分部积分法例题编练

Preface

利用分部积分法可以解决的常见积分类型

第一类

幂函数与指数函数或正余弦函数乘积的积分，分部积分后，幂函数被降次，直至没有
可以设 $u=x^{n}$:

• $\int x^{n} e^{ax} dx$
• $\int x^{n} \sin ax dx$
• $\int x^{n} \cos ax dx$

第二类

幂函数与对数函数或反三角函数乘积的积分，分部积分后，
将无法直接积分的对数函数或反三角函数转化为“可以对其进行微分计算”，
简化被积函数，求出原函数
可以设 $du=x^{n} dx$ ：

• $\int x^{n} \ln x dx$
• $\int x^{n} \arcsin x dx$
• $\int x^{n} \arctan x dx$

第三类

指数函数与正余弦函数乘积的积分，通过被积表达式的自身“复制”求出原函数
可以设 $du=e^{ax} dx$,
也可以设 $u=e^{ax}$ ：

• $\int e^{ax} \sin bx dx$
• $\int e^{ax} \cos bx dx$

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paradigm 0

$$\begin{align} \int x\cos xdx=? \\ \\ \int x(\sin x)^{\prime}dx \\ \\ 设:u=x, \quad q=\sin x \\ \\ \Rightarrow\int x\left(dx\cdot\sin x\right)= \int x d\left(\sin x\right) \\ \\ =x\sin x-\int\sin xdx \\ \\ =x\sin x-(-\cos x)+C \\ \\ =x\sin x+\cos x+C \end{align}$$

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paradigm 1

$$\begin{align} \int xe^xdx=? \\ \\ \because (e^{x})^{\prime}=e^{x}\ln e=e^{x} \\ \\ \Rightarrow\int xd(e^{x})=xe^{x}-\int e^{x}dx \\ \\ =xe^{x}-(e^{x}+C) \\ \\ =e^{x}\left(x-1\right)+C \end{align}$$

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paradigm 2

$$\begin{align} \int x^{2} \ln x dx=? \\ \\ \because\left[\left(\frac{1}{2+1}\right) x^{2+1}\right]^{\prime}=\left(\frac{1}{3} x^{3}\right)^{\prime}=x^{2} \\ \\ \Rightarrow \int\left(\frac{1}{3} x^{3}\right)^{\prime} \ln x d x \\ \\ \int \ln x d\left(\frac{1}{3} x^{3}\right) \\ \\ 设: \ln x=u, \quad \frac{1}{3} x^{3}=q \\ \\ = \ln x \cdot \frac{1}{3} x^{3}-\int \frac{1}{3} x^{3} d(\ln x) \\ \\ d (\ln x)=d x \cdot(\ln x)^{\prime}=\frac{1}{x} d x \\ \\ \Rightarrow \int \frac{1}{3} x^{3} \cdot \frac{1}{x} d x=\frac{1}{3} \int x^{2} d x \\ \\ = \frac{1}{3}\left(\frac{1}{3} x^{3}\right)+C=\frac{1}{9} x^{3}+C \\ \\ \Rightarrow \frac{x^{3} \ln x}{3}-\frac{x^{3}}{9}+C \end{align}$$

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Paradigm 3

$$\begin{align} \int \arctan x d x=? \\ \\ \int(x)^{\prime} \arctan x d x=x \arctan x-\int xd(\arctan x ) \\ \\ \int x d(\arctan x) \Rightarrow \int x \cdot dx \cdot \frac{1}{1+x^{2}} \\ \\ \int \frac{x}{1+x^{2}} d x, \quad \text { 设: } u=1+x^{2} \\ \\ =\frac{1}{2} \int \frac{1}{u} du=\frac{1}{2} \ln u+C \\ \\ \Rightarrow x \arctan x-\frac{1}{2} \ln \left(1+x^{2}\right)+C \end{align}$$

