本文接自上一篇《第四章 不定积分（一）》，继续记录总习题四。

总习题四

4.求下列不定积分（其中$a$a、$b$b为常数）：

（7）$\displaystyle\int\tan^4x\mathrm{d}x;$∫tan4xdx;

解
$\begin{aligned} \displaystyle\int\tan^4x\mathrm{d}x&=\displaystyle\int\tan^2x(\sec^2x-1)\mathrm{d}x\\ &=\displaystyle\int\tan^2x\mathrm{d}(\tan x)-\displaystyle\int(\sec^2x-1)\mathrm{d}x\\ &=\cfrac{1}{3}\tan^3x-\tan x+x+C. \end{aligned}$∫tan4xdx​=∫tan2x(sec2x−1)dx=∫tan2xd(tanx)−∫(sec2x−1)dx=31​tan3x−tanx+x+C.​
（这道题主要利用三角变换公式进行计算）

（10）$\displaystyle\int\sqrt{\cfrac{a+x}{a-x}}\mathrm{d}x;$∫a−xa+x​​dx;

解一
$\begin{aligned} \displaystyle\int\sqrt{\cfrac{a+x}{a-x}}\mathrm{d}x&=\displaystyle\int\cfrac{a+x}{\sqrt{a^2-x^2}}\mathrm{d}x=a\displaystyle\int\cfrac{1}{\sqrt{1-\left(\cfrac{x}{a}\right)^2}}\mathrm{d}\left(\cfrac{x}{a}\right)-\cfrac{1}{2}\displaystyle\int\cfrac{\mathrm{d}(a^2-x^2)}{\sqrt{a^2-x^2}}\\ &=a\arcsin\cfrac{x}{a}-\sqrt{a^2-x^2}+C. \end{aligned}$∫a−xa+x​​dx​=∫a2−x2​a+x​dx=a∫1−(ax​)2​1​d(ax​)−21​∫a2−x2​d(a2−x2)​=aarcsinax​−a2−x2​+C.​
解二  令$u=\sqrt{\cfrac{a+x}{a-x}}$u=a−xa+x​​，即$x=a\cfrac{u^2-1}{u^2+1}$x=au2+1u2−1​，则
$\begin{aligned} \displaystyle\int\sqrt{\cfrac{a+x}{a-x}}\mathrm{d}x&=\displaystyle\int u\cdot\cfrac{4au}{(1+u^2)^2}\mathrm{d}u=\displaystyle\int-2au\mathrm{d}\left(\cfrac{1}{1+u^2}\right)\\ &=-\cfrac{2au}{1+u^2}+\displaystyle\int\cfrac{2a}{1+u^2}\mathrm{d}u\\ &=-\cfrac{2au}{1+u^2}+2a\arctan u+C\\ &=(x-a)\sqrt{\cfrac{a+x}{a-x}}+2a\arctan\sqrt{\cfrac{a+x}{a-x}}+C. \end{aligned}$∫a−xa+x​​dx​=∫u⋅(1+u2)24au​du=∫−2aud(1+u21​)=−1+u22au​+∫1+u22a​du=−1+u22au​+2aarctanu+C=(x−a)a−xa+x​​+2aarctana−xa+x​​+C.​
（这道题可以利用换元积分或进行分式变换求解）

（11）$\displaystyle\int\cfrac{\mathrm{d}x}{\sqrt{x(x+1)}};$∫x(x+1)​dx​;

