[高等数学]换元积分法

一、知识点

(一) 第一类换元法

定理1

设 $f(u)$ 具有原函数，$u=\phi (x)$ 可导，则有换元公式:

$$\int f[\phi (x)]{\phi }^{\prime }(x)dx=[\int f(u)du{]}_{u=\phi (x)}.$$

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(二) 第二类换元法

定理2

设 $x=\phi (x)$ 是单调的、可导的函数，并且 ${\phi }^{\prime }(t)\ne 0$. 又设 $f[\phi (t)]{\phi }^{\prime }(t)$ 具有原函数，则有换元公式 $$\int f(x)dx=[\int f[\phi (t)]{\phi }^{\prime }(t)dt{]}_{t={\phi }^{-1}(x)}，$$ 其中，${\phi }^{-1}(x)$ 是 $x=\phi (t)$ 的反函数.

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二、练习题（求不定积分）

$\begin{array}{rl}(1)& \int e^{5t}dt\\ & =\frac{1}{5}\int e^{5t}d(5t)\\ & =\frac{1}{5}{e}^{5t}+C\end{array}$

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$\begin{array}{rl}(2)& \int (3-2x{)}^{3}dx\\ & =-\frac{1}{2}\int (3-2x{)}^{3}d(3-2x)\\ & =-\frac{1}{8}(3-2x{)}^4+C\end{array}$

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$\begin{array}{rl}(3)& \int \frac{dx}{1-2x}\\ & =-\frac{1}{2}\int \frac{d(1-2x)}{1-2x}\\ & =-\frac{1}{2}ln|1-2x|+C\end{array}$

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$\begin{array}{rl}(4)& \int \frac{dx}{\sqrt[3]{2-3x}}\\ & =-\frac{1}{3}\int \frac{d(2-3x)}{(2-3x{)}^{\frac{1}{3}}}\\ & =-\frac{1}{2}(2-3x{)}^{\frac{2}{3}}+C\end{array}$

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$\begin{array}{rl}(5)& \int (sinax-e^{\frac{x}{b}})dx\\ & =\frac{1}{a}\int sinaxd(ax)-b\int e^{\frac{x}{b}}d(\frac{x}{b})\\ & =-\frac{1}{a}cosax-b{e}^{\frac{x}{b}}+C\end{array}$

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$\begin{array}{rl}(6)& \int \frac{sin\sqrt{t}}{\sqrt{t}}dt\\ & =2\int sin\sqrt{t}d\sqrt{t}\\ & =-2cos\sqrt{t}+C\end{array}$

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$\begin{array}{rl}(7)& \int x{e}^{-x^2}dx\\ & =-\frac{1}{2}\int e^{-x^2}d(-x^2)\\ & =-\frac{1}{2}{e}^{-x^2}+C\end{array}$

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$\begin{array}{rl}(8)& \int xcos{x}^{2}dx\\ & =\frac{1}{2}\int cos{x}^{2}d(x^2)\\ & =\frac{1}{2}sin{x}^2+C\end{array}$

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$\begin{array}{rl}(9)& \int \frac{x}{\sqrt{2-3x^2}}dx\\ & =-\frac{1}{6}\int (2-3x^2{)}^{-\frac{1}{2}}d(2-3x^2)\\ & =-\frac{1}{3}\sqrt{2-3x^2}+C\end{array}$

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$\begin{array}{rl}(10)& \int \frac{3x^3}{1-x^4}dx\\ & =-\frac{3}{4}\int \frac{d(1-x^4)}{1-x^4}\\ & =-\frac{3}{4}ln|1-x^4|+C\end{array}$

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$\begin{array}{rl}(11)& \int \frac{x+1}{x^2+2x+5}dx\\ & =\frac{1}{2}\int \frac{d(x^2+2x+5)}{x^2+2x+5}\\ & =\frac{1}{2}ln(x^2+2x+5)+C\end{array}$

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$\begin{array}{rl}(12)& \int co{s}^2(\omega t+\phi )sin(\omega t+\phi )dt\\ & =-\frac{1}{\omega }\int co{s}^2(\omega t+\phi )d[cos(\omega t+\phi )]\\ & =-\frac{1}{3\omega }co{s}^3(\omega t+\phi )+C\end{array}$

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$\begin{array}{rl}(13)& \int \frac{sinx}{co{s}^{3}x}dx\\ & =-\int \frac{d(cosx)}{co{s}^{3}x}\\ & =\frac{1}{2}co{s}^{-2}x+C\end{array}$

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$\begin{array}{rl}(14)& \int \frac{sinx+cosx}{\sqrt[3]{sinx-cosx}}dx\\ & =\int (sinx-cosx{)}^{-\frac{1}{3}}d(sinx-cosx)\\ & =\frac{3}{2}(sinx-cosx{)}^{\frac{2}{3}}+C\end{array}$

