不定积分的计算

单变量函数的积分

张瑞
中国科学技术大学数学科学学院

不定积分的计算

第一换元

定理 1. (第一换元)
 设$g(t)$定义在$[\alpha,\beta]$上，$t=\phi(x)$在$[a,b]$上有连续导数，且$\phi(x)\in[\alpha,\beta], \forall x\in[a,b]$。记

\[f(x)=g(\phi(x))\cdot\phi'(x) , x\in [a,b] \]

若$g(t)$在$[\alpha,\beta]$上有原函数$G(t)$，则

\[\begin{aligned} \int f(x)dx =&\int g(\phi(x))\phi'(x)dx \\ =&\int g(t)dt=G(t)+c=G(\phi(x))+c \end{aligned} \]

证：

\[\begin{aligned} (G(\phi(x))+c)' =&G'(\phi(x))\phi'(x) \\ =&g(\phi(x))\phi'(x)=f(x) \end{aligned} \]

第二换元

定理 2. (第二换元)
 设$g(t)$定义在$[\alpha,\beta]$上，$t=\phi(x)$在$[a,b]$上有连续导数，$\phi'(x)\neq0,x\in[a,b]$， 且$\phi(x)\in[\alpha,\beta], \forall x\in[a,b]$。记

\[f(x)=g(\phi(x))\cdot\phi'(x) , x\in [a,b] \]

若$f(x)$在$[a,b]$上有原函数$F(x)$ 时，则

\[\begin{aligned} \int g(t)dt =&\int g(\phi(x))\phi'(x)dx \\ =&\int f(x)dx =F(x)+c=F(\phi^{-1}(t))+c \end{aligned} \]

证： $\phi'(x)\neq0, x\in[a,b]$，则$x=\phi^{-1}(t)$存在，且$x'(t)=\dfrac{1}{\phi'(x)}$

\[\begin{aligned} \dfrac{dF(\phi^{-1}(t)}{dt}=&F'(\phi^{-1}{t})x'(t) \\ =&F'(\phi^{-1}(t))\dfrac{1}{\phi'(x)} =f(x)\dfrac{1}{\phi'(x)} \\ =&g(\phi(x))\phi'(x)\dfrac{1}{\phi'(x)} \\ =&g(\phi(x))=g(t) \end{aligned} \]
  • 第一换元法，又叫做凑微分法。

    如，注意到$\cos xdx=d(\sin x)$

    \[\begin{aligned} \int \cos x\sin x dx =&\int \sin x d(\sin x) \\ (t=\sin x)\quad =&\int t dt=\frac{1}2t^2+c \\ =&\frac{1}2\sin^2x+c \end{aligned} \]
  • 第二换元法则是主动用表达式替换$x$。

    如令$t=\sin x$，则$dt=\cos x dx$，即$dx=\dfrac{1}{\cos x}dt$，则有

    \[\begin{aligned} \int \cos x \sin x dx =&\int \cos(x) t dx=\int t \cos x \dfrac{1}{\cos x}dt \\ =&\int t dt=\frac{1}2t^2+c \\ =&\frac{1}2\sin^2x+c \end{aligned} \]

换元表

\[\begin{aligned} \int f(ax+b)dx=&\dfrac1a f(ax+b)d(ax+b) \\ \int \cdot \cos xdx=\int \cdot d\sin x , &\int \cdot \sin xdx=-\int \cdot d\cos x \\ \int \cdot \dfrac1x dx= \int \cdot d\ln x, &\int \cdot e^x dx= \int \cdot de^x \\ \int \cdot x^{n-1}dx=\dfrac1n\int \cdot dx^n , &\int \cdot \dfrac1{x^{n+1}}dx=-\dfrac1n\int \cdot d\dfrac1{x^n} \\ \int \cdot \dfrac1{x^2}dx=-\dfrac1n\int \cdot d\dfrac1x , &\int \cdot \dfrac1{\sqrt x}dx=2\int \cdot d\sqrt{x} \\ \end{aligned} \]
\[\begin{aligned} \int \cdot \dfrac1{\cos^2x}dx=\int \cdot d\tan x , & \int \cdot \dfrac1{\sin^2x}dx=-\int \cdot d\cot x \\ \int \cdot \dfrac1{\sqrt{1-x^2}}dx=&\int \cdot d\arcsin x \\ \int \cdot \dfrac1{1+x^2}dx=&-\int \cdot d\arctan x \\ \end{aligned} \]
\[\begin{aligned} \int\dfrac{f'(x)}{f(x)}dx=&\ln|f(x)| \\ \int\dfrac1{\sqrt{x^2+a}}dx=&\ln|x+\sqrt{x^2+a}| \end{aligned} \]

分部积分

定理 3. (分部积分)
 设$u(x),v(x)$可导，$\displaystyle\int u'(x)v(x)dx$存在，则 $\displaystyle \int u(x)v'(x)dx$也存在，且

\[\int u(x)v'(x)dx=u(x)v(x)-\int u'(x)v(x)dx \]

证：

\[\begin{aligned} (u(x)v(x))'=u'(x)v(x)+u(x)v'(x) \\ u(x)v'(x)=(u(x)v(x))'-u'(x)v(x) \end{aligned} \]

由$\displaystyle\int u'(x)v(x)dx$存在，则

\[\begin{aligned} \int u(x)v'(x)dx =&\int(u(x)v(x))'dx-\int u'(x)v(x)dx \\ =&u(x)v(x)-\int u'(x)v(x)dx \end{aligned} \]

算例

换元

例 1. [例4.2.10] $\displaystyle\int \dfrac{1}{\sqrt{4+(ax+b)^2}}dx$

例 2. $\displaystyle\int\dfrac{1}{(1+x)\sqrt{x}}dx$

例 3. [例4.2.12] $\displaystyle\int\dfrac{1}{\sqrt{1+e^{2x}}}dx$

例 4. $\displaystyle\int\dfrac{1}{\sin x}dx$, $\displaystyle\int\dfrac{1}{\cos x}dx$

1.

\[\begin{aligned} &\int \dfrac{1}{\sqrt{4+(ax+b)^2}}dx \\ =&\int\dfrac{\frac1a}{\sqrt{4+(ax+b)^2}}d(ax+b) \\ =&\dfrac1a\ln(ax+b+\sqrt{4+(ax+b)^2})+c \end{aligned} \]
\[\begin{aligned} &\displaystyle\int\dfrac{1}{(1+x)\sqrt{x}}dx \\ =&2\int\dfrac{1}{1+x}d\sqrt x \\ =&2\int\dfrac1{1+(\sqrt x)^2}d\sqrt x \\ =&2\arctan\sqrt x+c \end{aligned} \]
\[\begin{aligned} &\displaystyle\int\dfrac{1}{\sqrt{1+e^{2x}}}dx \\ =&\int\dfrac{1}{e^x\sqrt{1+e^{-2x}}}dx \\ =&\int\dfrac{-1}{\sqrt{1+e^{-2x}}}d(e^{-x}) \\ =&-\ln(e^{-x}+\sqrt{1+e^{-2x}})+c \end{aligned} \]

令$e^x=t$

$\displaystyle\int\dfrac{1}{\sin x}dx$, $\displaystyle\int\dfrac{1}{\cos x}dx$

\[\begin{aligned} =&\int\dfrac{1}{2\sin\frac{x}2\cos \frac{x}2}dx =\int\dfrac1{2\tan\frac{x}2\cos^2\frac{x}2}dx \\ =&\int\dfrac1{\tan\frac{x}2}d(\tan\frac{x}2)=\ln|\tan\frac{x}2|+c \end{aligned} \]

或令$t=\tan\frac{x}2$

$\displaystyle\int\dfrac{1}{\sin x}dx$, $\displaystyle\int\dfrac{1}{\cos x}dx$

\[\begin{aligned} =&\int\dfrac{\sin x}{\sin^2x}dx=\int\dfrac1{\cos^2x-1}d\cos x \\ =&\frac{1}2\int(\dfrac{-1}{\cos x+1}+\dfrac1{\cos x-1})d\cos x \end{aligned} \]