解一
$\begin{aligned} \displaystyle\int\cfrac{\mathrm{d}x}{\sqrt{x(x+1)}}&=\displaystyle\int\cfrac{\mathrm{d}x}{\sqrt{\left(x+\cfrac{1}{2}\right)^2-\left(\cfrac{1}{2}\right)^2}}\\ &\xlongequal{x=-\cfrac{1}{2}+\cfrac{1}{2}\sec u}\displaystyle\int\sec u\mathrm{d}u=\ln|\sec u+\tan u|+C\\ &=\ln|2x+1+2\sqrt{x(x+1)}|+C. \end{aligned}$∫x(x+1)​dx​​=∫(x+21​)2−(21​)2​dx​x=−21​+21​secu∫secudu=ln∣secu+tanu∣+C=ln∣2x+1+2x(x+1)​∣+C.​
解二  当$x>0$x>0时，因为$\cfrac{1}{\sqrt{x(x+1)}}=\cfrac{1}{x}\sqrt{\cfrac{x}{1+x}}$x(x+1)​1​=x1​1+xx​​，故令$u=\sqrt{\cfrac{x}{1+x}}$u=1+xx​​，即$x=\cfrac{u^2}{1-u^2}$x=1−u2u2​，则
$\begin{aligned} \displaystyle\int\cfrac{\mathrm{d}x}{\sqrt{x(x+1)}}&=\displaystyle\int\cfrac{2}{1-u^2}\mathrm{d}u=\displaystyle\int\left(\cfrac{1}{1-u}+\cfrac{1}{1+u}\right)\mathrm{d}u\\ &=\ln|\cfrac{1+u}{1-u}|+C=\ln\left|\cfrac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}\right|+C\\ &=\ln|2x+1+2\sqrt{x(x+1)}|+C. \end{aligned}$∫x(x+1)​dx​​=∫1−u22​du=∫(1−u1​+1+u1​)du=ln∣1−u1+u​∣+C=ln∣∣∣∣∣​1+x​−x​1+x​+x​​∣∣∣∣∣​+C=ln∣2x+1+2x(x+1)​∣+C.​
  当$x<-1$x<−1时，同样可得$\displaystyle\int\cfrac{\mathrm{d}x}{\sqrt{x(x+1)}}=\ln|2x+1+2\sqrt{x(x+1)}|+C$∫x(x+1)​dx​=ln∣2x+1+2x(x+1)​∣+C。（这道题主要用分段或配方进行计算）

（15）$\displaystyle\int\cfrac{\mathrm{d}x}{x^2\sqrt{x^2-1}};$∫x2x2−1​dx​;

解  $\displaystyle\int\cfrac{\mathrm{d}x}{x^2\sqrt{x^2-1}}\xlongequal{x=\cfrac{1}{u}}-\displaystyle\int\cfrac{u\mathrm{d}u}{\sqrt{1-u^2}}=\sqrt{1-u^2}+C=\cfrac{\sqrt{x^2-1}}{x}+C,$∫x2x2−1​dx​x=u1​−∫1−u2​udu​=1−u2​+C=xx2−1​​+C,易知当$x<0$x<0和$x>0$x>0时的结果相同。（这道题主要利用换元积分计算）

（16）$\displaystyle\int\cfrac{\mathrm{d}x}{(a^2-x^2)^{\frac{5}{2}}};$∫(a2−x2)25​dx​;

解  设$x=a\sin u\left(-\cfrac{\pi}{2}<u<\cfrac{\pi}{2}\right)$x=asinu(−2π​<u<2π​)，则$\sqrt{a^2-x^2}=a\cos u,\mathrm{d}x=a\cos u\mathrm{d}u$a2−x2​=acosu,dx=acosudu，于是
$\begin{aligned} \displaystyle\int\cfrac{\mathrm{d}x}{(a^2-x^2)^{\frac{5}{2}}}&=\cfrac{1}{a^4}\displaystyle\int\sec^4u\mathrm{d}u=\cfrac{1}{a^4}\displaystyle\int(\tan^2u+1)\mathrm{d}(\tan u)\\ &=\cfrac{\tan^3u}{3a^4}+\cfrac{\tan u}{a^4}+C\\ &=\cfrac{1}{3a^4}\left[\cfrac{x^3}{\sqrt{(a^2-x^2)^3}}+\cfrac{3x}{\sqrt{a^2-x^2}}\right]+C. \end{aligned}$∫(a2−x2)25​dx​​=a41​∫sec4udu=a41​∫(tan2u+1)d(tanu)=3a4tan3u​+a4tanu​+C=3a41​[(a2−x2)3​x3​+a2−x2​3x​]+C.​
（这道题主要利用三角函数进行替换）

（17）$\displaystyle\int\cfrac{\mathrm{d}x}{x^4\sqrt{1+x^2}};$∫x41+x2​dx​;

