高等数学随记 - 利用双元法求不定积分

前言

双元法是近些年来网络上流传出的一种求解不定积分的新方式,其既有准确得值、计算简便的快捷性,又在传统的第一类、第二类换元积分法的基础上有所创新,给我们提供了另一个视角看待不定积分求解的思维历程. 尤其用在考研时该方法的正确使用亦可在作答试卷时起到事半功倍的效果,特此深入研究了一下该方法,以个人见解对其总结汇集于此.

例题引入

例1. 求\(\int\sqrt{x^2+a^2}\mathrm{d}x\).

解1. [第二类换元积分法]

\[\int\sqrt{x^2+a^2}\mathrm{d}x = a\int\sqrt{(\frac{x}{a})^2+1}\mathrm{d}x = a^2\int\sqrt{(\frac{x}{a})^2+1}\mathrm{d}(\frac{x}{a}). \]

利用三角换元,\(\mathrm{tan}\theta = \frac{x}{a}\),则\(\mathrm{sec}\theta = \sqrt{\mathrm{tan}^2\theta+1} = \frac{\sqrt{x^2+a^2}}{a}\).

\[\int\sqrt{x^2+a^2}\mathrm{d}x = a^2\int\mathrm{sec}\theta\mathrm{d}\mathrm{tan}\theta \]

\[= a^2\mathrm{sec}\theta\mathrm{tan}\theta-a^2\int\mathrm{tan}\theta\mathrm{d}\mathrm{sec}\theta \]

\[= a^2\mathrm{sec}\theta\mathrm{tan}\theta-a^2\int\mathrm{tan}^2\theta\mathrm{sec}\theta\mathrm{d}\theta \]

\[= a^2\mathrm{sec}\theta\mathrm{tan}\theta-a^2\int(\mathrm{sec}^2\theta-1)\mathrm{sec}\theta\mathrm{d}\theta \]

\[= a^2\mathrm{sec}\theta\mathrm{tan}\theta-a^2\int\mathrm{sec}^3\theta\mathrm{d}\theta+a^2\int\mathrm{sec}\theta\mathrm{d}\theta. \]

注意到

\[\int\sqrt{x^2+a^2}\mathrm{d}x = a^2\int\mathrm{sec}\theta\mathrm{d}\mathrm{tan}\theta \]

\[= a^2\int\mathrm{sec}^3\theta\mathrm{d}\theta, \]

故移项得

\[2a^2\int\mathrm{sec}\theta\mathrm{d}\mathrm{tan}\theta = a^2\mathrm{sec}\theta\mathrm{tan}\theta+a^2\int\mathrm{sec}\theta\mathrm{d}\theta, \]

从而

\[\int\sqrt{x^2+a^2}\mathrm{d}x = a^2\int\mathrm{sec}\theta\mathrm{d}\mathrm{tan}\theta = \frac{a^2}{2}\mathrm{sec}\theta\mathrm{tan}\theta+\frac{a^2}{2}\int\mathrm{sec}\theta\mathrm{d}\theta \]

\[= \frac{a^2}{2}\mathrm{sec}\theta\mathrm{tan}\theta+\frac{a^2}{2}\mathrm{ln}\left|\mathrm{tan}\theta+\mathrm{sec}\theta\right|+C \]

\[= \frac{a^2}{2}\cdot\frac{x}{a}\cdot\frac{\sqrt{x^2+a^2}}{a}+\frac{a^2}{2}\mathrm{ln}\left|\frac{x}{a}+\frac{\sqrt{x^2+a^2}}{a}\right|+C \]

\[= \frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\mathrm{ln}(x+\sqrt{x^2+a^2})+C.\]

(注意\(-\frac{a^2}{2}\mathrm{ln}\left|a\right|\)已并入常数项\(C\)中;由于\(\sqrt{x^2+a^2}+x\)恒为正,故去掉绝对值符号. )

解2. [双元虚圆换元法] 令\(y = \sqrt{x^2+a^2}\),易得\(y^2-x^2 = a^2\),从而\(y\mathrm{d}y=x\mathrm{d}x\).

