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。在被积表达式中，正好看到了某个微分式。

    如，注意到$\cos xdx=d(\sin x)$
    \[\begin{aligned} \int \cos x\sin x dx =&\int \sin x d(\sin x) \\ =&\int t dt=\frac{1}2t^2+c=\frac{1}2\sin^2x+c \end{aligned} \]

  • 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. $\displaystyle\int \dfrac{1}{\sqrt{4+(ax+b)^2}}dx$

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

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

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

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

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

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

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

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

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

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

递推

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

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

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

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

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

有

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

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

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

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

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

目录

本节读完

例 15. Thanks

15.