分部积分

例 5. $\displaystyle\int\ln xdx$, $\displaystyle\int\arctan x dx$, $\displaystyle\int\arcsin x dx$

例 6. $\displaystyle\int\sin x\ln(\tan x)dx$

例 7. $\displaystyle\int\left(\dfrac{\ln x}{x}\right)^2dx$

例 8. $\displaystyle\int x^2\sin(2x)dx$

$\displaystyle\int\ln xdx$, $\displaystyle\int\arctan x dx$, $\displaystyle\int\arcsin x dx$

\[\begin{aligned} =&x\ln x-\int xd\ln x=x\ln x-\int x\dfrac1x dx \\ =&x\ln x-x +c \end{aligned} \]

$t=\ln x$， 则$dx=e^t dt$

\[\begin{aligned} \int t e^t dt =&\int t de^t=t e^t-\int e^t dt \\ =&t e^t-e^t+c \end{aligned} \]

$\displaystyle\int(\dfrac{\ln x}{x})^2dx$

\[\begin{aligned} =&\int(\ln x)^2d\dfrac{-1}{x} =\dfrac{-1}{x}(\ln x)^2-\int\dfrac{-1}xd(\ln x)^2 \\ =&\dfrac{-1}{x}(\ln x)^2+\int\dfrac1x2\ln x\dfrac1xdx \\ =&\dfrac{-1}{x}(\ln x)^2+2\int\ln xd\dfrac{-1}x \\ =&\dfrac{-1}{x}(\ln x)^2-\dfrac2x\ln x+\int\dfrac2x\dfrac1xdx \\ =&\dfrac{-1}{x}(\ln x)^2-\dfrac2x\ln x-\dfrac2x+c \end{aligned} \]

$\displaystyle\int x^2\sin(2x)dx$

\[\begin{aligned} =&\int x^2d(\dfrac{-1}2\cos(2x)) =-\dfrac12x^2\cos(2x)+\int\cos(2x)\cdot xdx \\ =&-\dfrac12x^2\cos(2x)+\int xd(\frac{1}2\sin(2x)) \\ =&-\dfrac12x^2\cos(2x)+\frac{1}2x\sin(2x)-\frac{1}2\int\sin(2x)dx \\ =&-\dfrac12x^2\cos(2x)+\frac{1}2x\sin(2x)+\dfrac14\cos(2x)+c \end{aligned} \]

$\displaystyle\int\sin x\ln(\tan x)dx$

\[\begin{aligned} =&\int\ln(\tan x)d(-\cos x) \\ =&-\cos x\ln(\tan x)+\int\cos x\dfrac1{\tan x}\dfrac1{\cos^x}dx \\ =&-\cos x\ln(\tan x)+\int\dfrac1{\sin x}dx \\ =&-\cos x\ln(\tan x)+\ln|\tan\frac{x}2|+c \end{aligned} \]

对积分$\displaystyle\int p_n(x) f(x) dx$，其中$p_n(x)$为多项式，

  • $f(x)$为$\sin(x), \cos(x), e^x$，则可以采用降幂的方法
    \[\int p_n(x)\sin xdx=-\int p_n(x)d\cos x \]
  • $f(x)$为$\ln x,\arctan x, \arcsin x$，可以采用升幂的方法
    \[\int p_n(x)\ln xdx=\int \ln x dp_{n+1}(x) \]

循环

例 9. $\displaystyle\int x^2\sqrt{a^2+x^2}dx$

例 10. $\displaystyle\int \sqrt{x^2+A}dx$ （例4.2.10）

例 11. [例4.2.16] $\displaystyle\int e^{ax}\cos(bx)dx , a,b\neq 0$

$\displaystyle\int x^2\sqrt{a^2+x^2}dx$

\[\begin{aligned} =&\int\dfrac{x}3d(a^2+x^2)^{\frac32} \\ =&\dfrac{x}3(a^2+x^2)^{\frac32}-\dfrac13\int(a^2+x^2)^{\frac32}dx \\ =&\dfrac{x}3(a^2+x^2)^{\frac32}-\dfrac13\int(a^2+x^2)\sqrt{a^2+x^2}dx \\ =&\dfrac{x}3(a^2+x^2)^{\frac32}-\dfrac13\int x^2\sqrt{a^2+x^2}dx \\ & -\dfrac13\int a^2\sqrt{a^2+x^2}dx \end{aligned} \]
\[4\int x^2\sqrt{a^2+x^2}dx=x(a^2+x^2)^{\frac32}-a^2\int\sqrt{a^2+x^2}dx \]

$\displaystyle\int \sqrt{x^2+A}dx$

\[\begin{aligned} =&x\sqrt{x^2+A}-\int x d\sqrt{x^2+A} \\ =&x\sqrt{x^2+A}-\int x \dfrac{x}{\sqrt{x^2+A}}dx \\ =&x\sqrt{x^2+A}-\int \dfrac{x^2+A}{\sqrt{x^2+A}}dx+\int \dfrac{A}{\sqrt{x^2+A}}dx \\ =&x\sqrt{x^2+A}-\int \sqrt{x^2+A}dx+A\int \dfrac1{\sqrt{x^2+A}}dx \\ \end{aligned} \]

$\displaystyle\int e^{ax}\cos(bx)dx , a,b\neq 0$

\[\begin{aligned} =&\dfrac1a\int\cos(bx)d(e^{ax}) \\ =&\dfrac1a\cos(bx)e^{ax}+\dfrac1a\int e^{ax}b\sin(bx)dx \\ =&\dfrac1a\cos(bx)e^{ax}+\dfrac{b}{a}\int\sin(bx)\dfrac1a de^{ax} \\ =&\dfrac1a\cos(bx)e^{ax}+\dfrac{b}{a^2}\sin(bx)e^{ax}-\dfrac{b}{a^2}\int e^{ax}d(\sin(bx)) \\ =&\dfrac1a\cos(bx)e^{ax}+\dfrac{b}{a^2}\sin(bx)e^{ax}-\dfrac{b^2}{a^2}\int e^{ax}\cos(bx)dx \end{aligned} \]

递推

例 12. $\displaystyle\int \sec^n xdx=I_n$

例 13. $\displaystyle\int\tan^n xdx=I_n$

例 14. [例4.2.16] $\displaystyle\int\dfrac1{(x^2+a^2)^n}dx$

例 15. [例4.2.17] $\displaystyle\int \ln^n x dx$

$\displaystyle\int \sec^n xdx=I_n$

\[\begin{aligned} =&\int sec^{n-2}x d\tan x \\ =&\tan x\sec^{n-2}x-\int\tan x d\sec^{n-2}x \\ =&\tan x\sec^{n-2}x-\int\tan x(n-2)\sec^{n-3}x\sec x\tan xdx \\ =&\tan x\sec^{n-2}x-\int\tan^2x\sec^{n-2}x(n-2)dx \\ =&\tan x\sec^{n-2}x-\int(\sec^2-1)\sec^{n-2}x(n-2)dx \\ \end{aligned} \]
\[\begin{aligned} =&\tan x\sec^{n-2}x-(n-2)\int\sec^n xdx \\ &+(n-2)\int \sec^{n-2}xdx \end{aligned} \]

有

\[I_n=\tan x\sec^{n-2}x-(n-2)I_n+(n-2)I_{n-2} \]
\[I_n=\dfrac1{n-1}\tan x\sec^{n-2}x+\dfrac{n-2}{n-1}I_{n-2} \]
\[\begin{aligned} I_0=&x \\ I_1=&\ln|\sec x+\tan x|+c \end{aligned} \]

$\displaystyle\int\tan^n xdx=I_n$

\[\begin{aligned} \int\tan^nxdx+\int\tan^{n-2}xdx =&\int\tan^{n-2}x\sec^2xdx \\ =&\int \tan^{n-2}d\tan x \\ =&\dfrac1{n-1}\tan^{n-1}x+c \end{aligned} \]
\[I_n+I_{n-2}=\dfrac1{n-1}\tan^{n-1}x ,n=2,3,\cdots \]
\[\begin{aligned} I_0=&x \\ I_1=&\int\tan x dx=-\ln|\cos x|+c \end{aligned} \]

