不定积分形式待定法

定理

情况一：

对于 ${\displaystyle \int \frac{f(x)}{g^2(x)}\text{dd}x}$ 型，存在 $h(x)$ ，使得

$${\displaystyle \int \frac{f(x)}{g^2(x)}\text{dd}x=\frac{h(x)}{g(x)}+C}$$

其中 $f(x)=h^{\prime }(x)g(x)-h(x)g^{\prime }(x)$ ， $h(x)$ 要通过具体的等式结构另求出.

情况二：

对于 ${\displaystyle \int f(x){\mathrm{e}}^{g(x)}\text{dd}x}$ 型，存在 $h(x)$ ，使得

$${\displaystyle \int f(x){\mathrm{e}}^{g(x)}\text{dd}x=h(x){\mathrm{e}}^{g(x)}+C}$$

其中 $f(x)=h^{\prime }(x)+h(x)g^{\prime }(x)$ ，同样 $h(x)$ 要通过具体的等式结构另求出.

例题 1

求不定积分 ${\displaystyle \int \frac{{\mathrm{e}}^{-\sin x}\cdot \sin 2x}{(1-\sin x{)}^2}\text{dd}x.}$

解答

令 ${\displaystyle f(x)={\mathrm{e}}^{-\sin x}\cdot \sin 2x,g(x)=1-\sin x}$ ，由

$$f(x)=h^{\prime }(x)g(x)-h(x)g^{\prime }(x)=h^{\prime }(x)-h^{\prime }(x)\sin x+h(x)\cos x$$

设 $h(x)=A{\mathrm{e}}^{-\sin x}$ ，那么 $h^{\prime }(x)=-A\cos x{\mathrm{e}}^{-\sin x}$ ，所以

$$h^{\prime }(x)-h^{\prime }(x)\sin x+h(x)\cos x=A\sin x\cos x{\mathrm{e}}^{-\sin x}=\frac{A}{2}\sin 2x{\mathrm{e}}^{-\sin x}={\mathrm{e}}^{-\sin x}\sin 2x$$

解得 $A=2$ ，所以 $h(x)=2{\mathrm{e}}^{-\sin x}.$

$$\begin{array}{rl}\text{原式}& =\int \frac{2\sin x\cos x{\mathrm{e}}^{-\sin x}}{(1-\sin x{)}^2}\text{dd}x=\int \frac{2{\mathrm{e}}^{-\sin x}(-\cos x)(1-\sin x)+2\cos x{\mathrm{e}}^{-\sin x}}{(1-\sin x{)}^2}\text{dd}x\\ & =\int \frac{h^{\prime }(x)(1-\sin x)-(1-\sin x{)}^{\prime }h(x)}{(1-\sin x{)}^2}\text{dd}x=\int \text{dd}(\frac{h(x)}{1-\sin x})=\frac{2{\mathrm{e}}^{-\sin x}}{1-\sin x}+C.\end{array}$$

例题 2

${\displaystyle \int \frac{x^2\text{dd}x}{(x\sin x+\cos x{)}^2}.}$

解答：法一

注意到 $1={\sin }^{2}x+{\cos }^{2}x$ ，所以

$$\begin{array}{rl}\text{原式}& =\int \frac{x^2({\sin }^{2}x+{\cos }^{2}x)\text{dd}x}{(x\sin x+\cos x{)}^2}=\int {\displaystyle \frac{x^2{\cos }^{2}x-x\sin x\cos x}{{(x\sin x+\cos x)}^2}}\text{dd}x+\int {\displaystyle \frac{x^2{\sin }^{2}x+x\sin x\cos x}{{(x\sin x+\cos x)}^2}}\text{dd}x\\ & =\int {\displaystyle \frac{x\cos x(x\cos x-\sin x)}{{(x\sin x+\cos x)}^2}}\text{dd}x+\int {\displaystyle \frac{x\sin x\text{dd}x}{x\sin x+\cos x}}\\ & =\int (\sin x-x\cos x)\text{dd}({\displaystyle \frac{1}{x\sin x+\cos x}})+\int {\displaystyle \frac{x\sin xdx}{x\sin x+\cos x}}\\ & ={\displaystyle \frac{\sin x-x\cos x}{x\sin x+\cos x}}-\int {\displaystyle \frac{x\sin x\text{dd}x}{x\sin x+\cos x}}+\int {\displaystyle \frac{x\sin x\text{dd}x}{x\sin x+\cos x}}={\displaystyle \frac{\sin x-x\cos x}{x\sin x+\cos x}}+C.\end{array}$$

解答：法二

令 $f(x)=x^2,g(x)=x\sin x+\cos x$ ，由

$$f(x)=h^{\prime }(x)g(x)-h(x)g^{\prime }(x)=h^{\prime }(x)(x\sin x+\cos x)-h(x)x\cos x=x^2=x^2({\sin }^{2}x+{\cos }^{2}x)$$

