第一换元积分法(别称凑微分法)

eduction

$$\begin{align} \text {设} u=\varphi(x) 在点x可导 ,F(u) 在对应点u=\varphi(x) 可导 \\ 则F[\varphi(x)]在点x可导，设F[\varphi(x)]的导数为f[\varphi(x)]，有如下: \\ \\ \quad F^{\prime}(u)=f(u) \\ \\ \Rightarrow \int f(u) d x=F^{\prime}(u)+C, \quad(式0.0.0) \\ \\ 已知F(u)为复合函数，根据链式法则: \\ F^{\prime} (u)=F^{\prime}(u) \cdot(u)^{\prime}=f(u) \cdot(u)^{\prime} \\ \\ \Rightarrow \int f(u) \cdot(u)^{\prime} \cdot dx=F^{\prime}(u)+C, \quad(式1.1 .1) \\ \\ \quad \therefore(式0.0.0)=(式1.1 .1) \\ \\ \because(u)^{\prime} d x=d u \\ \Rightarrow \int f(u) d u=F^{\prime}(u)+C \\ \\ \\ 公式: \enspace \int f[\varphi(x)] \cdot \varphi^{\prime}(x) d x=\int f[\varphi(x)] d[\varphi(x)] =F^{\prime}[\varphi(x)]+C \end{align}$$

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Sample 0

$$\begin{eqnarray} \int \sec x d x =? \\ \int \sec x d x \Rightarrow \int \frac{1}{\cos x} d x \\ \\ \Rightarrow \int \frac{\cos x}{\cos ^{2} x} d x=\int \frac{1}{1-\sin ^{2} x} \cdot \cos x d x \\ \\ \text {设:} \enspace \cos x d x=d q =(q)^{\prime} d x \\ \\ \quad(q)^{\prime}=(\sin x)^{\prime}=\cos x \\ \\ \therefore \cos x dx=d \sin x \\ \\ \Rightarrow \int \frac{1}{1-\sin ^{2} x} d \sin x \\ \\ \because \quad 1-\sin ^{2} x=(1-\sin x)(1+\sin x) \\ \\ \frac{1}{1-\sin ^{2} x} \Rightarrow \frac{(1-\sin x)+(1+\sin x)}{2 \cdot(1-\sin x)(1+\sin x)} \\ \\ \Rightarrow \frac{1}{2} \int \frac{(1-\sin x)+(1+\sin x)}{(1-\sin x)(1+\sin x)} d \sin x \\ \\ \Rightarrow \frac{1}{2} \int\left(\frac{1}{1+\sin x}+\frac{1}{1-\sin x}\right) d \sin x \\ \\ \\ \frac{1}{2}\int\frac{1}{1+\sin x}d\sin x+\frac{1}{2}\int\frac{1}{1-\sin x}d\sin x \\ \\ \\ 设:u=1+\sin x,\enspace du=(1+\sin x)^{\prime}dx \\ \\ du=0+\cos xdx=(\sin x)^{\prime}dx \\ \\ \therefore(1+\sin x)^{\prime}dx=d\sin x \\ \\ \Rightarrow\frac{1}{2}\int\frac{1}{1+\sin x}d(1+\sin x) =\frac{1}{2}\ln|1+\sin x|+C \\ \\ \\ 设: \enspace b=1-\sin x,\enspace db=(1-\sin x)^{\prime}dx \\ \\ \enspace db=0-\cos xdx=-(\sin x)^{\prime}dx \\ \\ \therefore(1-\sin x)^{\prime}dx=-d\sin x \\ \\ \Rightarrow\frac{1}{2}\int\frac{1}{1-\sin x}\cdot-1\cdot d(1-\sin x) \\ \\ =-\frac{1}{2}\ln|1-\sin x|+C \\ \\ 综合:\quad \Rightarrow \frac{1}{2} \ln |1+\sin x|-\frac{1}{2} \ln |1-\sin x|+C \\ \\ \Rightarrow \frac{1}{2} \ln \left|\frac{1+\sin x}{1-\sin x}\right|+C \\ \\ \Rightarrow \ln \left|\left(\frac{1+\sin x}{1-\sin x}\right)^{\frac{1}{2}}\right|=\ln \left|\frac{\sqrt{1+\sin x}}{\sqrt{1-\sin x}}\right| \\ \\ 分子分母同乘以 \sqrt[]{1+\sin x}: \\ \\ \left|\frac{\sqrt{(1+\sin x)(1+\sin x)}}{\sqrt{1-\sin x)(1+\sin x)}}\right| =\ln \left|\frac{\sqrt{(1+\sin x)^{2}}} {\sqrt{1+\sin x-\sin x- \sin^{2} x}} \right| \\ \\ =\ln \left|\frac{1+\sin x}{\sqrt{1- \sin^{2} x}}\right| =\ln \left|\frac{1+\sin x}{\sqrt{\cos^{2} x}}\right| =\ln \left|\frac{1+\sin x}{\cos x}\right| \\ \\ =\ln \left|\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right| \\ \\ =\ln |\sec x+\tan x|+C \end{eqnarray}$$

