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} $ ：
- \(\int e^{ax} \sin bx dx\)
- \(\int e^{ax} \cos bx dx\)
paradigm 0
\[\begin{align}
\int x\cos xdx=?
\\ \\
\int x(\sin x)^{\prime}dx
\\ \\
设:u=x, \quad q=\sin x
\\ \\
\Rightarrow\int x\left(dx\cdot\sin x\right)=
\int x d\left(\sin x\right)
\\ \\
=x\sin x-\int\sin xdx
\\ \\
=x\sin x-(-\cos x)+C
\\ \\
=x\sin x+\cos x+C
\end{align}
\]
paradigm 1
\[\begin{align}
\int xe^xdx=?
\\ \\
\because (e^{x})^{\prime}=e^{x}\ln e=e^{x}
\\ \\
\Rightarrow\int xd(e^{x})=xe^{x}-\int e^{x}dx
\\ \\
=xe^{x}-(e^{x}+C)
\\ \\
=e^{x}\left(x-1\right)+C
\end{align}
\]
paradigm 2
\[\begin{align}
\int x^{2} \ln x dx=?
\\ \\
\because\left[\left(\frac{1}{2+1}\right) x^{2+1}\right]^{\prime}=\left(\frac{1}{3} x^{3}\right)^{\prime}=x^{2}
\\ \\
\Rightarrow \int\left(\frac{1}{3} x^{3}\right)^{\prime} \ln x d x
\\ \\
\int \ln x d\left(\frac{1}{3} x^{3}\right)
\\ \\
设: \ln x=u, \quad \frac{1}{3} x^{3}=q
\\ \\
= \ln x \cdot \frac{1}{3} x^{3}-\int \frac{1}{3} x^{3} d(\ln x)
\\ \\
d (\ln x)=d x \cdot(\ln x)^{\prime}=\frac{1}{x} d x
\\ \\
\Rightarrow \int \frac{1}{3} x^{3} \cdot \frac{1}{x} d x=\frac{1}{3} \int x^{2} d x
\\ \\
= \frac{1}{3}\left(\frac{1}{3} x^{3}\right)+C=\frac{1}{9} x^{3}+C
\\ \\
\Rightarrow \frac{x^{3} \ln x}{3}-\frac{x^{3}}{9}+C
\end{align}
\]
Paradigm 3
\[\begin{align}
\int \arctan x d x=?
\\ \\
\int(x)^{\prime} \arctan x d x=x \arctan x-\int xd(\arctan x )
\\ \\
\int x d(\arctan x) \Rightarrow \int x \cdot dx \cdot \frac{1}{1+x^{2}}
\\ \\
\int \frac{x}{1+x^{2}} d x, \quad \text { 设: } u=1+x^{2}
\\ \\
=\frac{1}{2} \int \frac{1}{u} du=\frac{1}{2} \ln u+C
\\ \\
\Rightarrow x \arctan x-\frac{1}{2} \ln \left(1+x^{2}\right)+C
\end{align}
\]
Paradigm 4
易错点: \((\ln\sqrt{x})^{\prime}=(\ln u)^{\prime}\cdot u^{\prime}=\frac{1}{2x}\)
不是: \(\frac{1}{\sqrt[]{x}}\)
\[\begin{align}
\int \arccos xdx=?
