$$$\sqrt{x^{2} - 1}$$$ 的积分
您的输入
求$$$\int \sqrt{x^{2} - 1}\, dx$$$。
解答
设$$$x=\cosh{\left(u \right)}$$$。
则$$$dx=\left(\cosh{\left(u \right)}\right)^{\prime }du = \sinh{\left(u \right)} du$$$(步骤见»)。
此外,可得$$$u=\operatorname{acosh}{\left(x \right)}$$$。
所以,
$$$\sqrt{x^{2} - 1} = \sqrt{\cosh^{2}{\left( u \right)} - 1}$$$
利用恒等式 $$$\cosh^{2}{\left( u \right)} - 1 = \sinh^{2}{\left( u \right)}$$$:
$$$\sqrt{\cosh^{2}{\left( u \right)} - 1}=\sqrt{\sinh^{2}{\left( u \right)}}$$$
假设$$$\sinh{\left( u \right)} \ge 0$$$,我们得到如下结果:
$$$\sqrt{\sinh^{2}{\left( u \right)}} = \sinh{\left( u \right)}$$$
因此,
$${\color{red}{\int{\sqrt{x^{2} - 1} d x}}} = {\color{red}{\int{\sinh^{2}{\left(u \right)} d u}}}$$
应用降幂公式 $$$\sinh^{2}{\left(\alpha \right)} = \frac{\cosh{\left(2 \alpha \right)}}{2} - \frac{1}{2}$$$,并令 $$$\alpha= u $$$:
$${\color{red}{\int{\sinh^{2}{\left(u \right)} d u}}} = {\color{red}{\int{\left(\frac{\cosh{\left(2 u \right)}}{2} - \frac{1}{2}\right)d u}}}$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \cosh{\left(2 u \right)} - 1$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$${\color{red}{\int{\left(\frac{\cosh{\left(2 u \right)}}{2} - \frac{1}{2}\right)d u}}} = {\color{red}{\left(\frac{\int{\left(\cosh{\left(2 u \right)} - 1\right)d u}}{2}\right)}}$$
逐项积分:
$$\frac{{\color{red}{\int{\left(\cosh{\left(2 u \right)} - 1\right)d u}}}}{2} = \frac{{\color{red}{\left(- \int{1 d u} + \int{\cosh{\left(2 u \right)} d u}\right)}}}{2}$$
应用常数法则 $$$\int c\, du = c u$$$,使用 $$$c=1$$$:
$$\frac{\int{\cosh{\left(2 u \right)} d u}}{2} - \frac{{\color{red}{\int{1 d u}}}}{2} = \frac{\int{\cosh{\left(2 u \right)} d u}}{2} - \frac{{\color{red}{u}}}{2}$$
设$$$v=2 u$$$。
则$$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (步骤见»),并有$$$du = \frac{dv}{2}$$$。
该积分可以改写为
$$- \frac{u}{2} + \frac{{\color{red}{\int{\cosh{\left(2 u \right)} d u}}}}{2} = - \frac{u}{2} + \frac{{\color{red}{\int{\frac{\cosh{\left(v \right)}}{2} d v}}}}{2}$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(v \right)} = \cosh{\left(v \right)}$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$:
$$- \frac{u}{2} + \frac{{\color{red}{\int{\frac{\cosh{\left(v \right)}}{2} d v}}}}{2} = - \frac{u}{2} + \frac{{\color{red}{\left(\frac{\int{\cosh{\left(v \right)} d v}}{2}\right)}}}{2}$$
双曲余弦的积分为 $$$\int{\cosh{\left(v \right)} d v} = \sinh{\left(v \right)}$$$:
$$- \frac{u}{2} + \frac{{\color{red}{\int{\cosh{\left(v \right)} d v}}}}{4} = - \frac{u}{2} + \frac{{\color{red}{\sinh{\left(v \right)}}}}{4}$$
回忆一下 $$$v=2 u$$$:
$$- \frac{u}{2} + \frac{\sinh{\left({\color{red}{v}} \right)}}{4} = - \frac{u}{2} + \frac{\sinh{\left({\color{red}{\left(2 u\right)}} \right)}}{4}$$
回忆一下 $$$u=\operatorname{acosh}{\left(x \right)}$$$:
$$\frac{\sinh{\left(2 {\color{red}{u}} \right)}}{4} - \frac{{\color{red}{u}}}{2} = \frac{\sinh{\left(2 {\color{red}{\operatorname{acosh}{\left(x \right)}}} \right)}}{4} - \frac{{\color{red}{\operatorname{acosh}{\left(x \right)}}}}{2}$$
因此,
$$\int{\sqrt{x^{2} - 1} d x} = \frac{\sinh{\left(2 \operatorname{acosh}{\left(x \right)} \right)}}{4} - \frac{\operatorname{acosh}{\left(x \right)}}{2}$$
使用公式 $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$,化简该表达式:
$$\int{\sqrt{x^{2} - 1} d x} = \frac{x \sqrt{x - 1} \sqrt{x + 1}}{2} - \frac{\operatorname{acosh}{\left(x \right)}}{2}$$
进一步化简:
$$\int{\sqrt{x^{2} - 1} d x} = \frac{x \sqrt{x - 1} \sqrt{x + 1} - \operatorname{acosh}{\left(x \right)}}{2}$$
加上积分常数:
$$\int{\sqrt{x^{2} - 1} d x} = \frac{x \sqrt{x - 1} \sqrt{x + 1} - \operatorname{acosh}{\left(x \right)}}{2}+C$$
答案
$$$\int \sqrt{x^{2} - 1}\, dx = \frac{x \sqrt{x - 1} \sqrt{x + 1} - \operatorname{acosh}{\left(x \right)}}{2} + C$$$A