$\displaystyle\int\dfrac1{(x^2+a^2)^n}dx$

\[\begin{aligned} &\int\dfrac1{(x^2+a^2)^{n-1}}dx \\ =&\dfrac{x}{(x^2+a^2)^{n-1}}+2(n-1)\int\dfrac{x^2}{(x^2+a^2)^{n}}dx \\ =&\dfrac{x}{(x^2+a^2)^{n-1}}+2(n-1)\int\dfrac{x^2+a^2-a^2}{(x^2+a^2)^{n}}dx \\ =&\dfrac{x}{(x^2+a^2)^{n-1}}+2(n-1)\int(\dfrac{1}{(x^2+a^2)^{n-1}}-\dfrac{a^2}{(x^2+a^2)^{n}})dx \\ \end{aligned} \]
\[I_{n-1}=\dfrac{x}{(x^2+a^2)^{n-1}}+2(n-1)(I_{n-1}-a^2I_n) \]

有理函数的不定积分

定义 1.
两个实系数多项式的商，称为有理函数，其一般形式为

\[R(x)=\frac{P(x)}{Q(x)}=\frac{a_0+a_1 x+\cdots+a_nx^n}{b_0+b_1x+\cdots+b_mx^m} \]

其中$m,n$为非负整数，$a_n\neq0$, $b_m\neq0$，$a_i$, $b_i$均为实数。 $n<m$时，称为有理真分式，否则，称$R(x)$为有理假分式。

任何一个有理真分式可以拆解一个有理真分式与一个多项式的和。

几个简单有理函数的积分

\[\begin{aligned} \displaystyle\int\dfrac{1}{1+x^2}dx=&\arctan x \\ \displaystyle\int\dfrac{x}{1+x^2}dx=&\int\dfrac{1}{1+x^2}\frac{1}2d(x^2)=\frac{1}2\ln(1+x^2) \\ \end{aligned} \]

当$b^2-4ac<0$时，

\[\begin{aligned} \int\dfrac{px+q}{ax^2+bx+c}dx=&\int\dfrac{p(x+d)+q-pd}{a((x+d)^2+e)}dx \\ =&\dfrac{q-pd}a\int\dfrac1{(x+d)^2+e}d(x+d) \\ &+\frac{p}{2a}\int\dfrac1{(x+d)^2+e}d((x+d)^2) \\ \end{aligned} \]

当$b^2-4ac\geq0$时，上面的积分是多少？

$\displaystyle\int\dfrac1{(x^2+a^2)^n}dx$由前面给出了递推关系计算。

\[\begin{aligned} \int\frac{px+q}{(ax^2+bx+c)^n}dx=&\int\dfrac{p(x+d)+q-pd}{a^n((x+d)^2+e)^n}dx \\ =&\frac{p}{2a^n}\int\frac{x+d}{((x+d)^2+e)^n}d(x+d) \\ &+\frac{q-pd}{2a^n}\int\frac{1}{((x+d)^2+e)^n}d(x+d) \\ \end{aligned} \]

例 16. $\displaystyle\int\dfrac{x^2}{1+x^2}dx$

例 17. $\displaystyle\int\dfrac{x^3}{1+x^2}dx$

例 18. $\displaystyle\int\dfrac{x-1}{x^2+2x+3}dx$

例 19. $\displaystyle\int\dfrac{x-1}{x^2+2x-3}dx$

$\displaystyle\int\dfrac{x^2}{1+x^2}dx$

\[\begin{aligned} \int\dfrac{x^2}{1+x^2}dx =&\int\dfrac{1+x^2-1}{1+x^2}dx \\ =&\int 1 dx-\int\dfrac{1}{1+x^2}dx \end{aligned} \]
\[\begin{aligned} \int\dfrac{x^3}{1+x^2}dx =&\int\dfrac{x(1+x^2)-x}{1+x^2}dx \\ =&\int x dx-\int\dfrac{x}{1+x^2}dx \end{aligned} \]
\[\begin{aligned} &\int\dfrac{x-1}{x^2+2x+3}dx\\ =&\int\dfrac{x+1}{x^2+2x+3}dx-\int\dfrac{2}{x^2+2x+3}dx \\ =&\int\dfrac{\frac12}{x^2+2x+3}d(x^2+2x+3)-2\int\dfrac{1}{(x+1)^2+2}d(x+1) \\ =&\frac{1}2\ln(x^2+2x+3)-\sqrt2\arctan\dfrac{x+1}{\sqrt 2}+c \end{aligned} \]

若$Q(x)=(x-a)^k(x^2+px+q)^m$，则

\[\begin{aligned} \frac{P(x)}{Q(x)}=& \dfrac{c_1}{x-a}+\dfrac{c_2}{(x-a)^2}+\cdots+\dfrac{c_k}{(x-a)^k} \\ &+\dfrac{M_1x+N_1}{x^2+px+q}+\dfrac{M_2x+N_2}{(x^2+px+q)^2}+\cdots+\dfrac{M_mx+N_m}{(x^2+px+q)^m} \end{aligned} \]

例 20. $\displaystyle\int\dfrac{x}{(x+1)(x+2)(x+3)}dx$

例 21. $\displaystyle\int\dfrac{1}{(x+1)(x+2)^2(x+3)^3}dx$

\[\int\dfrac{x}{(x+1)(x+2)(x+3)}dx \]

设

\[\dfrac{x}{(x+1)(x+2)^2(x+3)^3}=\dfrac{A}{x+1}+\dfrac{B}{x+2}+\dfrac{C}{x+3} \]

由

\[\begin{aligned} \dfrac{x}{(x+2)^2(x+3)^3}={A}+\dfrac{B(x+1)}{x+2}+\dfrac{C(x+1)}{x+3} \\ \end{aligned} \]

取$x=-1$，则有

\[\dfrac{-1}2=A+0+0 \]
\[\int\dfrac{1}{(x+1)(x+2)^2(x+3)^3}dx \]

设

\[\begin{aligned} \dfrac{1}{(x+1)(x+2)^2(x+3)^3} =\dfrac{A}{x+1}+\dfrac{B}{x+2}+\dfrac{C}{(x+2)^2} \\ +\dfrac{D}{x+3}+\dfrac{E}{(x+3)^2}+\dfrac{F}{(x+3)^3} \end{aligned} \]
  • 两边乘$(x+1)$，取$x=-1$，得$A=\dfrac18$
  • 两边乘$(x+2)^2$，取$x=-2$，得$C=-1$
  • 两边乘$(x+3)^3$，取$x=-3$，得$F=-\dfrac12$

将$A,C,F$代入，得到

\[\dfrac{-x^2+x+22}{8(x+2)(x+3)^2}=\dfrac{B}{x+2}+\dfrac{D}{x+3}+\dfrac{E}{(x+3)^2} \]
  • 两边乘$(x+2)$，取$x=-2$，得$B=2$
  • 两边乘$(x+3)^2$，取$x=-3$，得$E=-\dfrac54$

最后，可得$D=-\dfrac{17}8$

例 22. $\displaystyle\int\dfrac{1}{x^3+1}dx$

例 23. $\displaystyle\int\dfrac{1}{x^4+1}dx$

例 24. $\displaystyle\int\dfrac{1}{x^6+1}dx$

\[\begin{aligned} \int\dfrac{1}{x^3+1}dx=\int\dfrac1{(x+1)(x^2-x+1)}dx \\ =\int\dfrac{A}{x+1}+\dfrac{Bx+C}{x^2-x+1}dx \end{aligned} \]
  • 乘$x+1$，取$x=-1$，得$A=\dfrac13$
  • 然后，可得$B=-\dfrac13, C=\dfrac23$
\[\begin{aligned} \int\dfrac{1}{x^4+1}dx=&\int\dfrac{1}{(x^2+1)^2-2x^2}dx \\ =&\int\dfrac{ax+b}{x^2+\sqrt2x+1}+\dfrac{cx+d}{x^2-\sqrt2x+1}dx \end{aligned} \]