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example 0

$$\begin{align} 已知: \int f(x) d x=x^{2}+C \\ \int x f\left(1-x^{2}\right) d x=? \\ \\ \\ \Rightarrow\int f\left(1-x^{2}\right) \cdot x d x \\ \\ \Rightarrow\int f\left(1-x^{2}\right) \cdot-\frac{1}{2}(-2 x d x) \\ \\ \because-2 x=[F(x)+ C]^{\prime} \\ \\ \therefore-2 x=\left(1-x^{2}\right)^{\prime}, \quad-2 x d x=d\left(1-x^{2}\right) \\ \\ \Rightarrow-\frac{1}{2} \int f\left(1-x^{2}\right) \cdot d\left(1-x^{2}\right) \\ \\ \because\left(1-x^{2}\right) \Leftrightarrow x \\ \\ \therefore-\frac{1}{2} \int f\left(1-x^{2}\right) \cdot d\left(1-x^{2}\right)=-\frac{1}{2}\left(1-x^{2}\right)^{2}+C \end{align}$$

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example 1

$$\begin{align} \int\frac{1}{a^{2}+x^{2}}dx=? \\ \\ \int\frac{1}{a^{2}(1+\frac{x^{2}}{a^{2}})}dx=\int\frac{1}{a^{2}}\cdot\frac{1}{1+(\frac{x}{a})^{2}}dx, \\ \\ 设: \enspace u=\frac{x}{a}, \enspace 令u 取代 x ,\enspace 成为新的积分变量 \\ \\ =\int\frac{1}{a^{2}}\cdot\frac{1}{1+u^{2}}\cdot d(\frac{x}{a}) \\ \\ du=d(\frac{x}{a})=(\frac{x}{a})^{\prime}dx, \enspace du=\frac{1}{a}dx\Rightarrow adu=dx \\ \\ \int\frac{1}{a^{2}}\cdot\frac{1}{1+u^{2}}\cdot adu\Rightarrow\int\frac{1}{a}\frac{1}{1+u^{2}}du \\ \\ 据公式: \int\frac{1}{1+x^{2}}dx=\arctan x+C \\ \\ \therefore\int\frac{1}{a}\frac{1}{1+u^{2}}du=\frac{1}{a}\arctan(u)+C \\ \\ =\frac{1}{a}\arctan\frac{x}{a}+ C \end{align}$$

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example 2

$$\begin{align} \int\frac{1}{3+2x}dx=? \\ \\ 设:u=3+2x \\ \\ \int \frac{1}{u}du=\int \frac{1}{3+2x}d(3+2x) \\ \\ d(3+2x)=(u)^{\prime}dx=2dx, \enspace \frac{1}{2}du=dx \\ \\ 因为令u代替了x,所以\frac{1}{2}du就等价于更换积分变量后的dx \\ \\ \int\frac{1}{u}\cdot\frac{1}{2}du=\frac{1}{2}\ln|u|+C \\ =\frac{1}{2}\ln\vert3+2x\vert+C \end{align}$$

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example 3

$$\begin{align} \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=? \\ \\ \Rightarrow \int \frac{1}{\left(a^{2}-x^{2}\right)^{\frac{1}{2}}} d x \\ \\ \text {根据公式}: \int \frac{d x}{\left(1-x^{2}\right)^{\frac{1}{2}}}=\arcsin x+C \\ \\ \left(a^{2}-x^{2}\right)^{\frac{1}{2}}=\left[a^{2}\left(1-\frac{x^{2}}{a^{2}}\right)\right]^{\frac{1}{2}} \\ \\ \Rightarrow \int \frac{d x}{\left[a^{2}\left(1-\frac{x^{2}}{a^{2}}\right)\right]^{\frac{1}{2}}}=\int \frac{1}{a} \cdot \frac{d x}{\left[1-\left(\frac{x}{a}\right)^{2}\right]^{\frac{1}{2}}} \\ \\ \operatorname{设}: u=\frac{x}{a} , \enspace 令u成为新的积分变量 \\ \\ du=d\left(\frac{x}{a}\right)=\left(\frac{x}{a}\right)^{\prime} d x, \quad d x=a d u \\ \\ \Rightarrow \int \frac{1}{a} \cdot \frac{1}{(1-u^{2})^{\frac{1}{2}}} \cdot a d u= \arcsin (u)+C \\ \\ =\arcsin \frac{x}{a}+C \end{align}$$

