利用基本积分表与积分的性质，所能求解的不定积分非常有限，有必要进一步研究不定积分的求法.本节把复合函数的微分法反过来用于求不定积分，利用中间变量的代换得到复合函数的积分法，称为换元积分法，简称换元法.

换元积分法，即换元法就是复合函数的积分法.

第一类换元法

设$f(u)$具有原函数$F(u)$，即

$$F'\left( u \right) = f\left( u \right),\int {f\left( u \right){\text{d}}u} = F\left( u \right) + C$$

如果$u$是中间变量：$u=φ(x)$，且$φ(x)$可微，根据复合函数微分法，有

$$\frac{{{\text{d}}F(u)}}{{{\text{d}}x}} = f[\varphi (x)]\varphi '(x){\text{d}}x$$ $${\text{d}}F\left( u \right) = f[\varphi (x)]\varphi '(x){\text{d}}x$$

根据不定积分的定义，有

$$\int f[\varphi (x)]\varphi '(x){\text{d}}x = F\left[ {\varphi \left( x \right)} \right] + C $$ $$= {[\int {f(u)du} ]_{u = \varphi (x)}}$$

于是有下述定理：

定理1 设$f(u)$具有原函数，$u=φ(x)$可导，则有换元公式

$$ f[\varphi (x)]\varphi '(x){\text{d}}x = {[\int {f\left( u \right)\text{d}u} ]_{u = \varphi \left( x \right)}}$$

一般地，对于不定积分$\int {f\left( {ax + b} \right){\text{d}}x} $，总可作变换$u=ax+b$，把它化为

$$\int {f\left( {ax + b} \right){\text{d}}x} = \int {\frac{1}{a}f\left( {ax + b} \right){\text{d}}\left( {ax + b} \right)} $$ $$ = \frac{1}{a}{[\int {f(u)du} ]_{u = ax + b}}$$

 

例 求

$$\int {\frac{{{x^2}}}{{{{(x + 2)}^3}}}{\text{d}}x.} $$

解 令$u=x+2$，则$$\frac{{{\text{d}}u}}{{{\text{d}}x}} = 1 \to {\text{d}}u = {\text{d}}x,x=u-2$$

$$\eqalign{ & \int {\frac{{{x^2}}}{{{{\left( {x + 2} \right)}^3}}}{\text{d}}x} = \int {\frac{{{{\left( {u - 2} \right)}^2}}}{{{u^3}}}{\text{d}}u} \cr & = \int {\left( {{u^2} - 4u + 4} \right){u^{ - 3}}{\text{d}}u} \cr & = \int {\left( {{u^{ - 1}} - 4{u^{ - 2}} + 4{u^{ - 3}}} \right){\text{d}}u} \cr & = \int {\left( {\frac{1}{u} - 4{u^{ - 2}} + 4{u^{ - 3}}} \right){\text{d}}u} \cr & = \int {\frac{1}{u}{\text{d}}u} - \int {4{u^{ - 2}}{\text{d}}u} + \int {4{u^{ - 3}}{\text{d}}u} \cr & = \int {\frac{1}{u}{\text{d}}u} - 4\int {{u^{ - 2}}{\text{d}}u} + 4\int {{u^{ - 3}}{\text{d}}u} \cr & = \ln |u| + 4{u^{ - 1}} - 2{u^{ - 2}} + C \cr} $$

例求

$$\int {\sec x{\text{d}}x} $$

解

$$\eqalign{ & \int {\sec x\text{d}x} = \int {\frac{1}{{\cos x}}} \text{d}x = \int {\frac{{\cos x}}{{{{\cos }^2}x}}} \text{d}x \cr & = \int {\frac{1}{{{{\cos }^2}x}}} \text{d}\sin x = \int {\frac{1}{{1 - {{\sin }^2}x}}} \text{d}\sin x \cr} $$ $$\eqalign{ & = \int {\frac{1}{{1 - {{\sin }^2}x}}} \text{d}\sin x \cr & = \int {\frac{1}{2}} (\frac{1}{{1 - \sin x}} + \frac{1}{{1 + \sin x}})\text{d}\sin x \cr & = \frac{1}{2}\int {\frac{{ - 1}}{{1 - \sin x}}\text{d}(1 - \sin x)} \cr & + \frac{1}{2}\int {\frac{1}{{1 + \sin x}}} \text{d}(1 + \sin x) \cr & = \frac{1}{2}[\ln (1 + \sin x) - \ln (1 - \sin x)] + C \cr & = \frac{1}{2}\ln \frac{{1 + \sin x}}{{1 - \sin x}} + C = \ln {(\frac{{1 + \sin x}}{{\cos x}})^2} + C \cr & = \ln |\sec x + \tan x| + C \cr} $$

换元积分的出发点是将复杂函数的不定积分化简为简单函数的不定积分，化简过程中,通常的做法是：

1.多次方化简为低次方，可以利用三角形倍角公式

2.有分母的化简为$\frac{1}{{f\left( x \right)}}{\text{d}}f(x)$

3.化简其它三角函数为$\sin x$，$\cos x$，若$\sin x$，$\cos x$在分母，则将其化为2次方后利用$\tan x$，$\cot x$的积分公式求导

4.$\sqrt {f(x)} {\text{d}}x$化简为$\sqrt {f(x)} {\text{d}}f(x).$