解
$\begin{aligned} \displaystyle\int\cfrac{\mathrm{d}x}{x^4\sqrt{1+x^2}}&\xlongequal{x=\cfrac{1}{u}}\displaystyle\int\cfrac{-u^3\mathrm{d}u}{\sqrt{1+u^2}}=-\displaystyle\int\left(u\sqrt{1+u^2}-\cfrac{u}{\sqrt{1+u^2}}\right)\mathrm{d}u\\ &=-\cfrac{1}{3}(1+u^2)^{\frac{3}{2}}+\sqrt{1+u^2}+C\\ &=-\cfrac{1}{3}\cfrac{\sqrt{(1+x^2)^3}}{x^3}+\cfrac{\sqrt{1+x^2}}{x}+C. \end{aligned}$∫x41+x2​dx​​x=u1​∫1+u2​−u3du​=−∫(u1+u2​−1+u2​u​)du=−31​(1+u2)23​+1+u2​+C=−31​x3(1+x2)3​​+x1+x2​​+C.​
  易知当$x<0$x<0和$x>0$x>0时结果相同。（这道题用倒数换元再拆分因式计算）

（21）$\displaystyle\int\arctan\sqrt{x}\mathrm{d}x;$∫arctanx​dx;

解
$\begin{aligned} \displaystyle\int\arctan\sqrt{x}\mathrm{d}x&=\displaystyle\int\arctan\sqrt{x}\mathrm{d}(1+x)=(1+x)\arctan\sqrt{x}-\displaystyle\int\cfrac{1}{2\sqrt{x}}\mathrm{d}x\\ &=(1+x)\arctan\sqrt{x}-\sqrt{x}+C. \end{aligned}$∫arctanx​dx​=∫arctanx​d(1+x)=(1+x)arctanx​−∫2x​1​dx=(1+x)arctanx​−x​+C.​
（这道题在积分时凑出一个因式简化计算）

（22）$\displaystyle\int\cfrac{\sqrt{1+\cos x}}{\sin x}\mathrm{d}x;$∫sinx1+cosx​​dx;

解
$\begin{aligned} \displaystyle\int\cfrac{\sqrt{1+\cos x}}{\sin x}\mathrm{d}x&=\displaystyle\int\cfrac{\sqrt{2}|\cos\cfrac{x}{2}|}{2\sin\cfrac{x}{2}\cos\cfrac{x}{2}}\mathrm{d}x=\pm\sqrt{2}\displaystyle\int\csc\cfrac{x}{2}\mathrm{d}\left(\cfrac{x}{2}\right)\\ &=\pm\sqrt{2}\ln|\csc\cfrac{x}{2}-\cot\cfrac{x}{2}|+C. \end{aligned}$∫sinx1+cosx​​dx​=∫2sin2x​cos2x​2​∣cos2x​∣​dx=±2​∫csc2x​d(2x​)=±2​ln∣csc2x​−cot2x​∣+C.​
  上式当$\cos\cfrac{x}{2}>0$cos2x​>0时取正，当$\cos\cfrac{x}{2}<0$cos2x​<0时取负。
  当$\cos\cfrac{x}{2}>0$cos2x​>0时，
$\ln|\csc\cfrac{x}{2}-\cot\cfrac{x}{2}|=\ln\cfrac{1-\cos\cfrac{x}{2}}{|\sin\cfrac{x}{2}|}\\ =\ln\left(|\csc\cfrac{x}{2}|-|\cot\cfrac{x}{2}|\right).$ln∣csc2x​−cot2x​∣=ln∣sin2x​∣1−cos2x​​=ln(∣csc2x​∣−∣cot2x​∣).
  当$\cos\cfrac{x}{2}<0$cos2x​<0时，
$\ln|\csc\cfrac{x}{2}-\cot\cfrac{x}{2}|=\ln\cfrac{1-\cos\cfrac{x}{2}}{|\sin\cfrac{x}{2}|}\\ =\ln\left(|\csc\cfrac{x}{2}|+|\cot\cfrac{x}{2}|\right)=-\ln\left(|\csc\cfrac{x}{2}|-|\cot\cfrac{x}{2}|\right).$ln∣csc2x​−cot2x​∣=ln∣sin2x​∣1−cos2x​​=ln(∣csc2x​∣+∣cot2x​∣)=−ln(∣csc2x​∣−∣cot2x​∣).
  因此有
$\displaystyle\int\cfrac{\sqrt{1+\cos x}}{\sin x}\mathrm{d}x=\sqrt{2}\ln\left(|\csc\cfrac{x}{2}|-|\cot\cfrac{x}{2}|\right)+C.$∫sinx1+cosx​​dx=2​ln(∣csc2x​∣−∣cot2x​∣)+C.
（这道题主要利用了分段求解）