\[\int\sqrt{x^2+a^2}\mathrm{d}x = \int y\mathrm{d}x \]

\[= xy-\int x\mathrm{d}y \]

\[= xy-\int\frac{x^2\mathrm{d}x}{y} \]

\[= xy-\int\frac{(y^2-a^2)\mathrm{d}x}{y} \]

\[= xy-\int y\mathrm{d}x+a^2\int\frac{\mathrm{d}x}{y}, \]

移项得

\[2\int y\mathrm{d}x = xy+a^2\int\frac{\mathrm{d}x}{y}, \]

\[\int y\mathrm{d}x = \frac{xy}{2}+\frac{a^2}{2}\int\frac{\mathrm{d}x}{y}. \]

\(y\mathrm{d}y=x\mathrm{d}x\),故\(\frac{\mathrm{d}x}{y}=\frac{\mathrm{d}y}{x}\),由合比定理有\(\frac{\mathrm{d}x}{y}=\frac{\mathrm{d}y}{x}=\frac{\mathrm{d}x+\mathrm{d}y}{x+y}=\frac{\mathrm{d}(x+y)}{x+y}=\mathrm{d}\mathrm{ln}(x+y)\)(此处绝对值符号可去,原因同解1).

\[\int\sqrt{x^2+a^2}\mathrm{d}x = \frac{xy}{2}+\frac{a^2}{2}\int\frac{\mathrm{d}x}{y} \]

\[= \frac{xy}{2}+\frac{a^2}{2}\mathrm{ln}(x+y)+C. \]

\(y = \sqrt{x^2+a^2}\)代入得

\[\int\sqrt{x^2+a^2}\mathrm{d}x = \frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\mathrm{ln}(x+\sqrt{x^2+a^2})+C. \]

方法总结:不定积分的双元换元法

一般地,我们将解不定积分的双元换元法分为两种类型(其中\(C\)为常实数):

  • 类型1. 实圆双元:\(x^2+y^2=C \Leftrightarrow x\mathrm{d}x=-y\mathrm{d}y\).
  • 类型2. 虚圆双元:\(x^2-y^2=C \Leftrightarrow x\mathrm{d}x=y\mathrm{d}y\).

对于以上两类型双元换元,我们有以下公式(由于常数\(C\)已在上两式中使用,故接下来积分常数用\(C_0\)表示):

1. 双元第一公式

  • (1) 对于实圆双元,\(\int\frac{\mathrm{d}x}{y} = \mathrm{arctan}\frac{y}{x}+C_0\)
  • (2) 对于虚圆双元,\(\int\frac{\mathrm{d}x}{y} = \mathrm{ln}\left|x+y\right|+C_0\).

证明. (1) 由于\(y\mathrm{d}y=-x\mathrm{d}x\),故

\[y\mathrm{d}x-x\mathrm{d}y=y\mathrm{d}x-x\cdot\frac{x\mathrm{d}x}{y}=\frac{(y^2+x^2)\mathrm{d}x}{y} \]

\[\mathrm{d}x = \frac{y(y\mathrm{d}x-x\mathrm{d}y)}{x^2+y^2} \]

\[\int\frac{\mathrm{d}x}{y} = \int\frac{1}{y}\cdot\frac{y(y\mathrm{d}x-x\mathrm{d}y)}{x^2+y^2} \]

\[= \int\frac{y\mathrm{d}x-x\mathrm{d}y}{x^2+y^2}, \]

又由于

\[\mathrm{d}(\frac{x}{y}) = \frac{y\mathrm{d}x-x\mathrm{d}y}{y^2}, \]

\[\int\frac{\mathrm{d}x}{y} = \int\frac{y^2}{y^2+x^2}\cdot\mathrm{d}(\frac{x}{y}) \]

\[= \int\frac{1}{(\frac{x}{y})^2+1}\cdot\mathrm{d}(\frac{x}{y}) \]

\[= \mathrm{arctan}\frac{x}{y}+C_0. \]

(2) 因\(y\mathrm{d}y=x\mathrm{d}x\),故\(\frac{\mathrm{d}x}{y}=\frac{\mathrm{d}y}{x}\),由合比定理有\(\frac{\mathrm{d}x}{y}=\frac{\mathrm{d}y}{x}=\frac{\mathrm{d}x+\mathrm{d}y}{x+y}=\frac{\mathrm{d}(x+y)}{x+y}=\mathrm{d}\mathrm{ln}\left|x+y\right|\).

