$$$\sec^{5}{\left(x \right)}$$$ 的积分
您的输入
求$$$\int \sec^{5}{\left(x \right)}\, dx$$$。
解答
对于积分$$$\int{\sec^{5}{\left(x \right)} d x}$$$,使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。
设 $$$\operatorname{u}=\sec^{3}{\left(x \right)}$$$ 和 $$$\operatorname{dv}=\sec^{2}{\left(x \right)} dx$$$。
则 $$$\operatorname{du}=\left(\sec^{3}{\left(x \right)}\right)^{\prime }dx=3 \tan{\left(x \right)} \sec^{3}{\left(x \right)} dx$$$ (步骤见 »),并且 $$$\operatorname{v}=\int{\sec^{2}{\left(x \right)} d x}=\tan{\left(x \right)}$$$ (步骤见 »)。
所以,
$$\int{\sec^{5}{\left(x \right)} d x}=\sec^{3}{\left(x \right)} \cdot \tan{\left(x \right)}-\int{\tan{\left(x \right)} \cdot 3 \tan{\left(x \right)} \sec^{3}{\left(x \right)} d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} - \int{3 \tan^{2}{\left(x \right)} \sec^{3}{\left(x \right)} d x}$$
提出常数:
$$\tan{\left(x \right)} \sec^{3}{\left(x \right)} - \int{3 \tan^{2}{\left(x \right)} \sec^{3}{\left(x \right)} d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\tan^{2}{\left(x \right)} \sec^{3}{\left(x \right)} d x}$$
应用公式 $$$\tan^{2}{\left(x \right)} = \sec^{2}{\left(x \right)} - 1$$$:
$$\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\tan^{2}{\left(x \right)} \sec^{3}{\left(x \right)} d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\left(\sec^{2}{\left(x \right)} - 1\right) \sec^{3}{\left(x \right)} d x}$$
展开:
$$\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\left(\sec^{2}{\left(x \right)} - 1\right) \sec^{3}{\left(x \right)} d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\left(\sec^{5}{\left(x \right)} - \sec^{3}{\left(x \right)}\right)d x}$$
差的积分等于积分的差:
$$\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\left(\sec^{5}{\left(x \right)} - \sec^{3}{\left(x \right)}\right)d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} + 3 \int{\sec^{3}{\left(x \right)} d x} - 3 \int{\sec^{5}{\left(x \right)} d x}$$
因此,我们得到关于该积分的以下简单线性方程:
$${\color{red}{\int{\sec^{5}{\left(x \right)} d x}}}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} + 3 \int{\sec^{3}{\left(x \right)} d x} - 3 {\color{red}{\int{\sec^{5}{\left(x \right)} d x}}}$$
求解可得
$$\int{\sec^{5}{\left(x \right)} d x}=\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \int{\sec^{3}{\left(x \right)} d x}}{4}$$
对于积分$$$\int{\sec^{3}{\left(x \right)} d x}$$$,使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。
设 $$$\operatorname{u}=\sec{\left(x \right)}$$$ 和 $$$\operatorname{dv}=\sec^{2}{\left(x \right)} dx$$$。
则 $$$\operatorname{du}=\left(\sec{\left(x \right)}\right)^{\prime }dx=\tan{\left(x \right)} \sec{\left(x \right)} dx$$$ (步骤见 »),并且 $$$\operatorname{v}=\int{\sec^{2}{\left(x \right)} d x}=\tan{\left(x \right)}$$$ (步骤见 »)。
因此,
$$\int{\sec^{3}{\left(x \right)} d x}=\sec{\left(x \right)} \cdot \tan{\left(x \right)}-\int{\tan{\left(x \right)} \cdot \tan{\left(x \right)} \sec{\left(x \right)} d x}=\tan{\left(x \right)} \sec{\left(x \right)} - \int{\tan^{2}{\left(x \right)} \sec{\left(x \right)} d x}$$
应用公式 $$$\tan^{2}{\left(x \right)} = \sec^{2}{\left(x \right)} - 1$$$:
$$\tan{\left(x \right)} \sec{\left(x \right)} - \int{\tan^{2}{\left(x \right)} \sec{\left(x \right)} d x}=\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{2}{\left(x \right)} - 1\right) \sec{\left(x \right)} d x}$$
展开:
$$\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{2}{\left(x \right)} - 1\right) \sec{\left(x \right)} d x}=\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{3}{\left(x \right)} - \sec{\left(x \right)}\right)d x}$$
差的积分等于积分的差:
$$\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{3}{\left(x \right)} - \sec{\left(x \right)}\right)d x}=\tan{\left(x \right)} \sec{\left(x \right)} + \int{\sec{\left(x \right)} d x} - \int{\sec^{3}{\left(x \right)} d x}$$
因此,我们得到关于该积分的以下简单线性方程:
$${\color{red}{\int{\sec^{3}{\left(x \right)} d x}}}=\tan{\left(x \right)} \sec{\left(x \right)} + \int{\sec{\left(x \right)} d x} - {\color{red}{\int{\sec^{3}{\left(x \right)} d x}}}$$
求解可得
$$\int{\sec^{3}{\left(x \right)} d x}=\frac{\tan{\left(x \right)} \sec{\left(x \right)}}{2} + \frac{\int{\sec{\left(x \right)} d x}}{2}$$
因此,
$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 {\color{red}{\int{\sec^{3}{\left(x \right)} d x}}}}{4} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 {\color{red}{\left(\frac{\tan{\left(x \right)} \sec{\left(x \right)}}{2} + \frac{\int{\sec{\left(x \right)} d x}}{2}\right)}}}{4}$$
将正割改写为 $$$\sec\left(x\right)=\frac{1}{\cos\left(x\right)}$$$:
$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\sec{\left(x \right)} d x}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}}}{8}$$
使用公式$$$\cos\left(x\right)=\sin\left(x + \frac{\pi}{2}\right)$$$将余弦用正弦表示,然后使用二倍角公式$$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$将正弦改写。:
$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}}{8}$$
将分子和分母同时乘以 $$$\sec^2\left(\frac{x}{2} + \frac{\pi}{4} \right)$$$:
$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}}{8}$$
设$$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$。
则$$$du=\left(\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2} dx$$$ (步骤见»),并有$$$\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} dx = 2 du$$$。
积分变为
$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{u} d u}}}}{8}$$
$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{u} d u}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8}$$
回忆一下 $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$:
$$\frac{3 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} = \frac{3 \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}}}\right| \right)}}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8}$$
因此,
$$\int{\sec^{5}{\left(x \right)} d x} = \frac{3 \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8}$$
加上积分常数:
$$\int{\sec^{5}{\left(x \right)} d x} = \frac{3 \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8}+C$$
答案
$$$\int \sec^{5}{\left(x \right)}\, dx = \left(\frac{3 \ln\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right|\right)}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8}\right) + C$$$A