（23）$\displaystyle\int\cfrac{x^3}{(1+x^8)^2}\mathrm{d}x;$∫(1+x8)2x3​dx;

解
$\displaystyle\int\cfrac{x^3}{(1+x^8)^2}\mathrm{d}x=\cfrac{1}{4}\displaystyle\int\cfrac{1}{(1+x^8)^2}\mathrm{d}(x^4)\xlongequal{u=x^4}\cfrac{1}{4}\displaystyle\int\cfrac{1}{(1+u^2)^2}\mathrm{d}u.$∫(1+x8)2x3​dx=41​∫(1+x8)21​d(x4)u=x441​∫(1+u2)21​du.
  设$u=\tan t\left(-\cfrac{\pi}{2}<t<\cfrac{\pi}{2}\right)$u=tant(−2π​<t<2π​)，则$1+u^2=\sec^2t,\mathrm{d}u=\sec^2t\mathrm{d}t$1+u2=sec2t,du=sec2tdt，于是
$\begin{aligned} \text{原式}&=\cfrac{1}{4}\displaystyle\int\cos^2t\mathrm{d}t=\cfrac{2t+\sin2t}{16}+C\\ &=\cfrac{\arctan x^4}{8}+\cfrac{x^4}{8(1+x^8)}+C. \end{aligned}$原式​=41​∫cos2tdt=162t+sin2t​+C=8arctanx4​+8(1+x8)x4​+C.​
（这道题利用了两次换元求解）

（26）$\displaystyle\int\cfrac{\sin x}{1+\sin x}\mathrm{d}x;$∫1+sinxsinx​dx;

解一
  令$u=\tan\cfrac{x}{2}$u=tan2x​，得
$\begin{aligned} \displaystyle\int\cfrac{\sin x}{1+\sin x}\mathrm{d}x&=\displaystyle\int\cfrac{4u}{(1+u)^2(1+u^2)}\mathrm{d}u=\displaystyle\int\left[\cfrac{-2}{(1+u)^2}+\cfrac{2}{1+u^2}\right]\mathrm{d}u\\ &=\cfrac{2}{1+u}+2\arctan u+C=\cfrac{2}{1+\tan\cfrac{x}{2}}+x+C. \end{aligned}$∫1+sinxsinx​dx​=∫(1+u)2(1+u2)4u​du=∫[(1+u)2−2​+1+u22​]du=1+u2​+2arctanu+C=1+tan2x​2​+x+C.​
解二
$\begin{aligned} \displaystyle\int\cfrac{\sin x}{1+\sin x}\mathrm{d}x&=\displaystyle\int\cfrac{\sin x(1-\sin x)}{\cos^2x}\mathrm{d}x\\ &=-\displaystyle\int\cfrac{1}{\cos^2x}\mathrm{d}(\cos x)-\displaystyle\int(\sec^2x-1)\mathrm{d}x\\ &=\sec x-\tan x+x+C. \end{aligned}$∫1+sinxsinx​dx​=∫cos2xsinx(1−sinx)​dx=−∫cos2x1​d(cosx)−∫(sec2x−1)dx=secx−tanx+x+C.​
（这道题主要利用了换元或三角变换求解）

（27）$\displaystyle\int\cfrac{x+\sin x}{1+\cos x}\mathrm{d}x;$∫1+cosxx+sinx​dx;

解
$\begin{aligned} \displaystyle\int\cfrac{x+\sin x}{1+\cos x}\mathrm{d}x&=\displaystyle\int\cfrac{x}{2}\sec^2\cfrac{x}{2}\mathrm{d}x+\displaystyle\int\tan\cfrac{x}{2}\mathrm{d}x\\ &=\displaystyle\int x\mathrm{d}(\tan\cfrac{x}{2})+\displaystyle\int\tan\cfrac{x}{2}\mathrm{d}x\\ &=x\tan\cfrac{x}{2}+C. \end{aligned}$∫1+cosxx+sinx​dx​=∫2x​sec22x​dx+∫tan2x​dx=∫xd(tan2x​)+∫tan2x​dx=xtan2x​+C.​
（这道题主要倍角公式求解）