\[\int\frac{\mathrm{d}x}{y} = \int\mathrm{d}\mathrm{ln}\left|x+y\right| \]

\[= \mathrm{ln}\left|x+y\right|+C_0. \]

2. 双元第三公式

  • (1) 对于实圆双元,\(\int\frac{\mathrm{d}x}{y^3} = \frac{1}{y^2+x^2}\cdot\frac{x}{y}+C_0 = \frac{x}{Cy}+C_0\)
  • (2) 对于虚圆双元,\(\int\frac{\mathrm{d}x}{y^3} = \frac{1}{y^2-x^2}\cdot\frac{x}{y}+C_0 = -\frac{x}{Cy}+C_0\).

证明. (1) 类似双元第一公式的证法,

\[\int\frac{\mathrm{d}x}{y^3} = \int\frac{1}{y^2}\cdot\frac{\mathrm{d}x}{y} \]

\[= \int\frac{1}{y^2}\cdot\frac{y\mathrm{d}x-x\mathrm{d}y}{x^2+y^2} \]

\[= \int\frac{1}{x^2+y^2}\cdot\frac{y\mathrm{d}x-x\mathrm{d}y}{y^2} \]

\[= \int\frac{1}{x^2+y^2}\cdot\mathrm{d}(\frac{x}{y}) \]

\[= \frac{1}{x^2+y^2}\cdot\frac{x}{y}+C_0 \]

\[= \frac{x}{Cy}+C_0. \]

(2) 由于

\[x\mathrm{d}x=y\mathrm{d}y, \]

\[\mathrm{d}y=\frac{x\mathrm{d}x}{y}, \]

\[y\mathrm{d}x-x\mathrm{d}y = y\mathrm{d}x-x\cdot\frac{x\mathrm{d}x}{y} \]

\[= \frac{(y^2-x^2)\mathrm{d}x}{y}, \]

\[\mathrm{d}x = \frac{y(y\mathrm{d}x-x\mathrm{d}y)}{y^2-x^2}, \]

\[\int\frac{\mathrm{d}x}{y^3} = \int\frac{1}{y^3}\cdot\frac{y(y\mathrm{d}x-x\mathrm{d}y)}{y^2-x^2} \]

\[= \int\frac{y\mathrm{d}x-x\mathrm{d}y}{y^2(y^2-x^2)} \]

\[= \int\frac{1}{y^2-x^2}\mathrm{d}(\frac{x}{y}) \]

\[= \frac{1}{y^2-x^2}\cdot\frac{x}{y}+C_0 \]

\[= -\frac{x}{Cy}+C_0. \]

3. 双元第二公式

注:此公式的形式已在例1中解2的过程中给出,此处总结一般情形.

  • (1) 对于实圆双元,\(\int y\mathrm{d}x = \frac{xy}{2}+\frac{x^2+y^2}{2}\int\frac{\mathrm{d}x}{y} = \frac{xy}{2}+\frac{C}{2}\int\frac{\mathrm{d}x}{y}\)
  • (2) 对于虚圆双元,\(\int y\mathrm{d}x = \frac{xy}{2}-\frac{x^2-y^2}{2}\int\frac{\mathrm{d}x}{y} = \frac{xy}{2}-\frac{C}{2}\int\frac{\mathrm{d}x}{y}\).

这里每一个结论均有两种证法,其中分部积分法即为例1中解2的过程所用形式.