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example 4

$$\begin{align} \int x \sqrt{1-x^{2}} d x=? \\ \\ \Rightarrow \int x\left(1-x^{2}\right)^{\frac{1}{2}} d x \\ \\ \text { 设: } u=\left(1-x^{2}\right) \\ \\ d u=d\left(1-x^{2}\right)=\left(1-x^{2}\right)^{\prime} d x \\ \\ -2 x d x=d u, \quad \frac{d u}{-2 x}=d x \\ \\ \Rightarrow \int x \cdot (u)^{\frac{1}{2}} \cdot \frac{d u}{-2 x} \\ \\ \Rightarrow \int-\frac{1}{2}(u)^{\frac{1}{2}} d u =-\frac{1}{2}\left(\frac{2}{3} u^{\frac{3}{2}}\right)^{\prime}+C \\ \\ =-\frac{1}{3}(u)^{\frac{3}{2}}+ C \\ \\ =-\frac{1}{3}\left(1-x^{2}\right)^{\frac{3}{2}}+C \end{align}$$

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example 5

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

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example 6

$$\begin{align} \int \tan x d x=? \\ \\ \Rightarrow \int \frac{\sin x}{\cos x} d x \\ \\ 设: u=\cos x \\ \\ \Rightarrow \int \sin x \cdot \frac{1}{u} \cdot d(\cos x) \\ \\ d(\cos x)=(\cos x)^{\prime} d x, \quad d x=\frac{d u}{-\sin x} \\ \\ \Rightarrow \int \sin x \cdot \frac{1}{u} \cdot \frac{d u}{-\sin x} \\ \\ = -\int \frac{1}{u} d u=-\ln |u|+C \\ \\ = -\ln |\cos x|+C \end{align}$$

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example 7

$$\begin{align} \int\cos3x\sin2xdx=? \\ \\ 据积化和差: \enspace \cos2\sin\beta=\frac{1}{2}[\sin(\alpha+\beta)-\cos(\alpha-\beta)] \\ \\ \Rightarrow\int\frac{1}{2}[\sin(3x+2x)-\sin(3x-2x)]dx \\ \\ \\ =\int\frac{1}{2}(\sin 5x-\sin x)dx \\ \\ =\frac{1}{2}\int\sin 5xdx-\frac{1}{2}\int\sin xdx \\ \\ 首先:\frac{1}{2}\int\sin xdx=\frac{1}{2}\cdot-\cos x+C \\ \\ \\ 设:u=5x, \enspace 得: \frac{1}{2}\int\sin u \cdot d(5x) \\ \\ \Rightarrow d(5x)=(5x)^{\prime}dx,\quad dx=\frac{du}{5} \\ \\ =\frac{1}{2}\int\sin u\frac{1}{5}du=\frac{1}{2}\cdot\frac{1}{5}\cdot-\cos u+c \\ \\ \\ 最终: \enspace -\frac{1}{10}\cos5x+\frac{1}{2}\cos x+C \end{align}$$

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example 8

$$\begin{align} \int\frac{x}{1+x^{2}}dx=? \\ \\ 设:u=1+x^{2}\\du=d(1+x^{2})=(1+x^{2})^{\prime}dx=2xdx \\ \\ dx=\frac{1}{2x}du \\ \\ \Rightarrow\int\frac{x}{u}\cdot\frac{1}{2x}du=\frac{1}{2}\int\frac{1}{u}du \\ \\ =\frac{1}{2}\ln|u|+C \\ \\ =\frac{1}{2}\ln|1+x^{2}|+C \end{align}$$

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example 9

$$\begin{eqnarray} \text { 据微分公式:} \\ y=f(x) \\ 则:\boxed{ dy=f^{\prime}(x) d x } \\ \\ \int \cos ^{5} x \sin x d x=? \\ \\ \text { 设: } u=\cos x \\ d u=d(\cos x)=(\cos x)^{'} d x=-\sin x d x \\ \\ \int \cos ^{5} x \sin x d x \Rightarrow \int u^{5} d(-\cos x) \Rightarrow \int u^{5} d(-u) \\ \\ d(-u)=-(-\sin x)d x=\sin x d x=-d u \\ \\ \Rightarrow-\int u^{5} d u=-[\frac{1}{1+5} u^{5+1}]+C \\ \\ =-[\frac{1}{6} u^{6}]+C \\ \\ =-[\frac{1}{6} \cos ^{6} x]+C \end{eqnarray}$$

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