English · PDF 687
中文 · PDF 687
Figure 8.61
用于求值的换元法 $ff(x)dx$ .
| If contains | try | |
|---|---|---|
| $\sqrt{(a^2 - x^2)}$ | $x = a \sin \theta,$ or $x = a \tanh u,$ |
$\frac{dx}{d\theta} = a \cos \theta$ $\frac{dx}{du} = a \sec^2 u$ |
| $\sqrt{(a^2 + x^2)}$ | $x = a \sinh u,$ or $x = a \tan \theta,$ |
$\frac{dx}{du} = a \cosh u$ $\frac{dx}{d\theta} = a \sec^2 \theta$ |
| $\sqrt{(x^2 - a^2)}$ | $x = a \cosh u$ or $x = a \sec \theta$ |
$\frac{dx}{du} = a \sinh u$ $\frac{dx}{d\theta} = a \sec \theta \tan \theta$ |
| 圆函数 | $s = \sin x,$ or $c = \cos x,$ or $t = \tan \frac{1}{2}x,$ |
$\frac{ds}{dx} = \cos x$ $\frac{dc}{dx} = -\sin x$ $\left( \sin x = \frac{2t}{1+t^2}, \quad \cos x = \frac{1-t^2}{1+t^2}, \quad \frac{dx}{dt} = \frac{2}{1+t^2} \right)$ |
| 双曲函数 | $u = e^x,$ or $s = \sinh x,$ or $c = \cosh x,$ or $t = \tanh \frac{1}{2}x,$ |
$\frac{du}{dx} = e^x$ $\frac{ds}{dx} = \cosh x$ $\frac{dc}{dx} = \sinh x$ $\frac{dt}{dx} = \frac{1}{2} \sec^2 \frac{1}{2}x$ |
Example 8.62
利用代换 $u = \sqrt{x+2}$ , 计算定积分
$$\int_{-2}^2 \frac{\sqrt{x+2}}{x+6} dx$$
Solution
Setting $u = \sqrt{x+2}$ , or $u^2 = x+2$ , gives $2u du = dx$ . 关于极限, when $x = -2$ , $u = 0$ 且当 $x = 2$ , $u = \sqrt{4} = 2$ .