（28）$\displaystyle\int e^{\sin x}\cfrac{x\cos^3x-\sin x}{\cos^2x}\mathrm{d}x;$∫esinxcos2xxcos3x−sinx​dx;

解
$\begin{aligned} \displaystyle\int e^{\sin x}\cfrac{x\cos^3x-\sin x}{\cos^2x}\mathrm{d}x&=\displaystyle\int xe^{\sin x}\cos x\mathrm{d}x-\displaystyle\int e^{\sin x}\tan x\sec x\mathrm{d}x\\ &=\displaystyle\int x\mathrm{d}(e^{\sin x})-\displaystyle\int e^{\sin x}\mathrm{d}(\sec x)\\ &=xe^{\sin x}-\displaystyle\int e^{\sin x}\mathrm{d}x-(\sec xe^{\sin x}-\displaystyle\int e^{\sin x}\mathrm{d}x)\\ &=(x-\sec x)e^{\sin x}+C. \end{aligned}$∫esinxcos2xxcos3x−sinx​dx​=∫xesinxcosxdx−∫esinxtanxsecxdx=∫xd(esinx)−∫esinxd(secx)=xesinx−∫esinxdx−(secxesinx−∫esinxdx)=(x−secx)esinx+C.​
（这道题利用分部积分法化简至有相同部分可以相互抵消求解）

（31）$\displaystyle\int\cfrac{e^{3x}+e^x}{e^{4x}-e^{2x}+1}\mathrm{d}x;$∫e4x−e2x+1e3x+ex​dx;

解
$\begin{aligned} \displaystyle\int\cfrac{e^{3x}+e^x}{e^{4x}-e^{2x}+1}\mathrm{d}x&=\displaystyle\int\cfrac{e^{x}+e^{-x}}{e^{2x}+e^{-2x}-1}\mathrm{d}x=\displaystyle\int\cfrac{\mathrm{d}(e^{x}-e^{-x})}{(e^{x}-e^{-x})^2+1}\\ &=\arctan(e^{x}-e^{-x})+C. \end{aligned}$∫e4x−e2x+1e3x+ex​dx​=∫e2x+e−2x−1ex+e−x​dx=∫(ex−e−x)2+1d(ex−e−x)​=arctan(ex−e−x)+C.​
（这道题利用分式约分求解）

（33）$\displaystyle\int\ln^2(x+\sqrt{1+x^2})\mathrm{d}x;$∫ln2(x+1+x2​)dx;

解
$\begin{aligned} \displaystyle\int\ln^2(x+\sqrt{1+x^2})\mathrm{d}x&=x\ln^2(x+\sqrt{1+x^2})-\displaystyle\int\cfrac{2x\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm{d}x\\ &=x\ln^2(x+\sqrt{1+x^2})-\displaystyle\int2\ln(x+\sqrt{1+x^2})\mathrm{d}(\sqrt{1+x^2})\\ &=x\ln^2(x+\sqrt{1+x^2})-2\sqrt{1+x^2}\ln(x+\sqrt{1+x^2})+2x+C. \end{aligned}$∫ln2(x+1+x2​)dx​=xln2(x+1+x2​)−∫1+x2​2xln(x+1+x2​)​dx=xln2(x+1+x2​)−∫2ln(x+1+x2​)d(1+x2​)=xln2(x+1+x2​)−21+x2​ln(x+1+x2​)+2x+C.​
（这道题主要利用$(\ln(x+\sqrt{1+x^2}))'=\cfrac{1}{\sqrt{1+x^2}}$(ln(x+1+x2​))′=1+x2​1​进行化简求解）

（34）$\displaystyle\int\cfrac{\ln x}{(1+x^2)^{\frac{3}{2}}}\mathrm{d}x;$∫(1+x2)23​lnx​dx;

