$$$\csc{\left(2 x \right)}$$$ 的积分
您的输入
求$$$\int \csc{\left(2 x \right)}\, dx$$$。
解答
设$$$u=2 x$$$。
则$$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (步骤见»),并有$$$dx = \frac{du}{2}$$$。
积分变为
$${\color{red}{\int{\csc{\left(2 x \right)} d x}}} = {\color{red}{\int{\frac{\csc{\left(u \right)}}{2} d u}}}$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \csc{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$${\color{red}{\int{\frac{\csc{\left(u \right)}}{2} d u}}} = {\color{red}{\left(\frac{\int{\csc{\left(u \right)} d u}}{2}\right)}}$$
将余割改写为$$$\csc\left( u \right)=\frac{1}{\sin\left( u \right)}$$$:
$$\frac{{\color{red}{\int{\csc{\left(u \right)} d u}}}}{2} = \frac{{\color{red}{\int{\frac{1}{\sin{\left(u \right)}} d u}}}}{2}$$
使用二倍角公式 $$$\sin\left( u \right)=2\sin\left(\frac{ u }{2}\right)\cos\left(\frac{ u }{2}\right)$$$ 改写正弦:
$$\frac{{\color{red}{\int{\frac{1}{\sin{\left(u \right)}} d u}}}}{2} = \frac{{\color{red}{\int{\frac{1}{2 \sin{\left(\frac{u}{2} \right)} \cos{\left(\frac{u}{2} \right)}} d u}}}}{2}$$
将分子和分母同时乘以 $$$\sec^2\left(\frac{ u }{2} \right)$$$:
$$\frac{{\color{red}{\int{\frac{1}{2 \sin{\left(\frac{u}{2} \right)} \cos{\left(\frac{u}{2} \right)}} d u}}}}{2} = \frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{u}{2} \right)}}{2 \tan{\left(\frac{u}{2} \right)}} d u}}}}{2}$$
设$$$v=\tan{\left(\frac{u}{2} \right)}$$$。
则$$$dv=\left(\tan{\left(\frac{u}{2} \right)}\right)^{\prime }du = \frac{\sec^{2}{\left(\frac{u}{2} \right)}}{2} du$$$ (步骤见»),并有$$$\sec^{2}{\left(\frac{u}{2} \right)} du = 2 dv$$$。
因此,
$$\frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{u}{2} \right)}}{2 \tan{\left(\frac{u}{2} \right)}} d u}}}}{2} = \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2}$$
$$$\frac{1}{v}$$$ 的积分为 $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:
$$\frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2} = \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{2}$$
回忆一下 $$$v=\tan{\left(\frac{u}{2} \right)}$$$:
$$\frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} = \frac{\ln{\left(\left|{{\color{red}{\tan{\left(\frac{u}{2} \right)}}}}\right| \right)}}{2}$$
回忆一下 $$$u=2 x$$$:
$$\frac{\ln{\left(\left|{\tan{\left(\frac{{\color{red}{u}}}{2} \right)}}\right| \right)}}{2} = \frac{\ln{\left(\left|{\tan{\left(\frac{{\color{red}{\left(2 x\right)}}}{2} \right)}}\right| \right)}}{2}$$
因此,
$$\int{\csc{\left(2 x \right)} d x} = \frac{\ln{\left(\left|{\tan{\left(x \right)}}\right| \right)}}{2}$$
加上积分常数:
$$\int{\csc{\left(2 x \right)} d x} = \frac{\ln{\left(\left|{\tan{\left(x \right)}}\right| \right)}}{2}+C$$
答案
$$$\int \csc{\left(2 x \right)}\, dx = \frac{\ln\left(\left|{\tan{\left(x \right)}}\right|\right)}{2} + C$$$A