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Paradigm 4

易错点: $(\ln\sqrt{x})^{\prime}=(\ln u)^{\prime}\cdot u^{\prime}=\frac{1}{2x}$

不是: $\frac{1}{\sqrt[]{x}}$

$$\begin{align} \int \arccos xdx=? \\ \\ \Rightarrow\int(x)^{\prime}\arcsin xdx \\ \\ \Rightarrow x\arccos x-\int xd(\cos ar(x) \\ \\ \int xd(\cos arcx)\Rightarrow\int x\cdot dx\cdot(\arccos x) \\ \\ =\int\frac{x}{\sqrt{1-x^{2}}}dx \\ \\ 设:u=1-x^{2} \\ \\ \int\frac{x}{\sqrt{u}}d(1-x^{2}) \\ \\ d(1-x^{2})=(1-x^{2})^{\prime}dx=-2xdx=du \\ \\ \int\frac{x}{\sqrt{u}}\frac{du}{-2x}=-\int\frac{1}{\sqrt{u}}\cdot-\frac{1}{2}\cdot du =\frac{1}{2}\int\frac{1}{\sqrt{u}}du \\ \\ \frac{1}{2}\int\frac{1}{\sqrt{u}}du\Rightarrow\frac{1}{2}\int u^{-\frac{1}{2}}du \\ \\ =\frac{1}{2}(2u^{\frac{1}{2}})+C \\ \\ =\sqrt{1-x^{2}}+C \\ \\ 最终:\quad x\arccos x-\sqrt{1-x^{2}}+C \end{align}$$

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Paradigm 5

• 前置: $(\tan x)'=\sec^{2} x$

• 前置: $(\sec x)'=\sec x\tan x$

• 前置: $1+\tan^{2} x=\sec^{2} x$

• 前置: 援引题:Sample 0

$$\begin{align} \int\sec^{3}xdx=? \\ \\ \int\sec x\cdot\sec^{2}xdx=\int\sec xd(\tan x) \\ \\ =\sec x\tan x-\int\tan xd(\sec x) \\ \\ \int\tan xd(\sec x)=\int\tan xdx\cdot\sec x\tan x \\ \\ =\int\tan^{2}x\sec xdx \\ \\ =\int(\sec^{2}x-1)\sec xdx \\ \\ =\int\left(\sec^{3}x-\sec x\right)dx \\ \\ =\int\sec^{3}xdx-\int\sec xdx \\ \\ \int\sec^{3}xdx=\sec x\tan x-\int\sec^{3}xdx +\int\sec xdx \\ \\ 2\int\sec^{3}xdx=\sec x\tan x+\int\sec xdx \\ \\ \int\sec^{3}xdx =\frac{1}{2}\left(\sec x\tan x+\ln\left|\sec x+\tan x\right|\right)+C \end{align}$$

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paradigm 6

$$\begin{align} \int e^{x}\sin xdx=? \\ \\ \begin{aligned}\int e^{x}\left(-\cos x\right)^{\prime}dx \\ \\ \Rightarrow-\cos xe^{x}+\int\cos xd\left(e^{x}\right)\end{aligned} \\ \\ \int\cos xd\left(e^{x}\right)=\int\cos xdxex=\int\left(\sin x\right)'e^{x}dx \\ \\ =\int(\sin x)^{\prime}e^{x}dx=e^{x}\sin x-\int\sin xd\left(e^{x}\right) \\ \\ =e^{x}\sin x-\int\sin xe^{x}dx \\ \\ \int e^{x}\sin xdx=-\cos xe^{x}+e^{x}\sin x-\int\sin xe^{x}dx \\ \\ 2\int e^{x}\sin xdx=e^{x}\sin x-\cos xe^{x} \\ \\ \int e^{x}\sin xdx=\frac{1}{2}(e^{x}\sin x-\cos xe^{x})+C \\ \\ =\frac{1}{2}e^{x}\left(\sin x-\cos x\right)+C \end{align}$$

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Paradigm 7

$$\begin{align} \int x^{2} e^{x} d x=? \\ \\ \Rightarrow \int x^{2} d x\left(e^{x}\right)^{\prime} = \int x^{2} d\left(e^{x}\right) \\ \\ =x^{2} e^{x}-\int e^{x} d\left(x^{2}\right) \\ \\ \int e^{x} d\left(x^{2}\right)=\int e^{x} \cdot d x \cdot\left(x^{2}\right)^{\prime} \\ \\ \Rightarrow \int e^{x} 2 x d x=2 \int x e^{x} d x \\ \\ \int x e^{x} d x=\int x d\left(e^{x}\right) \\ \\ \Rightarrow x e^{x}-\int e^{x} d(x)=x e^{x}-e^{x}+C \\ \\ \Rightarrow x^{2} e^{x}-2\left(x e^{x}-e^{x}\right)+C \\ \\ x^{2} e^{x}-2 x e^{x}+2 e^{x}+C \\ \\ e^{x}\left(x^{2}-2 x+2\right)+C \end{align}$$

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