第二类换元法

不定积分

$$\int {f(x){\text{d}}x} $$

若$x=φ(t)$，且$x=φ(t)$可导，即${\text{d}}x = \varphi '(t){\text{d}}t$，则

$$\int {f(x){\text{d}}x} = \int {f\left[ {\varphi (t)} \right]\varphi '(t){\text{d}}t} $$

若等式右边的不定积分存在，即${f\left[ {\varphi (t)} \right]\varphi '(t){\text{d}}t}$有原函数且比左边的不定积分易于求解，则可以利用上述代换求解；其次，右端求出的原函数是以$t$为自变量的，需要通过$x=φ(t)$的反函数$t=φ^{-1} (x)$代换为以$x$为自变量的函数，这就要求$x=φ(t)$具有反函数，只有单射或一一映射才具有反函数，因此$x=φ(t)$还需要是单调的，因此，$x=φ(t)$必须是单调的，可导的.

再来讨论为什么$x=φ(t)$必须是单调，可导以外，还需要$φ'(t)≠0$.

设$f[φ(t)] φ' (t)$的原函数为$Φ(t)$，因为$x=φ(t)$单调可导，则$Φ(t)=Φ[φ^{-1} (x)]$，记$Φ[φ^{-1} (x)]=F(x)$，则

$$\frac{{{\text{d}}\Phi \left( t \right)}}{{{\text{d}}t}} = \frac{{{\text{d}}F(x)}}{{{\text{d}}x}}$$

根据复合函数的求导法则，

$$\eqalign{ & F'\left( x \right) = \frac{{{\text{d}}\Phi \left( t \right)}}{{{\text{d}}t}} \cdot \frac{{{\text{d}}t}}{{{\text{d}}x}} \cr & = f\left[ {\varphi \left( t \right)} \right]\varphi '\left( t \right) \cdot \frac{1}{{\varphi '\left( t \right)}} = f\left[ {\varphi \left( t \right)} \right] = f(x) \cr} $$

因为$F' (x)$是存在的，因此要求$φ'(t)≠0$.所以有

$$\int {f(x){\text{d}}x} = F\left( x \right) + C = \Phi \left[ {{\varphi ^{ - 1}}\left( x \right)} \right] + C $$ $$= {[\int {f\left[ {\varphi \left( t \right)} \right]\varphi '\left( t \right){\text{d}}t} ]_{t = {\varphi ^{-1}}(x)}}$$

例 求

$$\int {\sqrt {{a^2} - {x^2}} \text{d}x} ,(a > 0)$$

解 因为${\text{si}}{{\text{n}}^2}t + {\text{co}}{{\text{s}}^2}t = 1$，设$x = a\sin t, - \frac{\pi }{2} < t < \frac{\pi }{2}$，那么$$\sqrt {{a^2} - {x^2}} = \sqrt {{a^2} - {a^2}{\text{si}}{{\text{n}}^2}t} = {\text{acos}}t,$$ $${\text{d}}x = {\text{acos}}t{\text{d}}t$$，于是

$$\eqalign{ & \int {\sqrt {{a^2} - {x^2}} {\text{d}}x} = \int {{\text{acos}}t \cdot {\text{acos}}t{\text{d}}t} \cr & = {a^2}\int {{\text{co}}{{\text{s}}^2}t{\text{d}}t} \cr & = {a^2}\int {\frac{{1 + \cos 2t}}{2}{\text{d}}t} \cr & = {a^2}[\frac{1}{2}\left( {\int {{\text{d}}t} + \int {\cos 2t{\text{d}}t} } \right)] \cr & = {a^2}[\frac{1}{2}\int {{\text{d}}t} + \frac{1}{4}\int {\cos 2t{\text{d}}\left( {2t} \right)} ] \cr & = \frac{{{a^2}}}{2}t + \frac{{{a^2}}}{2}\sin t\cos t + C. \cr} $$

由于$x = a\sin t, - \frac{\pi }{2} < t < \frac{\pi }{2}$，所以

$$t = \arcsin \frac{x}{a},\cos t = \sqrt {1 - {\text{si}}{{\text{n}}^2}t} $$ $$= \sqrt {1 - {{(\frac{x}{a})}^2}} = \frac{{\sqrt {{a^2} - {x^2}} }}{a}$$

于是

$$\int {\sqrt {{a^2} - {x^2}} {\text{d}}x} = \frac{{{a^2}}}{2}\arcsin \frac{x}{a} $$ $$+ \frac{1}{2}x\sqrt {{a^2} - {x^2}} + C.$$

\begin{align} \int {\tan x{\text{d}}x} = - \ln \left| {\cos x} \right| + C \tag{16} \end{align} \begin{align} \int {\cot x{\text{d}}x} = - \ln \left| {\sin x} \right| + C \tag{17} \end{align} \begin{align} \begin{gathered} \int {\sec x} {\text{d}}x = \hfill \\ \ln \left| {\sec x + \tan x} \right| + C \hfill \\ \end{gathered} \tag{18} \end{align} \begin{align} \begin{gathered} \int {\csc x} {\text{d}}x = \hfill \\ \ln \left| {\csc x - \cot x} \right| + C \hfill \\ \end{gathered} \tag{19} \end{align} \begin{align} \int {\frac{{{\text{d}}x}}{{{a^2} + {x^2}}}} = \frac{1}{a}\arctan \frac{x}{a} + C \tag{20} \end{align} \begin{align} \int {\frac{{{\text{d}}x}}{{{x^2} - {a^2}}}} = \frac{1}{{2a}}\ln |\frac{{{\text{x}} - a}}{{{\text{x}} + a}}| + C \tag{21} \end{align} \begin{align} \int {\frac{{{\text{d}}x}}{{\sqrt {{a^2} - {x^2}} }}} = \arcsin \frac{x}{a} + C \tag{22} \end{align} \begin{align} \begin{gathered} \int {\frac{{{\text{d}}x}}{{\sqrt {{x^2} + {a^2}} }}} = \hfill \\ \ln (x + \sqrt {{x^2} + {a^2}} ) + C \hfill \\ \end{gathered} \tag{23} \end{align} \begin{align} \begin{gathered} \int {\frac{{{\text{d}}x}}{{\sqrt {{x^2} - {a^2}} }}} = \hfill \\ \ln |x + \sqrt {{x^2} - {a^2}} | + C \hfill \\ \end{gathered} \tag{24} \end{align}