第一换元积分法(别称凑微分法)
eduction
\[
\begin{align}
\text {设} u=\varphi(x) 在点x可导
,F(u) 在对应点u=\varphi(x) 可导
\\ 则F[\varphi(x)]在点x可导，设F[\varphi(x)]的导数为f[\varphi(x)]，有如下:
\\ \\
\quad F^{\prime}(u)=f(u)
\\ \\
\Rightarrow \int f(u) d x=F^{\prime}(u)+C, \quad(式0.0.0)
\\ \\
已知F(u)为复合函数，根据链式法则:
\\
F^{\prime} (u)=F^{\prime}(u) \cdot(u)^{\prime}=f(u) \cdot(u)^{\prime}
\\ \\
\Rightarrow \int f(u) \cdot(u)^{\prime} \cdot dx=F^{\prime}(u)+C,
\quad(式1.1 .1)
\\ \\
\quad \therefore(式0.0.0)=(式1.1 .1)
\\ \\
\because(u)^{\prime} d x=d u
\\
\Rightarrow \int f(u) d u=F^{\prime}(u)+C
\\ \\ \\
公式: \enspace
\int f[\varphi(x)] \cdot \varphi^{\prime}(x) d x=\int f[\varphi(x)] d[\varphi(x)]
=F^{\prime}[\varphi(x)]+C
\end{align}
\]
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}
\]
example 0
\[\begin{align}
已知: \int f(x) d x=x^{2}+C
\\
\int x f\left(1-x^{2}\right) d x=?
\\ \\ \\
\Rightarrow\int f\left(1-x^{2}\right) \cdot x d x
\\ \\
\Rightarrow\int f\left(1-x^{2}\right) \cdot-\frac{1}{2}(-2 x d x)
\\ \\
\because-2 x=[F(x)+ C]^{\prime}
\\ \\
\therefore-2 x=\left(1-x^{2}\right)^{\prime}, \quad-2 x d x=d\left(1-x^{2}\right)
\\ \\
\Rightarrow-\frac{1}{2} \int f\left(1-x^{2}\right) \cdot d\left(1-x^{2}\right)
\\ \\
\because\left(1-x^{2}\right) \Leftrightarrow x
\\ \\
\therefore-\frac{1}{2} \int f\left(1-x^{2}\right)
\cdot d\left(1-x^{2}\right)=-\frac{1}{2}\left(1-x^{2}\right)^{2}+C
\end{align}
\]
example 1
\[\begin{align}
\int\frac{1}{a^{2}+x^{2}}dx=?
\\ \\
\int\frac{1}{a^{2}(1+\frac{x^{2}}{a^{2}})}dx=\int\frac{1}{a^{2}}\cdot\frac{1}{1+(\frac{x}{a})^{2}}dx,
\\ \\
设: \enspace u=\frac{x}{a}, \enspace 令u 取代 x ,\enspace 成为新的积分变量
\\ \\
=\int\frac{1}{a^{2}}\cdot\frac{1}{1+u^{2}}\cdot d(\frac{x}{a})
\\ \\
du=d(\frac{x}{a})=(\frac{x}{a})^{\prime}dx, \enspace du=\frac{1}{a}dx\Rightarrow adu=dx
\\ \\
\int\frac{1}{a^{2}}\cdot\frac{1}{1+u^{2}}\cdot adu\Rightarrow\int\frac{1}{a}\frac{1}{1+u^{2}}du
\\ \\
据公式: \int\frac{1}{1+x^{2}}dx=\arctan x+C
\\ \\
\therefore\int\frac{1}{a}\frac{1}{1+u^{2}}du=\frac{1}{a}\arctan(u)+C
\\ \\
=\frac{1}{a}\arctan\frac{x}{a}+ C
\end{align}
\]
example 2
\[\begin{align}
\int\frac{1}{3+2x}dx=?
\\ \\
设:u=3+2x
\\ \\
\int \frac{1}{u}du=\int \frac{1}{3+2x}d(3+2x)
\\ \\
d(3+2x)=(u)^{\prime}dx=2dx, \enspace \frac{1}{2}du=dx
\\ \\
因为令u代替了x,所以\frac{1}{2}du就等价于更换积分变量后的dx
\\ \\
\int\frac{1}{u}\cdot\frac{1}{2}du=\frac{1}{2}\ln|u|+C
\\
=\frac{1}{2}\ln\vert3+2x\vert+C
\end{align}
\]
example 3
\[\begin{align}
\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=?
\\ \\
\Rightarrow \int \frac{1}{\left(a^{2}-x^{2}\right)^{\frac{1}{2}}} d x
\\ \\
\text {根据公式}: \int \frac{d x}{\left(1-x^{2}\right)^{\frac{1}{2}}}=\arcsin x+C
\\ \\
\left(a^{2}-x^{2}\right)^{\frac{1}{2}}=\left[a^{2}\left(1-\frac{x^{2}}{a^{2}}\right)\right]^{\frac{1}{2}}
\\ \\
\Rightarrow \int \frac{d x}{\left[a^{2}\left(1-\frac{x^{2}}{a^{2}}\right)\right]^{\frac{1}{2}}}=\int \frac{1}{a} \cdot \frac{d x}{\left[1-\left(\frac{x}{a}\right)^{2}\right]^{\frac{1}{2}}}
\\ \\
\operatorname{设}: u=\frac{x}{a} , \enspace 令u成为新的积分变量
\\ \\
du=d\left(\frac{x}{a}\right)=\left(\frac{x}{a}\right)^{\prime} d x, \quad d x=a d u
\\ \\
\Rightarrow \int \frac{1}{a} \cdot \frac{1}{(1-u^{2})^{\frac{1}{2}}}
\cdot a d u= \arcsin (u)+C
\\ \\
=\arcsin \frac{x}{a}+C
\end{align}
\]
example 4
\[\begin{align}
\int x \sqrt{1-x^{2}} d x=?