解
$\begin{aligned} \displaystyle\int\cfrac{\ln x}{(1+x^2)^{\frac{3}{2}}}&\xlongequal{x=\cfrac{1}{u}}\displaystyle\int\cfrac{u\ln u}{(1+u^2)^{\frac{3}{2}}}\mathrm{d}u=-\displaystyle\int\ln u\mathrm{d}((1+u^2)^{-\frac{1}{2}})\\ &=-\cfrac{\ln u}{\sqrt{1+u^2}}+\displaystyle\int\cfrac{\mathrm{d}u}{u\sqrt{1+u^2}}\\ &=x\cfrac{\ln x}{\sqrt{1+x^2}}-\displaystyle\int\cfrac{\mathrm{d}x}{\sqrt{1+x^2}}\\ &=x\cfrac{\ln x}{\sqrt{1+x^2}}-\ln(x+\sqrt{1+x^2})+C. \end{aligned}$∫(1+x2)23​lnx​​x=u1​∫(1+u2)23​ulnu​du=−∫lnud((1+u2)−21​)=−1+u2​lnu​+∫u1+u2​du​=x1+x2​lnx​−∫1+x2​dx​=x1+x2​lnx​−ln(x+1+x2​)+C.​
（这道题利用倒数换元求解）

（35）$\displaystyle\int\sqrt{1-x^2}\arcsin x\mathrm{d}x;$∫1−x2​arcsinxdx;

解  设$x=\sin u\left(-\cfrac{\pi}{2}<u<\cfrac{\pi}{2}\right)$x=sinu(−2π​<u<2π​)，则$\sqrt{1-x^2}=\cos u,\mathrm{d}x=\cos u\mathrm{d}u$1−x2​=cosu,dx=cosudu，于是
$\begin{aligned} \displaystyle\int\sqrt{1-x^2}\arcsin x\mathrm{d}x&=\displaystyle\int u\cos^2u\mathrm{d}u=\cfrac{1}{2}\displaystyle\int u(1+\cos2u)\mathrm{d}u\\ &=\cfrac{1}{4}\displaystyle\int u\mathrm{d}(2u+\sin2u)\\ &=\cfrac{u(2u+\sin2u)}{4}-\cfrac{1}{4}\displaystyle\int(2u+\sin2u)\mathrm{d}u\\ &=\cfrac{u^2}{4}+\cfrac{u}{4}\sin2u-\cfrac{\sin^2u}{4}+C\\ &=\cfrac{(\arcsin x)^2}{4}+\cfrac{x}{2}\sqrt{1-x^2}\arcsin x-\cfrac{x^2}{4}+C. \end{aligned}$∫1−x2​arcsinxdx​=∫ucos2udu=21​∫u(1+cos2u)du=41​∫ud(2u+sin2u)=4u(2u+sin2u)​−41​∫(2u+sin2u)du=4u2​+4u​sin2u−4sin2u​+C=4(arcsinx)2​+2x​1−x2​arcsinx−4x2​+C.​
（这道题利用三角函数换元求解）

（36）$\displaystyle\int\cfrac{x^3\arccos x}{\sqrt{1-x^2}}\mathrm{d}x;$∫1−x2​x3arccosx​dx;

解  设$x=\cos u\left(0<u<\pi\right)$x=cosu(0<u<π)，则$\sqrt{1-x^2}=\sin u,\mathrm{d}x=-\sin u\mathrm{d}u$1−x2​=sinu,dx=−sinudu，于是
$\begin{aligned} \displaystyle\int\cfrac{x^3\arccos x}{\sqrt{1-x^2}}\mathrm{d}x&=-\displaystyle\int u\cos^3u\mathrm{d}u=-\displaystyle\int u\mathrm{d}(\sin u-\cfrac{1}{3}\sin^3u)\\ &=-u(\sin u-\cfrac{1}{3}\sin^3u)+\displaystyle\int (\sin u-\cfrac{1}{3}\sin^3u)\mathrm{d}u\\ &=-u(\sin u-\cfrac{1}{3}\sin^3u)-\cfrac{1}{3}\displaystyle\int(2+\cos^2u)\mathrm(\cos u)\\ &=-u(\sin u-\cfrac{1}{3}\sin^3u)-\cfrac{2}{3}\cos u-\cfrac{1}{9}\cos^3u+C\\ &=-\cfrac{1}{3}\sqrt{1-x^2}(2+x^2)\arccos x-\cfrac{1}{9}(6+x^2)+C. \end{aligned}$∫1−x2​x3arccosx​dx​=−∫ucos3udu=−∫ud(sinu−31​sin3u)=−u(sinu−31​sin3u)+∫(sinu−31​sin3u)du=−u(sinu−31​sin3u)−31​∫(2+cos2u)(cosu)=−u(sinu−31​sin3u)−32​cosu−91​cos3u+C=−31​1−x2​(2+x2)arccosx−91​(6+x2)+C.​
（这道题主要用三角函数换元再分部积分求解）