$b=d=\frac{1}2$， $a=-c=\dfrac{\sqrt2}4$

\[\begin{aligned} \int\dfrac{1}{x^6+1}dx=\int\dfrac1{(x^2+1)((x^2+1)^2-3x^2)}dx \\ =\int\dfrac{1}{(x^2+1)(x^2+1-\sqrt3x)(x^2+1+\sqrt3x)}dx \end{aligned} \]

若$Q(x)=(x-a)^k(x^2+px+q)^{m-1}$，则

\[\begin{aligned} \dfrac{P(x)}{Q(x)}=&\left(\dfrac{P_1(x)}{Q_1(x)}\right)'+\dfrac{P_2(x)}{Q_2(x)} \\ =&\left(\dfrac{P_1(x)}{(x-a)^{k-1}(x^2+px+q)^{m-1}}\right)' \\ &+\dfrac{P_2(x)}{(x-a)(x^2+px+q)} \end{aligned} \]

例 25. $\displaystyle\int\dfrac{4x^5-1}{(x^5+x+1)^2}dx$

\[\int\dfrac{4x^5-1}{(x^5+x+1)^2}dx \]
\[\begin{aligned} \dfrac{4x^5-1}{(x^5+x+1)^2}=&(\dfrac{Ax^4+Bx^3+Cx^2+Dx+E}{x^5+x+1})' \\ &+\dfrac{A_1x^4+B_1x^3+C_1x^2+D_1x+E_1}{x^5+x+1} \\ =&(\dfrac{-x}{x^5+x+1})' \end{aligned} \]

三角函数有理式的不定积分

由三角函数与常数经过有限次四则运算所构成的式子，为三角函数有理式。 三角函数都可以由$\sin x, \cos x$表示，因此三角函数有理式可以记为$R(\sin x, \cos x)$

对$I=\displaystyle\int R(\sin x, \cos x)dx$用万能公式$t=\tan\frac{x}2$，则

\[\begin{aligned} \sin x=\dfrac{2t}{1+t^2}, \cos x=\dfrac{1-t^2}{1+t^2}, \\ dx=d(2\arctan t)=\dfrac2{1+t^2}dt \end{aligned} \]

则 $\displaystyle~I=\int R\left(\dfrac{1-t^2}{1+t^2},\dfrac{2t}{1+t^2}\right)\dfrac{2}{1+t^2}dt$

例 26. $\displaystyle\int\dfrac1{2\sin x-\cos x+5}dx$

例 27. $\displaystyle\int\dfrac{\sin x\cos x}{\sin x+\cos x}dx$

$\displaystyle\int\dfrac1{2\sin x-\cos x+5}dx$

$t=\tan\dfrac{x}2$，则

\[\begin{aligned} \int\dfrac1{2\sin x-\cos x+5}dx =&\int\dfrac{\frac2{1+t^2}}{2\frac{2t}{1+t^2}-\frac{1-t^2}{1+t^2}+5}dt \\ =&\int\dfrac2{4t-(1-t^2)+5(1+t^2)}dt \\ =&\int\dfrac1{3t^2+2t+2}dt \\ =&\dfrac13\int\dfrac1{(t+\frac13)^2+\frac59}dt \end{aligned} \]

$\displaystyle\int\dfrac{\sin x\cos x}{\sin x+\cos x}dx$

$t=\tan\dfrac{x}2$，则

\[=\int\dfrac{4t(1-t^2)}{(1+t^2)^2(-t^2+2t+1)}dt \]
\[\dfrac{4t(1-t^2)}{(1+t^2)^2(-t^2+2t+1)} =(\dfrac{t-1}{1+t^2})'+\dfrac1{t^2-2t-1} \]

万能代换会将有理式的幂次提高，计算量通常会比较大。可以考虑如下的代换：

  • 若$R(-\sin x, \cos x)=-R(\sin x ,\cos x)$， 令$t=\cos x$;
  • 若$R(\sin x, -\cos x)=-R(\sin x ,\cos x)$， 令$t=\sin x$;
  • 若$R(-\sin x, -\cos x)=R(\sin x ,\cos x)$， 令$t=\tan x$;

例 28. [$t=\tan x$] $\displaystyle\int\dfrac1{\sin^4x+\cos^4x}dx$

例 29. [$t=\sin x$] $\displaystyle\int\sin^2x\cos^3xdx$

例 30. [$t=\cos x$] $\displaystyle\int\dfrac{\sin^5x}{\cos^4x}dx$

例 31. [$t=\tan x$] $\displaystyle\int\dfrac1{\sin^4x\cos^2x}dx$

\[\int\dfrac1{\sin^4x+\cos^4x}dx=\int\dfrac{\sec^4x}{1+\tan^4x}dx \]

令$t=\tan x$，则有

\[\begin{aligned} =&\int\dfrac{1+t^2}{1+t^4}dt =\int\dfrac{1+\frac1{t^2}}{t^2+\frac1{t^2}}dt \\ =&\int\dfrac1{(t-\frac1t)^2+2}d(t-\frac1t) \end{aligned} \]

$\displaystyle\int\sin^2x\cos^3xdx$ 令$t=\sin x$，则

\[=\int t^2(1-t^2)dt=\dfrac{t^3}3-\dfrac{t^5}t+c \]

$\displaystyle\int\dfrac{\sin^5x}{\cos^4x}dx$ 令$t=\cos x$，则

\[=-\int\dfrac{t^4-2t^2+1}{t^4}dt=-t-\dfrac2t+\dfrac1{3t^3}+c \]

$\displaystyle\int\dfrac1{\sin^4x\cos^2x}dx$ 令$t=\tan x$，

\[=\int\dfrac{(1+t^2)^2}{t^4}dt=t-\dfrac2t-\dfrac1{3t^3}+c \]

例 32. [$t=\tan x$] $AC-B^2>0$,

\[\displaystyle\int\dfrac{1}{A\cos^2x+B\sin x\cos x+C\sin^2x}dx \]

例 33. [$t=\tan x$] $\displaystyle\int\dfrac{\sin^2 x\cos x}{\sin x+\cos x}dx$

例 34. [$t=\cos x$] $\displaystyle\int\dfrac1{\sin x\cos 2x}$

$\displaystyle\int\dfrac{1}{A\cos^2x+B\sin x\cos x+C\sin^2x}dx$

$t=\tan x$

\[\begin{aligned} =&\int\dfrac1{A+2Bt+Ct^2}dt \\ =&\dfrac1{\sqrt{AC-B^2}}\arctan\dfrac{C\tan x+B}{\sqrt{AC-B^2}}+c \end{aligned} \]

$\displaystyle\int\dfrac{\sin^2 x\cos x}{\sin x+\cos x}dx$

$t=\tan x$

$\displaystyle\int\dfrac1{\sin x\cos 2x}=\int\dfrac1{\sin x(2\cos^2x-1)}dx$

$t=\cos x$

\[\begin{aligned} =&\int\dfrac1{(1-t^2)(1-2t^2)}dt \\ =&\dfrac1{\sqrt 2}\ln|\dfrac{1+t\sqrt 2}{1-t\sqrt 2}|+\ln|\dfrac{1-t}{1+t}|+c \end{aligned} \]

若$R(\sin(x),\cos(x))$不满足上述的三个条件，则可以对函数$R(\cdot,\cdot)$做如下分解

\[\begin{aligned} R(u,v)=&\frac{R(u,v)-R(-u,v)}2+\frac{R(-u,v)-R(-u,-v)}2 \\ &+\frac{R(-u,-v)+R(u,v)}2 \end{aligned} \]

分解出的三个部分, 分别满足三个条件

解. 例如：

\[\begin{aligned} \int\dfrac{\sin x\cos x}{\sin x+\cos x}dx =&\int\dfrac{\sin x\cos x(-\sin x+\cos x)}{\cos^2x-\sin^2x}dx \\ =&\int\dfrac{\cos^2x\sin x}{2\cos^2x-1}dx+\int\dfrac{-\sin^2x\cos x}{1-2\sin^2x}dx \\ =&\int\dfrac{-\cos^2x}{2\cos^2x-1}d(\cos x) \\ &+\int\dfrac{-\sin^2x}{1-2\sin^2x}d(\sin x) \\ \end{aligned} \]