\\ \\
\Rightarrow \int x\left(1-x^{2}\right)^{\frac{1}{2}} d x
\\ \\
\text { 设: } u=\left(1-x^{2}\right)
\\ \\
d u=d\left(1-x^{2}\right)=\left(1-x^{2}\right)^{\prime} d x
\\ \\
-2 x d x=d u, \quad \frac{d u}{-2 x}=d x
\\ \\
\Rightarrow \int x \cdot (u)^{\frac{1}{2}} \cdot \frac{d u}{-2 x}
\\ \\
\Rightarrow \int-\frac{1}{2}(u)^{\frac{1}{2}} d u
=-\frac{1}{2}\left(\frac{2}{3} u^{\frac{3}{2}}\right)^{\prime}+C
\\ \\
=-\frac{1}{3}(u)^{\frac{3}{2}}+ C
\\ \\
=-\frac{1}{3}\left(1-x^{2}\right)^{\frac{3}{2}}+C
\end{align}
\]
example 5
\[\begin{align}
\int \frac{e^{x}}{1+e^{2 x}} d x=?
\\ \\
\text { 设: } \quad e^{x}=u
\\ \\
\Rightarrow \int u \frac{1}{1+u^{2}} d x
\\ \\
\Rightarrow \int u \frac{1}{1+u^{2}} d\left(e^{x}\right)
\\ \\
d\left(e^{x}\right)=\left(e^{x}\right)^{\prime} d x=e^{x} \ln e d x=e^{x} d x
\\ \\
d u=e^{x} d x, \quad d x=\frac{d u}{e^{x}}
\\ \\
\Rightarrow \int u \frac{1}{1+u^{2}} \cdot \frac{1}{e^{x}} d u=\int \frac{1}{1+u^{2}} d u
\\ \\
\text {根据公式: } \int \frac{1}{1+x^{2}} d x=\arctan x+C
\\ \\
=\arctan (u)+C
\\ \\
=\arctan e^{x}+C
\end{align}
\]
example 6
\[
\begin{align}
\int \tan x d x=?
\\ \\
\Rightarrow \int \frac{\sin x}{\cos x} d x
\\ \\
设: u=\cos x
\\ \\
\Rightarrow \int \sin x \cdot \frac{1}{u} \cdot d(\cos x)
\\ \\
d(\cos x)=(\cos x)^{\prime} d x, \quad d x=\frac{d u}{-\sin x}
\\ \\
\Rightarrow \int \sin x \cdot \frac{1}{u} \cdot \frac{d u}{-\sin x}
\\ \\
= -\int \frac{1}{u} d u=-\ln |u|+C
\\ \\
= -\ln |\cos x|+C
\end{align}
\]
example 7
\[\begin{align}
\int\cos3x\sin2xdx=?
\\ \\
据积化和差: \enspace
\cos2\sin\beta=\frac{1}{2}[\sin(\alpha+\beta)-\cos(\alpha-\beta)]
\\ \\
\Rightarrow\int\frac{1}{2}[\sin(3x+2x)-\sin(3x-2x)]dx
\\
\\ \\
=\int\frac{1}{2}(\sin 5x-\sin x)dx
\\ \\
=\frac{1}{2}\int\sin 5xdx-\frac{1}{2}\int\sin xdx
\\ \\
首先:\frac{1}{2}\int\sin xdx=\frac{1}{2}\cdot-\cos x+C
\\
\\ \\
设:u=5x, \enspace 得: \frac{1}{2}\int\sin u \cdot d(5x)
\\ \\
\Rightarrow d(5x)=(5x)^{\prime}dx,\quad dx=\frac{du}{5}
\\ \\
=\frac{1}{2}\int\sin u\frac{1}{5}du=\frac{1}{2}\cdot\frac{1}{5}\cdot-\cos u+c
\\ \\ \\
最终: \enspace -\frac{1}{10}\cos5x+\frac{1}{2}\cos x+C
\end{align}
\]
example 8
\[\begin{align}
\int\frac{x}{1+x^{2}}dx=?
\\ \\
设:u=1+x^{2}\\du=d(1+x^{2})=(1+x^{2})^{\prime}dx=2xdx
\\ \\
dx=\frac{1}{2x}du
\\ \\
\Rightarrow\int\frac{x}{u}\cdot\frac{1}{2x}du=\frac{1}{2}\int\frac{1}{u}du
\\ \\
=\frac{1}{2}\ln|u|+C
\\ \\
=\frac{1}{2}\ln|1+x^{2}|+C
\end{align}
\]
