5\\ C=0.5 \end{cases}\\ \therefore\begin{aligned} \int\frac{\mathrm dx}{1+\tan x} &=\int\frac{\mathrm du}{(1+u)\left(1+u^2\right)}\\ &=\int\frac{-0.5u+0.5}{1+u^2}+\frac{0.5}{1+u}\mathrm du\\ &=\frac{1}{2}\int\frac{-u+1}{1+u^2}\mathrm du+\frac{1}{2}\int\frac{\mathrm du}{1+u}\\ &=\frac{1}{2}\ln|u+1|+\frac{1}{2}\int\frac{-u}{1+u^2}\mathrm du+\frac{1}{2}\int\frac{\mathrm du}{1+u^2}\\ &=\frac{1}{2}\ln|u+1|-\frac{1}{2}\int\frac{u}{1+u^2}\mathrm du+\frac{1}{2}\int\frac{\mathrm du}{1+u^2}\\ &=\frac{1}{2}\ln|u+1|+\frac{1}{2}\arctan u-\frac{1}{2}\int\frac{1}{1+u^2}\mathrm d\left(\frac{1}{2}u^2\right)\\ &=\frac{1}{2}\ln|u+1|+\frac{1}{2}\arctan u-\frac{1}{4}\int\frac{\mathrm du^2}{1+u^2}\\ &=\frac{1}{2}\ln|u+1|+\frac{1}{2}\arctan u-\frac{1}{4}\ln\left|1+u^2\right|+C\\ &=\frac{1}{2}\ln|\tan x+1|+\frac{1}{2}x-\frac{1}{4}\ln\left|\sec^2x\right|+C\\ &=\frac{1}{2}\ln|\tan x+1|-\frac{1}{2}\ln\left|\sec x\right|+\frac{1}{2}x+C\\ &=\frac{1}{2}\ln\left|\frac{\tan x+1}{\sec x}\right|+\frac{1}{2}x+C\\ &=\frac{1}{2}\ln\left|\sin x+\cos x\right|+\frac{1}{2}x+C\\ \end{aligned} (29)∫1+tanxdx=(∫1+udarctanu)x=arctanu=∫(1+u)(1+u2)du令(1+u)(1+u2)1=1+u2Au+B+1+uC=(1+u)(1+u2)(A+C)u2+(A+B)u2+(B+C)∴⎩⎪⎨⎪⎧A+C=0A+B=0B+C=1解得⎩⎪⎨⎪⎧A=−0.5B=0.5C=0.5∴∫1+tanxdx=∫(1+u)(1+u2)du=∫1+u2−0.5u+0.5+1+u0.5du=21∫1+u2−u+1du+21∫1+udu=21ln∣u+1∣+21∫1+u2−udu+21∫1+u2du=21ln∣u+1∣−21∫1+u2udu+21∫1+u2du=21ln∣u+1∣+21arctanu−21∫1+u21d(21u2)=21ln∣u+1∣+21arctanu−41∫1+u2du2=21ln∣u+1∣+21arctanu−41ln∣∣1+u2∣∣+C=21ln∣tanx+1∣+21x−41ln∣∣sec2x∣∣+C=21ln∣tanx+1∣−21ln∣secx∣+21x+C=21ln∣∣∣∣secxtanx+1∣∣∣∣+21x+C=21ln∣sinx+cosx∣+21x+C

(30)∫dxx+x+1=∫x+1−xdx=∫x+1d(x+1)−∫xdx=23(x+1)32−23x32+C\begin{aligned} (30)\int\frac{\mathrm dx}{\sqrt x+\sqrt {x+1}} &=\int\sqrt{x+1}-\sqrt x\mathrm dx\\ &=\int\sqrt{x+1}\mathrm d(x+1)-\int\sqrt x\mathrm dx\\ &=\frac{2}{3}(x+1)^\frac{3}{2}-\frac{2}{3}x^\frac{3}{2}+C \end{aligned} (30)∫x+x+1dx=∫x+1−xdx=∫x+1d(x+1)−∫xdx=32(x+1)23−32x23+C

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自修不定积分：吴传生《经济数学 微积分》第四版 总习题五（二）

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自修不定积分：吴传生《经济数学微积分》第四版总习题五（二）

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