（38）$\displaystyle\int\cfrac{\mathrm{d}x}{\sin^3x\cos x};$∫sin3xcosxdx​;

解
$\begin{aligned} \displaystyle\int\cfrac{\mathrm{d}x}{\sin^3x\cos x}&=-\displaystyle\int\cot x\sec^2x\mathrm{d}(\cot x)\xlongequal{u=\cot x}-\displaystyle\int u\left(1+\cfrac{1}{u^2}\right)\mathrm{d}u\\ &=-\cfrac{u^2}{2}-\ln|u|+C=-\cfrac{\cot^2x}{2}-\ln|\cot x|+C. \end{aligned}$∫sin3xcosxdx​​=−∫cotxsec2xd(cotx)u=cotx−∫u(1+u21​)du=−2u2​−ln∣u∣+C=−2cot2x​−ln∣cotx∣+C.​
（这道题主要利用了三角函数换元）

（39）$\displaystyle\int\cfrac{\mathrm{d}x}{(2+\cos x)\sin x};$∫(2+cosx)sinxdx​;

解
$\begin{aligned} \displaystyle\int\cfrac{\mathrm{d}x}{(2+\cos x)\sin x}&=\displaystyle\int\cfrac{\mathrm{d}(\cos x)}{(2+\cos x)(\cos^2x+1)}\\ &\xlongequal{u=\cos x}\displaystyle\int\cfrac{\mathrm{d}u}{(2+u)(u^2+1)}\\ &=\displaystyle\int\left[\cfrac{1}{6(u-1)}-\cfrac{1}{2(u+1)}+\cfrac{1}{3(u+2)}\right]\mathrm{d}u\\ &=\cfrac{1}{6}\ln|u-1|-\cfrac{1}{2}\ln|u+1|+\cfrac{1}{3}\ln|u+2|+C\\ &=\cfrac{1}{6}\ln(1-\cos x)-\cfrac{1}{2}\ln(1+\cos x)+\cfrac{1}{3}\ln(2+\cos x)+C. \end{aligned}$∫(2+cosx)sinxdx​​=∫(2+cosx)(cos2x+1)d(cosx)​u=cosx∫(2+u)(u2+1)du​=∫[6(u−1)1​−2(u+1)1​+3(u+2)1​]du=61​ln∣u−1∣−21​ln∣u+1∣+31​ln∣u+2∣+C=61​ln(1−cosx)−21​ln(1+cosx)+31​ln(2+cosx)+C.​
（这道题主要利用三角函数换元和因式分解积分）

（40）$\displaystyle\int\cfrac{\sin x\cos x}{\sin x+\cos x}\mathrm{d}x;$∫sinx+cosxsinxcosx​dx;

