$$$\operatorname{acos}{\left(y \right)}$$$ 的积分
您的输入
求$$$\int \operatorname{acos}{\left(y \right)}\, dy$$$。
解答
对于积分$$$\int{\operatorname{acos}{\left(y \right)} d y}$$$,使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。
设 $$$\operatorname{u}=\operatorname{acos}{\left(y \right)}$$$ 和 $$$\operatorname{dv}=dy$$$。
则 $$$\operatorname{du}=\left(\operatorname{acos}{\left(y \right)}\right)^{\prime }dy=- \frac{1}{\sqrt{1 - y^{2}}} dy$$$ (步骤见 »),并且 $$$\operatorname{v}=\int{1 d y}=y$$$ (步骤见 »)。
所以,
$${\color{red}{\int{\operatorname{acos}{\left(y \right)} d y}}}={\color{red}{\left(\operatorname{acos}{\left(y \right)} \cdot y-\int{y \cdot \left(- \frac{1}{\sqrt{1 - y^{2}}}\right) d y}\right)}}={\color{red}{\left(y \operatorname{acos}{\left(y \right)} - \int{\left(- \frac{y}{\sqrt{1 - y^{2}}}\right)d y}\right)}}$$
设$$$u=1 - y^{2}$$$。
则$$$du=\left(1 - y^{2}\right)^{\prime }dy = - 2 y dy$$$ (步骤见»),并有$$$y dy = - \frac{du}{2}$$$。
积分变为
$$y \operatorname{acos}{\left(y \right)} - {\color{red}{\int{\left(- \frac{y}{\sqrt{1 - y^{2}}}\right)d y}}} = y \operatorname{acos}{\left(y \right)} - {\color{red}{\int{\frac{1}{2 \sqrt{u}} d u}}}$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$$y \operatorname{acos}{\left(y \right)} - {\color{red}{\int{\frac{1}{2 \sqrt{u}} d u}}} = y \operatorname{acos}{\left(y \right)} - {\color{red}{\left(\frac{\int{\frac{1}{\sqrt{u}} d u}}{2}\right)}}$$
应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=- \frac{1}{2}$$$:
$$y \operatorname{acos}{\left(y \right)} - \frac{{\color{red}{\int{\frac{1}{\sqrt{u}} d u}}}}{2}=y \operatorname{acos}{\left(y \right)} - \frac{{\color{red}{\int{u^{- \frac{1}{2}} d u}}}}{2}=y \operatorname{acos}{\left(y \right)} - \frac{{\color{red}{\frac{u^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}}}{2}=y \operatorname{acos}{\left(y \right)} - \frac{{\color{red}{\left(2 u^{\frac{1}{2}}\right)}}}{2}=y \operatorname{acos}{\left(y \right)} - \frac{{\color{red}{\left(2 \sqrt{u}\right)}}}{2}$$
回忆一下 $$$u=1 - y^{2}$$$:
$$y \operatorname{acos}{\left(y \right)} - \sqrt{{\color{red}{u}}} = y \operatorname{acos}{\left(y \right)} - \sqrt{{\color{red}{\left(1 - y^{2}\right)}}}$$
因此,
$$\int{\operatorname{acos}{\left(y \right)} d y} = y \operatorname{acos}{\left(y \right)} - \sqrt{1 - y^{2}}$$
加上积分常数:
$$\int{\operatorname{acos}{\left(y \right)} d y} = y \operatorname{acos}{\left(y \right)} - \sqrt{1 - y^{2}}+C$$
答案
$$$\int \operatorname{acos}{\left(y \right)}\, dy = \left(y \operatorname{acos}{\left(y \right)} - \sqrt{1 - y^{2}}\right) + C$$$A