证明. (1) [证法一 - 分部积分法] 由于

\[\int y\mathrm{d}x = xy-\int x\mathrm{d}y \]

\[= xy-\int x\cdot(-\frac{x\mathrm{d}x}{y}) \]

\[= xy+\int \frac{x^2\mathrm{d}x}{y} \]

\[= xy+\int \frac{C-y^2}{y}\mathrm{d}x \]

\[= xy+C\int \frac{\mathrm{d}x}{y}-\int y\mathrm{d}x, \]

故移项得

\[2\int y\mathrm{d}x = xy+C\int \frac{\mathrm{d}x}{y}, \]

\[\int y\mathrm{d}x = \frac{xy}{2}+\frac{C}{2}\int \frac{\mathrm{d}x}{y} \]

\[= \frac{xy}{2}+\frac{x^2+y^2}{2}\int \frac{\mathrm{d}x}{y}. \]

[证法二 - 二分裂项法]

\[\int y\mathrm{d}x = \frac{1}{2}\int y\mathrm{d}x-\frac{1}{2}\int(-y\mathrm{d}x) \]

\[= \frac{1}{2}\int(x\mathrm{d}y+y\mathrm{d}x)-\frac{1}{2}\int(x\mathrm{d}y-y\mathrm{d}x) \]

\[= \frac{1}{2}\int\mathrm{d}(xy)+\frac{1}{2}\int(y\mathrm{d}x-x\mathrm{d}y) \]

\[= \frac{1}{2}xy+\frac{1}{2}\int(y\mathrm{d}x-x(-\frac{x\mathrm{d}y}{y})) \]

\[= \frac{1}{2}xy+\frac{1}{2}\int\frac{x^2+y^2}{y}\mathrm{d}x \]

\[= \frac{xy}{2}+\frac{x^2+y^2}{2}\int\frac{\mathrm{d}x}{y} \]

\[= \frac{xy}{2}+\frac{C}{2}\int\frac{\mathrm{d}x}{y}. \]

(2) [证法一 - 分部积分法] 由于

\[\int y\mathrm{d}x = xy-\int x\mathrm{d}y \]

\[= xy-\int x\cdot\frac{x\mathrm{d}x}{y} \]

\[= xy-\int \frac{x^2\mathrm{d}x}{y} \]

\[= xy-\int \frac{C+y^2}{y}\mathrm{d}x \]

\[= xy-C\int \frac{\mathrm{d}x}{y}-\int y\mathrm{d}x, \]

故移项得

\[2\int y\mathrm{d}x = xy-C\int \frac{\mathrm{d}x}{y}, \]

\[\int y\mathrm{d}x = \frac{xy}{2}-\frac{C}{2}\int \frac{\mathrm{d}x}{y} \]

\[= \frac{xy}{2}-\frac{x^2-y^2}{2}\int \frac{\mathrm{d}x}{y}. \]

[证法二 - 二分裂项法]

\[\int y\mathrm{d}x = \frac{1}{2}\int y\mathrm{d}x-\frac{1}{2}\int(-y\mathrm{d}x) \]

\[= \frac{1}{2}\int(x\mathrm{d}y+y\mathrm{d}x)-\frac{1}{2}\int(x\mathrm{d}y-y\mathrm{d}x) \]

\[= \frac{1}{2}\int\mathrm{d}(xy)+\frac{1}{2}\int(y\mathrm{d}x-x\mathrm{d}y) \]

\[= \frac{1}{2}xy+\frac{1}{2}\int(y\mathrm{d}x-x\cdot\frac{x\mathrm{d}y}{y}) \]

\[= \frac{1}{2}xy+\frac{1}{2}\int\frac{y^2-x^2}{y}\mathrm{d}x \]

\[= \frac{xy}{2}+\frac{y^2-x^2}{2}\int\frac{\mathrm{d}x}{y} \]

\[= \frac{xy}{2}-\frac{x^2-y^2}{2}\int\frac{\mathrm{d}x}{y} \]

\[= \frac{xy}{2}+\frac{C}{2}\int\frac{\mathrm{d}x}{y}. \]

4. 双元点火公式

注:本公式的用途类似于 Wallis 公式,用于降低被积函数的次数以最终化为双元第一公式或双元第三公式的形式并得出待求式的结果.

对于实圆双元与虚圆双元,均有\((1+n)\int x^n\mathrm{d}y = x^ny+Cn\int x^{n-2}\mathrm{d}y\).