递推式

例 35. $\displaystyle\int\dfrac1{\cos^5x}dx$

例 36. $\displaystyle\int {\cos^5x}dx$

例 37. $\displaystyle\int \frac{\cos^4x}{\sin^3x}dx$

\[\begin{aligned} \int\frac1{\cos^3}dx =&\int \frac1{\cos^4x}d(\sin x) \\ =&\frac{\sin x}{\cos^4x}-\int \sin x \frac{-4}{\cos^5}(-\sin x)dx\\ =&\frac{\sin x}{\cos^4x}-\int (1-\cos^2x) \frac{4}{\cos^5}dx\\ \end{aligned} \]

得到

\[\int\dfrac1{\cos^5x}dx=\dfrac{\sin x}{4\cos^4x}+\dfrac34\int\dfrac1{\cos^3x}dx \]
\[\int\dfrac1{\cos^3x}dx=\dfrac{\sin x}{2\cos^2x}+\dfrac12\int\dfrac1{\cos x}dx \]
\[\begin{aligned} \int \cos^5dx =&\int \cos^4x d(\sin x) \\ =&\sin x \cos^4x-\int (\sin x)(4\cos^3x)(-\sin x)dx \\ =&\sin x \cos^4x+\int (1-\cos^2x)4\cos^3x dx\\ \end{aligned} \]

得到

\[5 \int \cos^5xdx = \sin x\cos^4x +4\int \cos^3x dx \]
\[\begin{aligned} \int\frac{\cos^4x}{\sin^3x}dx =-\frac{\cos^5x}{2\sin^2x}-\frac32\int\frac{\cos^4x}{\sin x}dx \end{aligned} \]
\[\begin{aligned} \int\frac{\cos^4x}{\sin x}dx =&-\frac13{\cos^3x}+\int\frac{\cos^2x}{\sin x}dx \\ =&\frac13\cos^3x+\cos x+\ln\left|\tan\frac{x}2\right|+C \end{aligned} \]
\[\int\frac{\cos^4x}{\sin^3x}dx =-\frac{\cos x}{2\sin^2x}-\cos x-\frac32\ln\left|\tan\frac{x}2\right|+C \]

例 38. [复习题]$\displaystyle\int\sin^px\cos^qxdx$ 形态

注意到

\[\begin{aligned} (\sin^px\cos^qx)' =&p\sin^{p-1}x\cos x\cos^qx+\sin^pxq\cos^{q-1}x(-\sin x) \\ =&p\sin^{p-1}x\cos^{q+1}x-q\sin^{p+1}x\cos^{q-1}x \\ {\color{blue}=}&p\sin^{p-1}x\cos^{q-1}x(\cos^2x)-q\sin^{p+1}x\cos^{q-1}x\\ (\sin^px\cos^qx)'=&p\sin^{p-1}x\cos^{q-1}x-(p+q)\sin^{p+1}x\cos^{q-1}x \\ {\color{blue}=}&p\sin^{p-1}x\cos^{q+1}x-q\sin^{p-1}x\cos^{q-1}x(\sin^2x)\\ (\sin^px\cos^qx)'=&(p+q)\sin^{p-1}x\cos^{q+1}x-q\sin^{p-1}x\cos^{q-1}x \\ \end{aligned} \]

上式两边积分后，可以得到$\displaystyle I_{p,q}=\int\sin^p(x)\cos^qxdx$的递推公式

  • $q\neq-1$ ,
    \[I_{p,q}=-\dfrac{\sin^{p+1}x\cos^{q+1}x}{q+1}+\dfrac{p+q+2}{q+1}I_{p,q+2}%\displaystyle\int\sin^px\cos^{q+2}xdx \]
  • $p\neq-1$,
    \[I_{p,q}=\dfrac{\sin^{p+1}x\cos^{q+1}x}{p+1}+\dfrac{p+q+2}{p+1}I_{p+2,q} %\displaystyle\int\sin^{p+2}x\cos^{q}xdx \]
  • $p+q\neq0$,
    \[\begin{aligned} I_{p,q}= \dfrac{\sin^{p+1}x\cos^{q-1}x}{p+q}+\dfrac{q-1}{p+q}I_{p,q-2}%\displaystyle\int\sin^{p}x\cos^{q-2}xdx \\ I_{p,q}= -\dfrac{\sin^{p-1}x\cos^{q+1}x}{p+q}+\dfrac{p-1}{p+q}I_{p-2,q}%\displaystyle\int\sin^{p-2}x\cos^{q}xdx \end{aligned} \]

当$p,q$为整数时，递推后，最终化为以下积分

\[\begin{aligned} &\int \cos xdx=\sin x ,&& \int \sin xdx=-\cos x \\ &\int \dfrac1{\cos x}dx=\ln\left|\tan(\frac{x}2+\dfrac{\pi}4)\right| ,&& \int \dfrac1{\sin x}dx=\ln\left|\tan(\frac{x}2)\right| \\ &\int \sin x\cos xdx=\dfrac{\sin^2x}2 ,&& \int\dfrac{1}{\sin x\cos x}dx=\ln|\tan x| \\ &\int \dfrac{\sin x}{\cos x}dx=-\ln|\cos x| ,&& \int \dfrac{\cos x}{\sin x}dx=\ln|\sin x| \end{aligned} \]

解. 利用递推公式，

\[\begin{aligned} \int\frac{\cos^4x}{\sin^3x}dx =-\frac{\cos^5x}{2\sin^2x}-\frac32\int\frac{\cos^4x}{\sin x}dx \end{aligned} \]
\[\begin{aligned} \int\frac{\cos^4x}{\sin x}dx =&-\frac13{\cos^3x}+\int\frac{\cos^2x}{\sin x}dx \\ =&\frac13\cos^3x+\cos x+\ln\left|\tan\frac{x}2\right|+C \end{aligned} \]
\[\int\frac{\cos^4x}{\sin^3x}dx =-\frac{\cos x}{2\sin^2x}-\cos x-\frac32\ln\left|\tan\frac{x}2\right|+C \]

例 39. [例4.2.28] $\displaystyle\int\dfrac{\sin x}{a\cos x+b\sin x}dx$,$\displaystyle\int\dfrac{\cos x}{a\cos x+b\sin x}dx$

例 40. $\displaystyle\int\dfrac{\sin x}{a\cos x+b\sin x+c}dx$,$\displaystyle\int\dfrac{\cos x}{a\cos x+b\sin x+c}dx$,

例 41. $\displaystyle\int\dfrac{1}{a+b\cos x}dx$,$\displaystyle\int\dfrac{1}{a+b\sin x}dx$,

例 42. $\displaystyle\int\dfrac{1}{a+b\cos x+c\sin x}dx$,

$T_1=\displaystyle\int\dfrac{\sin x}{a\cos x+b\sin x}dx$,$T_2=\displaystyle\int\dfrac{\cos x}{a\cos x+b\sin x}dx$,

\[\begin{aligned} aT_1+bT_2&=\int 1 dx=x \\ -aT_1+bT_2&=\int\dfrac{-a\sin x+b\cos x}{a\cos x+b\sin x}dx \\ &=\int\dfrac1{a\cos x+b\sin x}d{(a\cos x+b\sin x)} \end{aligned} \]
\[\begin{aligned} T_1=\dfrac{1}{a^2+b^2}(bx-a\ln|a\cos x+b\sin x|)+c \\ T_2=\dfrac{1}{a^2+b^2}(ax+b\ln|a\cos x+b\sin x|)+c \\ \end{aligned} \]

$\displaystyle\int\dfrac{1}{a+b\cos x}dx$， 令 $t=\tan(\frac{x}2)$

\[\int\dfrac{1}{a+b\cos x}dx=\int\dfrac{2}{(a+b)+(a-b)t^2}dt \]
  • $a>b,a>0,b>0$或$a<b,a<0,b<0$时，即$|a|>|b|$，有$=\dfrac{2}{\sqrt{a^2-b^2}}\arctan\left(\sqrt{\dfrac{a-b}{a+b}}t\right)+c$
  • $|a|<|b|$，有$=\dfrac{1}{\sqrt{b^2-a^2}}\ln\left|\dfrac{\sqrt{b+a}+t\sqrt{b-a}}{\sqrt{b+a}-t\sqrt{b-a}}\right|+c$