\\ \\
\Rightarrow\int(x)^{\prime}\arcsin xdx
\\ \\
\Rightarrow x\arccos x-\int xd(\cos ar(x)
\\ \\
\int xd(\cos arcx)\Rightarrow\int x\cdot dx\cdot(\arccos x)
\\ \\
=\int\frac{x}{\sqrt{1-x^{2}}}dx
\\ \\
设:u=1-x^{2}
\\ \\
\int\frac{x}{\sqrt{u}}d(1-x^{2})
\\ \\
d(1-x^{2})=(1-x^{2})^{\prime}dx=-2xdx=du
\\ \\
\int\frac{x}{\sqrt{u}}\frac{du}{-2x}=-\int\frac{1}{\sqrt{u}}\cdot-\frac{1}{2}\cdot du
=\frac{1}{2}\int\frac{1}{\sqrt{u}}du
\\ \\
\frac{1}{2}\int\frac{1}{\sqrt{u}}du\Rightarrow\frac{1}{2}\int u^{-\frac{1}{2}}du
\\ \\
=\frac{1}{2}(2u^{\frac{1}{2}})+C
\\ \\
=\sqrt{1-x^{2}}+C
\\ \\
最终:\quad x\arccos x-\sqrt{1-x^{2}}+C
\end{align}
\]
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
\[
\begin{align}
\int\sec^{3}xdx=?
\\ \\
\int\sec x\cdot\sec^{2}xdx=\int\sec xd(\tan x)
\\ \\
=\sec x\tan x-\int\tan xd(\sec x)
\\ \\
\int\tan xd(\sec x)=\int\tan xdx\cdot\sec x\tan x
\\ \\
=\int\tan^{2}x\sec xdx
\\ \\
=\int(\sec^{2}x-1)\sec xdx
\\ \\
=\int\left(\sec^{3}x-\sec x\right)dx
\\ \\
=\int\sec^{3}xdx-\int\sec xdx
\\ \\
\int\sec^{3}xdx=\sec x\tan x-\int\sec^{3}xdx
+\int\sec xdx
\\ \\
2\int\sec^{3}xdx=\sec x\tan x+\int\sec xdx
\\ \\
\int\sec^{3}xdx
=\frac{1}{2}\left(\sec x\tan x+\ln\left|\sec x+\tan x\right|\right)+C
\end{align}
\]
paradigm 6
\[\begin{align}
\int e^{x}\sin xdx=?
\\ \\
\begin{aligned}\int e^{x}\left(-\cos x\right)^{\prime}dx
\\ \\
\Rightarrow-\cos xe^{x}+\int\cos xd\left(e^{x}\right)\end{aligned}
\\ \\
\int\cos xd\left(e^{x}\right)=\int\cos xdxex=\int\left(\sin x\right)'e^{x}dx
\\ \\
=\int(\sin x)^{\prime}e^{x}dx=e^{x}\sin x-\int\sin xd\left(e^{x}\right)
\\ \\
=e^{x}\sin x-\int\sin xe^{x}dx
\\ \\
\int e^{x}\sin xdx=-\cos xe^{x}+e^{x}\sin x-\int\sin xe^{x}dx
\\ \\
2\int e^{x}\sin xdx=e^{x}\sin x-\cos xe^{x}
\\ \\
\int e^{x}\sin xdx=\frac{1}{2}(e^{x}\sin x-\cos xe^{x})+C
\\ \\
=\frac{1}{2}e^{x}\left(\sin x-\cos x\right)+C
\end{align}
\]
Paradigm 7
\[\begin{align}
\int x^{2} e^{x} d x=?
\\ \\
\Rightarrow \int x^{2} d x\left(e^{x}\right)^{\prime} = \int x^{2} d\left(e^{x}\right)
\\ \\
=x^{2} e^{x}-\int e^{x} d\left(x^{2}\right)
\\ \\
\int e^{x} d\left(x^{2}\right)=\int e^{x} \cdot d x \cdot\left(x^{2}\right)^{\prime}
\\ \\
\Rightarrow \int e^{x} 2 x d x=2 \int x e^{x} d x
\\ \\
\int x e^{x} d x=\int x d\left(e^{x}\right)
\\ \\
\Rightarrow x e^{x}-\int e^{x} d(x)=x e^{x}-e^{x}+C
\\ \\
\Rightarrow x^{2} e^{x}-2\left(x e^{x}-e^{x}\right)+C
\\ \\
x^{2} e^{x}-2 x e^{x}+2 e^{x}+C
\\ \\
e^{x}\left(x^{2}-2 x+2\right)+C
\end{align}
\]