证明. 先证实圆双元.

\[\int x^n\mathrm{d}y = x^ny-\int y\mathrm{d}(x^n) \]

\[= x^ny-n\int yx^{n-1}\mathrm{d}x \]

\[= x^ny-n\int yx^{n-1}\cdot(-\frac{y\mathrm{d}y}{x}) \]

\[= x^ny+n\int y^2x^{n-2}\mathrm{d}y \]

\[= x^ny+n\int (C-x^2)x^{n-2}\mathrm{d}y \]

\[= x^ny+Cn\int x^{n-2}\mathrm{d}y-n\int x^n\mathrm{d}y, \]

移项即得

\[(1+n)\int x^n\mathrm{d}y = x^ny+Cn\int x^{n-2}\mathrm{d}y. \]

再证虚圆双元.

\[\int x^n\mathrm{d}y = x^ny-\int y\mathrm{d}(x^n) \]

\[= x^ny-n\int yx^{n-1}\mathrm{d}x \]

\[= x^ny-n\int yx^{n-1}\cdot\frac{y\mathrm{d}y}{x} \]

\[= x^ny-n\int y^2x^{n-2}\mathrm{d}y \]

\[= x^ny-n\int (x^2-C)x^{n-2}\mathrm{d}y \]

\[= x^ny+Cn\int x^{n-2}\mathrm{d}y-n\int x^n\mathrm{d}y, \]

移项即得

\[(1+n)\int x^n\mathrm{d}y = x^ny+Cn\int x^{n-2}\mathrm{d}y. \]

方法运用:运用双元换元法求解不定积分

例2. 求 \(\int\sqrt{\frac{x-a}{x-b}}\mathrm{d}x\),其中\(a,b\in\mathbb{R}\)为常数.

解. 令\(u = \sqrt{x-a}\)\(v = \sqrt{x-b}\),易得\(u^2-v^2=b-a=C\)为常数(虚元双元),

\[\int\sqrt{\frac{x-a}{x-b}}\mathrm{d}x = \int\frac{u}{v}\mathrm{d}(u^2+a) \]

\[= \int\frac{2u^2\mathrm{d}u}{v} \]

\[= 2\int\frac{C+v^2}{v}\mathrm{d}u \]

\[= 2C\int\frac{\mathrm{d}u}{v}+2\int v\mathrm{d}u, \]

依次代入虚元双元的双元第二、第一公式可得

\[\int\sqrt{\frac{x-a}{x-b}}\mathrm{d}x = 2C\int\frac{\mathrm{d}u}{v}+2(\frac{xy}{2}-\frac{C}{2}\int\frac{\mathrm{d}u}{v}) \]

\[= C\int \frac{\mathrm{d}u}{v}+uv \]

\[= C\mathrm{ln}\left|u+v\right|+uv+C_0 \]

\[= (b-a)\mathrm{ln}\left|\sqrt{x-a}+\sqrt{x-b}\right|+\sqrt{(x-a)(x-b)}+C_0. \]

例3. 求 \(\int\sqrt{\frac{x-a}{b-x}}\mathrm{d}x\),其中\(a,b\in\mathbb{R}\)为常数.

解. 令\(u = \sqrt{x-a}\)\(v = \sqrt{b-x}\),易得\(u^2+v^2=b-a=C\)为常数(实元双元),

\[\int\sqrt{\frac{x-a}{b-x}}\mathrm{d}x = \int\frac{u}{v}\mathrm{d}(u^2+a) \]

\[= \int\frac{2u^2\mathrm{d}u}{v} \]

\[= -2\int u(-\frac{u\mathrm{d}u}{v}) \]

\[= -2\int u\mathrm{d}v \]

\[= -2(uv-\int v\mathrm{d}u) \]

\[= 2\int v\mathrm{d}u-2uv, \]

依次代入实元双元的双元第二、第一公式可得

\[\int\sqrt{\frac{x-a}{b-x}}\mathrm{d}x = 2(\frac{uv}{2}+\frac{C}{2}\int\frac{\mathrm{d}u}{v})-2uv \]

\[= C\int\frac{\mathrm{d}u}{v}-uv \]

\[= C\arctan{\frac{u}{v}}-uv+C_0 \]

\[= (b-a)\arctan{\sqrt{\frac{x-a}{b-x}}}-\sqrt{(x-a)(b-x)}+C_0. \]

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