设 $h(x)=A{x}^a\sin x+B{x}^b\cos x$ ，那么 $h^{\prime }(x)=A_a{x}^{a-1}\sin x+A{x}^a\cos x+Bb{x}^{b-1}\cos x-B{x}^b\sin x$ ，
代入上式，对比系数得 $A=1,a=0,B=-1,b=1$ ，于是 $h(x)=\sin x-x\cos x$ ，

$$\begin{array}{rl}\text{原式}& =\int \frac{x^2{\sin }^{2}x+x^2{\cos }^{2}x}{(x\sin x+\cos x{)}^2}\text{dd}x=\int \frac{x\sin x(x\sin x+\cos x)-x\cos x(\sin x-x\cos x)}{(x\sin x+\cos x{)}^2}\text{dd}x\\ & =\int \frac{h^{\prime }(x)(x\sin x+\cos x)-h(x)\cdot {(x\sin x+\cos x)}^{\prime }}{(x\sin x+\cos x{)}^2}\text{dd}x=\int \text{dd}(\frac{h(x)}{x\sin x+\cos x})=\frac{\sin x-x\cos x}{x\sin x+\cos x}+C.\end{array}$$

例题 3

${\displaystyle \int {\displaystyle \frac{x^2+6}{(x\cos x-3\sin x{)}^2}}\text{dd}x.}$

解答：法一

令 $f(x)=x^2+6,g(x)=x\cos x-3\sin x$ ，由

$$\begin{array}{rl}f(x)=h^{\prime }(x)g(x)-h(x)g^{\prime }(x)& =h^{\prime }(x)(x\cos x-3\sin x)+h(x)(x\sin x+2\cos x)\\ (x^2+6)\text{qty}({\sin }^{2}x+{\cos }^{2}x)& =h^{\prime }(x)(x\cos x-3\sin x)+h(x)(x\sin x+2\cos x)\end{array}$$

可知， $h(x)$ 中存在 $x\sin x$ 项和 $3\cos x$ ; $h^{\prime }(x)$ 中存在 $x\cos x$ 项和 $-2\sin x$ ，于是不妨取
$h(x)=x\sin x+3\cos x$ ，
故原积分等于 ${\displaystyle \frac{x\sin x+3\cos x}{x\cos x-3\sin x}}+C$ ，经检验该函数是原积分的原函数.

解答：法二

运用 ${\sin }^{2}x+{\cos }^{2}x=1$ ，则有

$$\begin{array}{rl}I& =\int {\displaystyle \frac{\text{qty}(x^2+6)\text{qty}({\sin }^{2}x+{\cos }^{2}x)}{(x\cos x-3\sin x{)}^2}}\text{dd}x=\int \left|\begin{array}{cc}x& 3\\ -2& x\end{array}\right|\left|\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right|{\displaystyle \frac{\text{dd}x}{(x\cos x-3\sin x{)}^2}}\\ & =\int \left|\begin{array}{cc}x\cos x-3\sin x& x\sin x+3\cos x\\ -2\cos x-x\sin x& -2\sin x+x\cos x\end{array}\right|{\displaystyle \frac{\text{dd}x}{(x\cos x-3\sin x{)}^2}}\\ & =\int {\displaystyle \frac{-2\sin x+x\cos x}{x\cos x-3\sin x}}\text{dd}x+\int {\displaystyle \frac{(x\sin x+3\cos x)(2\cos x+\sin x)}{(x\cos x-3\sin x{)}^2}}\text{dd}x\end{array}$$

注意到 ${\displaystyle \text{dv}x(x\cos x-3\sin x)=-(2\cos x+x\sin x)}$ ，故可对上式第二个积分利用分部积分公式，得

$$\begin{array}{rl}{I}_1& =\int {\displaystyle \frac{(x\sin x+3\cos x)(2\cos x+\sin x)}{(x\cos x-3\sin x{)}^2}}\text{dd}x=\int (x\sin x+3\cos x)\text{dd}({\displaystyle \frac{1}{x\cos x-3\sin x}})\\ & ={\displaystyle \frac{x\sin x+3\cos x}{x\cos x-3\sin x}}-\int {\displaystyle \frac{\text{dd}(x\sin x+3\cos x)}{x\cos x-3\sin x}}={\displaystyle \frac{x\sin x+3\cos x}{x\cos x-3\sin x}}-\int {\displaystyle \frac{x\cos x-2\sin x}{x\cos x-3\sin x}}\text{dd}x\end{array}$$

所以 ${\displaystyle I=\int {\displaystyle \frac{x\cos x-2\sin x}{x\cos x-3\sin x}}+I_1={\displaystyle \frac{x\sin x+3\cos x}{x\cos x-3\sin x}}+C.}$