$\displaystyle\int\dfrac{1}{a+b\sin x}dx=\displaystyle\int\dfrac{1}{a+b\cos(\dfrac{\pi}2-x)}dx$

$\displaystyle\int\dfrac{1}{a+b\cos x+c\sin x}dx=\displaystyle\int\dfrac{1}{a+\sqrt{b^2+c^2}\cos(x-\alpha)}dx$

其中

\[\cos\alpha=\dfrac{b}{\sqrt{b^2+c^2}}, \sin\alpha=\dfrac{c}{\sqrt{b^2+c^2}}, \]

可有理化的初等函数的不定积分

仍然记$R(u,v)$为关于变量$u$, $v$的有理函数，即$R(u,v)$是$u$, $v$和常数经有限次四则运算得到的。

  • $R(x,\sqrt[m]{\dfrac{\alpha x+\beta}{\gamma x+\delta}})$ 型积分
  • 形如$R(x,\sqrt{ax^2+bx+c})$的表达式的积分，其中$ax^2+bx+c$没有相等的根。

对于$R(x,\sqrt[m]{\dfrac{\alpha x+\beta}{\gamma x+\delta}})$ 型积分，其中$m\in\mathbb N$，$\alpha,\beta,\gamma,\delta$为常数。

令$t=\sqrt[m]{\dfrac{\alpha x+\beta}{\gamma x+\delta}}$，则 $\displaystyle~x=\phi(t)=\dfrac{\delta t^m-\beta}{\alpha-\gamma t^m},$ 这样，

\[\int R(x,\sqrt[m]{\dfrac{\alpha x+\beta}{\gamma x+\delta}})dx=\int R(\phi(t),t)\phi'(t)dt \]

为关于$t$的有理函数积分。

例 43. $\displaystyle\int\dfrac{x}{\sqrt[4]{x^3(a-x)}}dx$

例 44. $\displaystyle\int\dfrac1{\sqrt[3]{(x-1)(x+1)^2}}dx$

例 45. $\displaystyle\int\dfrac{1}{x(1+2\sqrt x+\sqrt[3]x)}dx$

例 46. $\displaystyle\int x^{\frac13}(a+b x^{\frac25})^2dx$

解. 可以看出，上面是$R(x,\sqrt[15]{x})$形态的积分，令$t=\sqrt[15]{x}$即可。

例 47. $\displaystyle\int\dfrac{{\sqrt[3]{1+\sqrt[4]{x}}}}{\sqrt x}dx$，令$z=x^{\frac14}$，有$\displaystyle\int 4z(1+z)^\frac13 dz$

例 48. $\displaystyle\int\dfrac1{\sqrt[4]{1+x^4}}dx$，令$z=x^{4}$，有$\displaystyle\int \frac14z^{-\frac34}(1+z)^{-\frac14} dz$

例 49. $\displaystyle\int\dfrac1{x\sqrt[3]{1+x^5}}dx$，令$z=x^{5}$，有$\displaystyle\int \frac15z^{-1}(1+z)^{-\frac13} dz$

形如$R(x,\sqrt{ax^2+bx+c})$的表达式的积分，其中$ax^2+bx+c$没有相等的根。

可以用Euler代换:

  • $a>0$, $\sqrt{ax^2+bx+c}=\pm\sqrt a x+t$
  • $c>0$, $\sqrt{ax^2+bx+c}=xt\pm\sqrt c$
  • $\sqrt{a(x-x_1)(x-x_2)}=t(x-x_1)$

例 50. $\displaystyle\int\dfrac1{x+\sqrt{x^2-x+1}}dx$

例 51. $\displaystyle\int\dfrac1{1+\sqrt{1-2x-x^2}}dx$

例 52. $\displaystyle\int\dfrac{x-\sqrt{x^2+3x+2}}{x+\sqrt{x^2+3x+2}}dx$

双曲代换

  • $R(x,\sqrt{x^2-1})$型，可以用双曲余弦代换
    \[x=\cosh(u)=\frac{e^u+e^{-u}}2,\ \ dx=\sinh(u), \ \ \ \sqrt{x^2-1}=\sinh(u) \]
  • $R(x,\sqrt{x^2+1})$型，可以用双曲正弦代换
    \[x=\sinh(u)=\frac{e^u-e^{-u}}2,\ \ dx=\cosh(u), \ \ \ \sqrt{x^2+1}=\cosh(u) \]
    或令
    \[u=x+\sqrt{x^2+1} \]
\[\sinh(x)=\frac{e^x-e^{-x}}2 , \cosh(x)=\frac{e^x+e^{-x}}2 , \tanh(x)=\frac{\sinh(x)}{\cosh(x)} \]
\[\mbox{arcsinh}(x)=\ln(x+\sqrt{x^2+1}) , \mbox{arccosh}(x)=\ln(x+\sqrt{x^2-1}) , \]
\[\mbox{arctanh}(x)=\frac12\ln\frac{1+x}{1-x} , x\in(-1,1) \]
\[\sinh'(x)=\cosh(x), \cosh'(x)=\sinh(x),\tanh'(x)=\frac1{\cosh^2(x)} \]
\[\mbox{arcsinh}'(x)=\frac1{\sqrt{x^2+1}} , \mbox{arccosh}'(x)=\frac1{\sqrt{x^2-1}} , \]
\[\mbox{arctanh}'(x)=\frac1{1-x^2} , x\in(-1,1) \]

例 53. $\displaystyle\int\sqrt{a^2+x^2}dx$

例 54. $\displaystyle\int\frac{x^2}{\sqrt{a^2+x^2}}dx$

无理函数

$ \displaystyle\int\frac1{\sqrt{a^2-x^2}}dx=\arcsin\frac{x}{a}+C $

$\displaystyle\int\sqrt{a^2-x^2}dx=\frac{a^2}2\arcsin\frac{x}a+\frac12x\sqrt{a^2-x^2}+C $

$ \displaystyle\int\frac1{\sqrt{A+x^2}}dx=\ln|x+\sqrt{x^2+A}|+C $

$\displaystyle\int\sqrt{A+x^2}dx=\frac{A}2\ln(x+\sqrt{x^2+A})+\frac12x\sqrt{x^2+A}+C $

降幂

若$P_n(x)$为$n$次多项式，$y=\sqrt{ax^2+bx+c}$，则有

\[\int\dfrac{P_n(x)}y dx=Q_{n-1}(x)y+\lambda\int\dfrac1y dx \]

其中$Q_{n-1}(x)$为$n-1$次多项式，$\lambda$为常数

例 55. $\displaystyle\int\dfrac{x^3}{\sqrt{1+2x-x^2}}dx$

解. 令$y={\sqrt{1+2x-x^2}}$则

\[\begin{aligned} \dfrac{x^3}y =&((Ax^2+Bx+C)y)'+\dfrac{\lambda}{y} \\ =&(2Ax+B)y+(Ax^2+Bx+C)\dfrac{2-2x}{2y}+\dfrac{\lambda}y \\ =&\dfrac1y((2Ax+B)(1+2x-x^2) \\ &+(1-x)(Ax^2+Bx+C)+\lambda) \end{aligned} \]
\[\begin{cases} -3A=1 \\ 5A-2B=0 \\ 2A+3B-C=0 \\ B+C+\lambda=0 \end{cases} \Rightarrow \begin{cases} A=-\frac13 \\ B=-\frac56 \\ C=-\frac{19}6 \\ \lambda=4 \end{cases} \]

杂题

例 56. $\displaystyle\int\dfrac{e^x\cos x-e^x\sin x}{e^{2x}}dx$

例 57. $\displaystyle\int\dfrac{x\cos x-\sin x}{x^2}dx$

例 58. $\displaystyle\int\dfrac{x\cos x-\sin x}{x^2\sqrt{1-\frac{\sin^2x}{x^2}}}dx$

\[\begin{aligned} \int\dfrac{e^x\cos x-e^x\sin x}{e^{2x}}dx=\int d\dfrac{\sin x}{e^x} \\ =\dfrac{\sin x}{e^x}+c \end{aligned} \]
\[\int\dfrac{x\cos x-\sin x}{x^2}dx=\int d\dfrac{\sin x}{x} \]
\[\begin{aligned} \int\dfrac{x\cos x-\sin x}{x^2\sqrt{1-\frac{\sin^2x}{x^2}}}dx =\int\dfrac1{\sqrt{1-(\frac{\sin x}{x})^2}}d\dfrac{\sin x}{x} \\ =\arcsin\dfrac{\sin x}x+c \end{aligned} \]