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$\begin{array}{rl}(15)& \int ta{n}^{10}xse{c}^{2}xdx\\ & =\int ta{n}^{10}d(tanx)\\ & =\frac{1}{11}ta{n}^{11}x+C\end{array}$

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$\begin{array}{rl}(16)& \int \frac{dx}{x(lnx)ln(lnx)}dx\\ & =\int \frac{d(lnx)}{(lnx)ln(lnx)}\\ & =\int \frac{d[ln(lnx)]}{ln(lnx)}\\ & =ln|ln(lnx)|+C(x>1)\end{array}$

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$\begin{array}{rl}(17)& \int \frac{dx}{(arcsinx{)}^2\sqrt{1-x^2}}\\ & =\int \frac{d(arcsinx)}{(arcsinx{)}^2}\\ & =-\frac{1}{arcsinx}+C\end{array}$

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$\begin{array}{rl}(18)& \int \frac{10^{2arccosx}}{\sqrt{1-x^2}}\\ & =-\frac{1}{2}\int 10^{2arccosx}d(2arccosx)\\ & =-\frac{10^{2arccosx}}{2ln10}+C\end{array}$

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$\begin{array}{rl}(19)& \int tan\sqrt{1+x^2}\cdot \frac{xdx}{\sqrt{1+x^2}}\\ & =\int tan\sqrt{1+x^2}d(\sqrt{1+x^2})\\ & =-\int \frac{d(cos\sqrt{1+x^2})}{cos\sqrt{1+x^2}}\\ & =-ln|cos\sqrt{1+x^2}|+C\end{array}$

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$\begin{array}{rl}(20)& \int \frac{arctan\sqrt{x}}{\sqrt{x}(1+x)}dx\\ & =2\int \frac{arctan\sqrt{x}}{1+(\sqrt{x}{)}^2}d\sqrt{x}\\ & =2\int arctan\sqrt{x}d(arctan\sqrt{x})\\ & =(arctan\sqrt{x}{)}^2+C\end{array}$

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$\begin{array}{rl}(21)& \int \frac{1+lnx}{(xlnx{)}^2}dx\\ & =\int \frac{d(xlnx)}{(xlnx{)}^2}\\ & =-\frac{1}{xlnx}+C\end{array}$

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$\begin{array}{rl}(22)& \int \frac{dx}{sinxcosx}\\ & =\int \frac{se{c}^{2}x}{tanx}dx\\ & =\int \frac{d(tanx)}{tanx}\\ & =ln|tanx|+C\end{array}$

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$\begin{array}{rl}(23)& \int \frac{lntanx}{cosxsinx}dx\\ & =\int \frac{lntanx}{co{s}^{2}xtanx}dx\\ & =\int \frac{lntanx}{tanx}d(tanx)\\ & =\int lntanxd(lntanx)\\ & =\frac{1}{2}(lntanx{)}^2+C\end{array}$

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$\begin{array}{rl}(24)& \int co{s}^{3}xdx\\ & =\int (1-si{n}^{2}x)d(sinx)\\ & =sinx-\frac{1}{3}si{n}^{3}x+C\end{array}$

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$\begin{array}{rl}(25)& \int co{s}^2(\omega t+\phi )dt\\ & =\int \frac{1+cos2(\omega t+\phi )}{2}dt\\ & =\int \frac{dt}{2}+\frac{1}{4\omega }\int cos2(\omega t+\phi )d[2(\omega t+\phi )]\\ & =\frac{1}{2}t+\frac{1}{4\omega }sin2(\omega t+\phi )+C\end{array}$

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$\begin{array}{rl}(26)& \int sin2xcos3xdx\\ & =\int \frac{1}{2}(sin5x-sinx)dx\\ & =\frac{1}{2}cosx-\frac{1}{10}cos5x+C\end{array}$

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$\begin{array}{rl}(27)& \int cosxcos\frac{x}{2}dx\\ & =\int \frac{1}{2}(cos\frac{3}{2}x+cos\frac{x}{2})dx\\ & =\frac{1}{2}\times \frac{2}{3}\int cos\frac{3}{2}xd(\frac{3}{2}x)+\frac{1}{2}\times 2\int cos\frac{x}{2}d(\frac{x}{2})\\ & =\frac{1}{3}sin\frac{3}{2}x+sin\frac{x}{2}+C\end{array}$

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$\begin{array}{rl}(28)& \int sin5xsin7xdx\\ & =\int -\frac{1}{2}(cos12x-cos2x)dx\\ & =-\frac{1}{2}\int cos12xdx+\frac{1}{2}\int cos2xdx\\ & =\frac{1}{4}sin2x-\frac{1}{24}sin12x+C\end{array}$

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$\begin{array}{rl}(29)& \int ta{n}^{3}xsecxdx\\ & =\int ta{n}^{2}xd(secx)\\ & =\int (se{c}^{2}x-1)d(secx)\\ & =\frac{1}{3}se{c}^{3}x-secx+C\end{array}$