解一
$\begin{aligned} \displaystyle\int\cfrac{\sin x\cos x}{\sin x+\cos x}\mathrm{d}x&=\displaystyle\int\cfrac{\cfrac{1}{2}(\sin x+\cos x)^2-\cfrac{1}{2}}{\sin x+\cos x}\mathrm{d}x\\ &=\cfrac{1}{2}\displaystyle\int(\sin x+\cos x)\mathrm{d}x-\cfrac{1}{2}\displaystyle\int\cfrac{1}{\sin x+\cos x}\mathrm{d}x\\ &=\cfrac{1}{2}(-\cos x+\sin x)-\cfrac{1}{2}\displaystyle\int\cfrac{1}{\sin x+\cos x}\mathrm{d}x. \end{aligned}$∫sinx+cosxsinxcosx​dx​=∫sinx+cosx21​(sinx+cosx)2−21​​dx=21​∫(sinx+cosx)dx−21​∫sinx+cosx1​dx=21​(−cosx+sinx)−21​∫sinx+cosx1​dx.​
  令$u=\tan\cfrac{x}{2}$u=tan2x​，则$\sin x=\cfrac{2u}{1+u^2},\cos x=\cfrac{1-u^2}{1+u^2},\mathrm{d}x=\cfrac{2}{1+u^2}\mathrm{d}u$sinx=1+u22u​,cosx=1+u21−u2​,dx=1+u22​du，故有
$\begin{aligned} \displaystyle\int\cfrac{1}{\sin x+\cos x}\mathrm{d}x&=\displaystyle\int\cfrac{2}{2u+1-u^2}\mathrm{d}u=-\displaystyle\int\cfrac{2}{(u-1)^2-(\sqrt{2})^2}\mathrm{d}u\\ &=-\cfrac{1}{\sqrt{2}}\displaystyle\int\cfrac{2}{u-1-\sqrt{2}}\mathrm{d}u+\cfrac{1}{\sqrt{2}}\displaystyle\int\cfrac{2}{u-1+\sqrt{2}}\mathrm{d}u\\ &=\cfrac{1}{\sqrt{2}}\ln\left|\cfrac{u-1+\sqrt{2}}{u-1-\sqrt{2}}\right|+C'. \end{aligned}$∫sinx+cosx1​dx​=∫2u+1−u22​du=−∫(u−1)2−(2​)22​du=−2​1​∫u−1−2​2​du+2​1​∫u−1+2​2​du=2​1​ln∣∣∣∣∣​u−1−2​u−1+2​​∣∣∣∣∣​+C′.​
  因此有
$\displaystyle\int\cfrac{\sin x\cos x}{\sin x+\cos x}\mathrm{d}x=\cfrac{1}{2}(\sin x-\cos x)-\cfrac{1}{2\sqrt{2}}\ln\left|\cfrac{\tan\cfrac{x}{2}-1+\sqrt{2}}{\tan\cfrac{x}{2}-1-\sqrt{2}}\right|+C.$∫sinx+cosxsinxcosx​dx=21​(sinx−cosx)−22​1​ln∣∣∣∣∣∣∣∣​tan2x​−1−2​tan2x​−1+2​​∣∣∣∣∣∣∣∣​+C.
解二
$\begin{aligned} \displaystyle\int\cfrac{\sin x\cos x}{\sin x+\cos x}\mathrm{d}x&=\displaystyle\int\cfrac{\sin x\cos x}{\sqrt{2}\sin\left(x+\cfrac{\pi}{4}\right)}\mathrm{d}x\xlongequal{u=x+\cfrac{\pi}{4}}\displaystyle\int\cfrac{2\sin^2u-1}{2\sqrt{2}\sin u}\mathrm{d}u\\ &=\cfrac{1}{\sqrt{2}}\displaystyle\int\sin u\mathrm{d}u-\cfrac{1}{2\sqrt{2}}\displaystyle\int\csc u\mathrm{d}u\\ &=-\cfrac{\cos\left(x+\cfrac{\pi}{4}\right)}{\sqrt{2}}-\cfrac{1}{2\sqrt{2}}\ln\left|\csc\left(x+\cfrac{\pi}{4}\right)-\cot\left(x+\cfrac{\pi}{4}\right)\right|+V. \end{aligned}$∫sinx+cosxsinxcosx​dx​=∫2​sin(x+4π​)sinxcosx​dxu=x+4π​∫22​sinu2sin2u−1​du=2​1​∫sinudu−22​1​∫cscudu=−2​cos(x+4π​)​−22​1​ln∣∣∣∣∣​csc(x+4π​)−cot(x+4π​)∣∣∣∣∣​+V.​
（这道题主要利用万能公式或三角函数换元积分求解）

写在最后

  如果觉得文章不错就点个赞吧。另外，如果有不同的观点，欢迎留言或私信。
   欢迎非商业转载，转载请注明出处。
  另，参考的积分表及公式见附录。

• 点赞
• 收藏
• 分享
• • 文章举报

古月忻 发布了6 篇原创文章 · 获赞 0 · 访问量 155 私信 关注