例 59. $\displaystyle\int x(1-x)^{10}dx$

例 60. $\displaystyle\int\dfrac{x^2}{(1-x)^{100}}dx$

例 61. $\displaystyle\int x^3(1-5x^2)^{10}dx$

例 62. $\displaystyle\int x\sqrt{2-5x}dx$

\[\begin{aligned} \int x(1-x)^{10}dx=\int (-(1-x)+1)(1-x)^{10}dx \\ =-\int(1-x)dx+\int(1-x)^{10}dx \end{aligned} \]

令$t=(1-x)$，则

\[\begin{aligned} \int x(1-x)^{10}dx=\int (1-t)t^{10}dx \\ =-\int t^{10}dt+\int t^{11}dx \end{aligned} \]

$\displaystyle\int\dfrac{x^2}{(1-x)^{100}}dx$

令$t=x-1$，则

\[\int\dfrac{x^2}{(1-x)^{100}}dx=\int\dfrac{(t+1)^2}{t^{100}}dt \]

$\displaystyle\int x^3(1-5x^2)^{10}dx$

$\displaystyle\int x\sqrt{2-5x}dx$

$t=2-5x$

目录

本节读完

例 63. Thanks

63.

$\displaystyle\int\sin^px\cos^qxdx=\int\frac12\sin^{p-1}x(1-\sin^2x)^{\frac{p-1}2}2\sin x\cos xdx$

令$z=\sin^2x$，则$dz=2\sin x\cos xdx$，有

\[\int\sin^px\cos^qxdx=\frac12\int(1-z)^{\frac{p-1}2}z^{\frac{q-1}2}dz \]

利用前面的结论，可以得到如下递推关系

  • $q\neq-1$ ,
    \[=-\dfrac{\sin^{p+1}x\cos^{q+1}x}{q+1}+\dfrac{p+q+2}{q+1}\displaystyle\int\sin^px\cos^{q+2}xdx \]
  • $p\neq-1$,
    \[=\dfrac{\sin^{p+1}x\cos^{q+1}x}{p+1}+\dfrac{p+q+2}{p+1}\displaystyle\int\sin^{p+2}x\cos^{q}xdx \]
  • $p+q\neq0$,
    \[\begin{aligned} = \dfrac{\sin^{p+1}x\cos^{q-1}x}{p+q}+\dfrac{q-1}{p+q}\displaystyle\int\sin^{p}x\cos^{q-2}xdx \\ = -\dfrac{\sin^{p-1}x\cos^{q+1}x}{p+q}+\dfrac{p-1}{p+q}\displaystyle\int\sin^{p-2}x\cos^{q}xdx \end{aligned} \]
\[\int\dfrac{a_1\sin x+b_1\cos x}{a \sin x+b \cos x}dx=A x+B\ln|a\sin x+b\cos x|+c \]
\[\begin{aligned} \int\dfrac{a_1\sin x+b_1\cos x+c_1}{a \sin x+b \cos x+c}dx=A x+B\ln|a\sin x+b\cos x+c| \\ +C\int\dfrac{1}{a\sin x+b\cos x+c}dx \end{aligned} \]
\[\begin{aligned} \int\dfrac{a_1\sin^2 x+2b_1\sin x\cos x+c_1\cos^2 x}{a \sin x+b \cos x}dx=A\sin x+B\cos x \\ +C\int\dfrac1{a\sin x+b\cos x}dx \end{aligned} \]
\[\begin{aligned} \int\dfrac1{(a\sin x+b\cos x)^n}dx=\dfrac{A\sin x+B\cos x}{(a\sin x+b\cos x)^{n-1}} \\ +C\int\dfrac1{(a\sin x+b\cos x)^{n-2}}dx \end{aligned} \]
\[\begin{aligned} \int\dfrac1{(a+b\cos x)^n}dx=\dfrac{A\sin x}{(a+b\cos x)^{n-1}}+B\int\dfrac1{(a+b\cos x)^{n-1}}dx \\ +C\int\dfrac{1}{(a+b\cos x)^{n-2}}dx \end{aligned} \]

例 64. $\displaystyle\int\dfrac{\sin(2x)}{\sqrt{2-\sin^4x}}dx$

例 65. $\displaystyle\int\cos^5x\sqrt{\sin x}dx$

三角函数化

例 66. $\displaystyle\int\dfrac1{(1-x^2)^{\frac32}}dx$

例 67. $\displaystyle\int\dfrac1{(1+x^2)\sqrt{1-x^2}}dx$

$\displaystyle\int\cos^5x\sqrt{\sin x}dx$

令$t=1-5x^2$，则$-10xdx=dt$

\[\begin{aligned} \int\cos^5x\sqrt{\sin x}dx=\int x^2x(1-5x^2)^{10}\dfrac1{-10x}dt \\ =\int\dfrac{1-t}5t^{10}\dfrac1{-10}dt \end{aligned} \]

$\displaystyle\int\dfrac{\sin^2x}{\cos^6x}dx$

令$t=\tan x$，则有

\[\begin{aligned} =\int\tan^2x(1+\tan^2x)d(\tan x) \\ =\int t^2(1+t^2)dt \end{aligned} \]

$\displaystyle\int\dfrac1{(1-x^2)^{\frac32}}dx$

$x=\sin t$， $dx=\cos t dt$，

\[\begin{aligned} \int\dfrac1{(1-x^2)^{\frac32}}dx=\int\dfrac{\cos t}{\cos^3 t}dt \\ =\int\dfrac1{\cos^2t}dt=\tan t+c \\ =\dfrac{x}{\sqrt{1-x^2}}+c \end{aligned} \]

$\displaystyle\int\dfrac1{(1-x^2)^{\frac32}}dx$

\[\begin{aligned} \int\dfrac1{(1-x^2)^{\frac32}}dx=\int\dfrac{1}{x^3(\frac1{x^2}-1)^{\frac32}}dt \\ =-\dfrac12\int\dfrac{1}{(\frac1{x^2}-1)^{\frac32}}d(\dfrac1{x^2}) =-\dfrac12\int\dfrac1{(\frac1{x^2}-1)^{\frac32}}d(\dfrac1{x^2}-1) \\ =(\dfrac1{x^2}-1)^{-\frac12}+c \end{aligned} \]
\[\begin{aligned} \int\dfrac{\sin(2x)}{\sqrt{2-\sin^4x}}dx =\int\dfrac{2\sin x\cos x}{\sqrt{2-\sin^4x}}dx \\ \end{aligned} \]
\[=\int\dfrac{1}{\sqrt{2-\sin^4x}}d(\sin^2 x) \]
\[\begin{aligned} =\int\dfrac{\sqrt 2}{\sqrt 2\sqrt{1-(\frac{\sin^2x}{\sqrt 2})^2}}d(\frac{\sin^2 x}{\sqrt 2}) \\ \end{aligned} \]
\[=\int\dfrac{1}{\sqrt{1-t^2}}dt=\arcsin\dfrac{\sin^2x}{\sqrt 2}+c \]
\[\begin{aligned} I_{p,q}=&\int \sin^px\cos^qxdx =\int \frac1{q+1}\sin^{p-1}x d\left(-\cos^{q+1}x\right) \\ =&-\frac{\sin^{p-1}x\cos^{q+1}x}{q+1}+\int \frac1{q+1}\cos^{q+1}xd\left(\sin^{p-1}x\right) \\ =&-\frac{\sin^{p-1}x\cos^{q+1}x}{q+1}+\int \frac{p-1}{q+1}\cos^{q+2}x\sin^{p-2}xdx \\ =&-\frac{\sin^{p-1}x\cos^{q+1}x}{q+1}+\int \frac{p-1}{q+1}\cos^{q}x(1-\sin^2x)\sin^{p-2}xdx \\ =&-\frac{\sin^{p-1}x\cos^{q+1}x}{q+1}+\frac{p-1}{q+1}I_{p-2,q}-\frac{p-1}{q+1}I_{p,q} \\ \end{aligned} \]