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$\begin{array}{rl}(30)& \int \frac{dx}{e^x+e^{-x}}\\ & =\int \frac{e^{x}dx}{e^{2x}+1}\\ & =\int \frac{d(e^x)}{1+(e^x{)}^2}\\ & =arctan(e^x)+C\end{array}$

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$\begin{array}{rl}(31)& \int \frac{1-x}{\sqrt{9-4x^2}}dx\\ & =\frac{1}{2}\int \frac{d(2x)}{\sqrt{3^2-(2x{)}^2}}+\frac{1}{8}\int (9-4x^2{)}^{-\frac{1}{2}}d(9-4x^2)\\ & =\frac{1}{2}arcsin\frac{2x}{3}+\frac{1}{4}\sqrt{9-4x^2}+C\end{array}$

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$\begin{array}{rl}(32)& \int \frac{x^3}{9+x^2}dx\\ & =\frac{1}{2}\int \frac{x^2}{9+x^2}d(x^2)\\ & =\frac{1}{2}\int \frac{x^2+9-9}{x^2+9}d(x^2)\\ & =\frac{1}{2}\int d(x^2)-\frac{9}{2}\int \frac{d(x^2+9)}{x^2+9}\\ & =\frac{x^2}{2}-\frac{9}{2}ln(x^2+9)+C\end{array}$

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$\begin{array}{rl}(33)& \int \frac{dx}{2x^2-1}\\ & =\int \frac{dx}{(\sqrt{2}x-1)(\sqrt{2}x+1)}\\ & =\frac{1}{2\sqrt{2}}\int (\frac{1}{\sqrt{2}x-1}-\frac{1}{\sqrt{2}x+1})d\sqrt{2}x\\ & =\frac{1}{2\sqrt{2}}(ln|\sqrt{2}x-1|-ln|\sqrt{2}x+1|)+C\\ & =\frac{1}{2\sqrt{2}}ln\left|\begin{array}{c}\frac{\sqrt{2}x-1}{\sqrt{2}x+1}\end{array}\right|+C\end{array}$

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$\begin{array}{rl}(34)& \int \frac{dx}{(x+1)(x-2)}\\ & =\frac{1}{3}\int (\frac{1}{x-2}-\frac{1}{x+1})dx\\ & =\frac{1}{3}ln\left|\begin{array}{c}\frac{x-2}{x+1}\end{array}\right|+C\end{array}$

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$\begin{array}{rl}(35)& \int \frac{x}{x^2-x-2}dx\\ & =\int \frac{x}{(x-2)(x+1)}dx\\ & =\frac{1}{3}\int (\frac{2}{x-2}+\frac{1}{x+1})dx\\ & =\frac{2}{3}ln|x-2|+\frac{1}{3}ln|x+1|+C\end{array}$

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$\begin{array}{rl}(36)& \int \frac{x^{2}dx}{\sqrt{a^2-x^2}}(a>0)\\ & =\int \frac{a^2(\frac{x}{a}{)}^2}{a\sqrt{1-(\frac{x}{a}{)}^2}}ad(\frac{x}{a})\\ & =a^2\int \frac{(\frac{x}{a}{)}^2}{\sqrt{1-(\frac{x}{a}{)}^2}}d(\frac{x}{a})\\ & (令t=arcsin(\frac{x}{a}))\\ & =a^2\int si{n}^{2}td(t)\\ & =a^2\int \frac{1-cos2t}{2}dt\\ & =\frac{a^{2}t}{2}-\frac{a^{2}sin2t}{4}+C\\ & =\frac{a^{2}arcsin\frac{x}{a}}{2}-\frac{x\sqrt{a^2-x^2}}{a^2}+C\end{array}$

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$\begin{array}{rl}(37)& \int \frac{dx}{x\sqrt{x^2-1}}\end{array}$

当 $x>1$ 时：

$\begin{array}{rl}& \int \frac{dx}{x\sqrt{x^2-1}}\\ & =\int \frac{dx}{x^2\sqrt{1-(\frac{1}{x}{)}^2}}\\ & =\int \frac{-d(\frac{1}{x})}{\sqrt{1-(\frac{1}{x}{)}^2}}\\ & =arccos\frac{1}{x}+C\end{array}$

当 $x<-1$ 时：

$\begin{array}{rl}& \int \frac{dx}{x\sqrt{x^2-1}}\\ & =\int \frac{-dx}{x^2\sqrt{1-(\frac{1}{x}{)}^2}}\\ & =\int \frac{-d(-\frac{1}{x})}{\sqrt{1-(-\frac{1}{x}{)}^2}}\\ & =arccos\frac{-1}{x}+C\end{array}$

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学习资料：《高等数学（第六版）》 ，同济大学数学系 编

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