移项后，有

\[\left(1+\frac{p-1}{q+1}\right)I_{p,q} =-\frac{\sin^{p-1}x\cos^{q+1}x}{q+1}+\frac{p-1}{q+1}I_{p-2,q} \]

从而，当$p+q\neq 0$时，有

\[\begin{aligned} \int \sin^px\cos^qxdx =&-\frac{\sin^{p-1}x\cos^{q+1}x}{p+q} \\ &+\frac{p-1}{p+q}\int \sin^{p-2}x\cos^qxdx \end{aligned} \]

当$p+q=0$时，有

\[\int \sin^{-q-2}x\cos^qxdx=-\frac1{q+1}\frac{\cos^{q+1}x}{\sin^{q+1}x} \]

$p+q\neq0$，且$p\neq -1$或$q\neq -1$时，有

\[\begin{aligned} \int \sin^px\cos^qxdx =& \dfrac{\sin^{p+1}x\cos^{q-1}x}{p+q} \\ &+\dfrac{q-1}{p+q}\displaystyle\int\sin^{p}x\cos^{q-2}xdx \\ \int \sin^px\cos^qxdx =& -\dfrac{\sin^{p-1}x\cos^{q+1}x}{p+q} \\ &+\dfrac{p-1}{p+q}\displaystyle\int\sin^{p-2}x\cos^{q}xdx \end{aligned} \]

2. 若$P(x)$是一个$n$次的多项式， 记$Y=ax^2+bx+c$，则有

\[\int\frac{P(x)}{\sqrt{ax^2+bx+c}}dx=Q(x)\sqrt{Y}+\lambda\int\frac1{\sqrt{Y}}dx \]

其中$Q(x)$是$n-1$次多项式，$\lambda$是数。

例 68. $\displaystyle\int\frac{x^3}{\sqrt{1+2x-x^2}}dx$

例 69. $\displaystyle\int\frac{x^3-x+1}{\sqrt{2+2x+x^2}}dx$

例 70. $\displaystyle\int\dfrac{2x^4-4x^3+24x^2-40x+20}{(x-1)(x^2-2x+2)^3}dx$

例 71. $\displaystyle\int\dfrac{x-1}{\sqrt{x^2+2x+3}}dx$

例 72. $\displaystyle\int\dfrac{1}{1+\sqrt x+\sqrt{1+x}}dx$

例 73. $\displaystyle\int\dfrac{x^2}{\sqrt{x^2+x+1}}dx$

例 74. $\displaystyle\int\dfrac{\sqrt{x^2+2x+2}}{x}dx$

\[\begin{aligned} &\int\dfrac{x-1}{\sqrt{x^2+2x+3}}dx \\ =&\int\dfrac{\frac12}{\sqrt{x^2+2x+3}}d({x^2+2x+3})-\int\dfrac2{\sqrt{x^2+2x+3}}dx \\ =&\sqrt{x^2+2x+3}-\int\dfrac2{\sqrt{(x+1)^2+2}}d(x+1) \\ =&\sqrt{x^2+2x+3}-\ln(x+1+{\sqrt{(x+1)^2+2}})+c \end{aligned} \]
\[\begin{aligned} \int\dfrac{t^2}{t^4+1}dt =&\frac{1}2\int\dfrac{t^2-1}{t^4+1}dt+\frac{1}2\int\dfrac{t^2+1}{t^4+1}dt \\ =&\frac{1}2\int\dfrac{1-\frac1{t^2}}{t^2+\frac1{t^2}}dt+ \frac{1}2\int\dfrac{1+\frac1{t^2}}{t^2+\frac1{t^2}}dt \\ =&\frac{1}2\int\dfrac{1}{(t-\frac1{t})^2+2}d(t-\dfrac1t) \\ & +\frac{1}2\int\dfrac{1}{(t+\frac1{t})^2-2}d(t+\dfrac1t) \\ \end{aligned} \]
\[\begin{aligned} &\int\dfrac{1}{1+\sqrt x+\sqrt{1+x}}dx \\ =&\int\dfrac{1+\sqrt x-\sqrt{1+x}}{(1+\sqrt x-\sqrt{1+x})(1+\sqrt x+\sqrt{1+x})}dx \\ =&\int\dfrac{1+\sqrt x-\sqrt{1+x}}{2\sqrt x}dx \\ =&\sqrt x+\frac{x}2-\frac{1}2\int\sqrt{\dfrac{1+x}x} dx \end{aligned} \]

令$t=\sqrt{\dfrac{1+x}x}$，$x=\dfrac1{t^2-1}$， $dx=-\dfrac{2t}{(t^2-1)^2}dt$

\[\begin{aligned} -\frac{1}2\int\sqrt{\dfrac{1+x}x} dx =&\int\dfrac{t^2}{(t^2-1)^2}dt \\ =&\int(\dfrac{-\frac12 t}{t^2-1})'+\dfrac{\frac12}{t^2-1}dt \end{aligned} \]

令$x=(\dfrac{u^2-1}{2u})^2$，可以把$\sqrt x, \sqrt{1+x}$同时有理化

\[\int\dfrac1{1+\dfrac{u^2-1}{2u}+\dfrac{u^2+1}{2u}}dx \]
\[\int\dfrac{x^2}{\sqrt{x^2+x+1}}dx=\int\dfrac{x^2}{\sqrt{(x+\frac12)^2+\frac34}}dx \]

$x+\frac12=t$

\[\begin{aligned} =&\int\dfrac{(t-\frac12)^2}{\sqrt{t^2+\frac34}}dt =\int\dfrac{t^2+\frac34-t-\frac12}{\sqrt{t^2+\frac34}}dt \\ =&\int\sqrt{t^2+\frac34}dt-\frac12\int\dfrac{1}{\sqrt{t^2+\frac34}}dt^2 -\frac12\int\dfrac1{\sqrt{t^2+\frac34}}dt \end{aligned} \]
\[\begin{aligned} \int\dfrac{\sqrt{x^2+2x+2}}{x}dx =&\int\dfrac{x^2+2x+2}{x\sqrt{x^2+2x+2}}dx \\ =&\int\dfrac{x+1}{\sqrt{(x+1)^2+1}}dx+\int\dfrac1{\sqrt{x^2+2x+2}}dx \\ &+\int\dfrac2{x\sqrt{x^2+2x+2}}dx \end{aligned} \]
\[\int\dfrac2{x\sqrt{x^2+2x+2}}dx =-\sqrt2\int\dfrac1{\sqrt{\frac1{x^2}+\frac1x+\frac12}}d\frac1x \]
\[\begin{aligned} &\int\dfrac{x^3}{\sqrt{1+2x-x^2}}dx \\ =&\int\dfrac{x^3}{\sqrt{2-(x-1)^2}}dx =\int\dfrac{(t+1)^3}{\sqrt{2-t^2}}dx \\ =&\int\dfrac{-t(2-t^2)}{\sqrt{2-t^2}}+\dfrac{-3(2-t^2)}{\sqrt{2-t^2}}+\dfrac{5t+7}{\sqrt{2-t^2}}dt \\ =&-\int t{\sqrt{2-t^2}}dt-3\int {\sqrt{2-t^2}}dt \\ &+5\int\dfrac{t}{\sqrt{2-t^2}}dt+7\int\dfrac1{\sqrt{2-t^2}}dt \\ \end{aligned} \]
\[\begin{aligned} &\int\dfrac{x^3}{\sqrt{1+2x-x^2}}dx \\ =&\dfrac13(2-t^2)^{\frac32} -3(\dfrac12t{\sqrt{2-t^2}}+\arcsin\dfrac{t}{\sqrt2}) \\ &-5{\sqrt{2-t^2}}+7\arcsin\dfrac{t}{\sqrt 2}+c \end{aligned} \]