$$$x^{2} \sec{\left(x^{3} \right)}$$$ 的积分

求$$$\int x^{2} \sec{\left(x^{3} \right)}\, dx$$$。

设$$$u=x^{3}$$$。

则$$$du=\left(x^{3}\right)^{\prime }dx = 3 x^{2} dx$$$ (步骤见»)，并有$$$x^{2} dx = \frac{du}{3}$$$。

该积分可以改写为

$${\color{red}{\int{x^{2} \sec{\left(x^{3} \right)} d x}}} = {\color{red}{\int{\frac{\sec{\left(u \right)}}{3} d u}}}$$

对 $$$c=\frac{1}{3}$$$ 和 $$$f{\left(u \right)} = \sec{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{\frac{\sec{\left(u \right)}}{3} d u}}} = {\color{red}{\left(\frac{\int{\sec{\left(u \right)} d u}}{3}\right)}}$$

将正割改写为 $$$\sec\left( u \right)=\frac{1}{\cos\left( u \right)}$$$:

$$\frac{{\color{red}{\int{\sec{\left(u \right)} d u}}}}{3} = \frac{{\color{red}{\int{\frac{1}{\cos{\left(u \right)}} d u}}}}{3}$$

使用公式$$$\cos\left( u \right)=\sin\left( u + \frac{\pi}{2}\right)$$$将余弦用正弦表示，然后使用二倍角公式$$$\sin\left( u \right)=2\sin\left(\frac{ u }{2}\right)\cos\left(\frac{ u }{2}\right)$$$将正弦改写。:

$$\frac{{\color{red}{\int{\frac{1}{\cos{\left(u \right)}} d u}}}}{3} = \frac{{\color{red}{\int{\frac{1}{2 \sin{\left(\frac{u}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{3}$$

将分子和分母同时乘以 $$$\sec^2\left(\frac{ u }{2} + \frac{\pi}{4} \right)$$$:

$$\frac{{\color{red}{\int{\frac{1}{2 \sin{\left(\frac{u}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{3} = \frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{3}$$

设$$$v=\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$。

则$$$dv=\left(\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}\right)^{\prime }du = \frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2} du$$$ (步骤见»)，并有$$$\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} du = 2 dv$$$。

因此，

$$\frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{3} = \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{3}$$

$$$\frac{1}{v}$$$ 的积分为 $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$\frac{{\color{red}{\int{\frac{1}{v} d v}}}}{3} = \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{3}$$

回忆一下 $$$v=\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$:

$$\frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{3} = \frac{\ln{\left(\left|{{\color{red}{\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}}}\right| \right)}}{3}$$

回忆一下 $$$u=x^{3}$$$:

$$\frac{\ln{\left(\left|{\tan{\left(\frac{\pi}{4} + \frac{{\color{red}{u}}}{2} \right)}}\right| \right)}}{3} = \frac{\ln{\left(\left|{\tan{\left(\frac{\pi}{4} + \frac{{\color{red}{x^{3}}}}{2} \right)}}\right| \right)}}{3}$$

因此，

$$\int{x^{2} \sec{\left(x^{3} \right)} d x} = \frac{\ln{\left(\left|{\tan{\left(\frac{x^{3}}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{3}$$

加上积分常数：

$$\int{x^{2} \sec{\left(x^{3} \right)} d x} = \frac{\ln{\left(\left|{\tan{\left(\frac{x^{3}}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{3}+C$$

$$$\int x^{2} \sec{\left(x^{3} \right)}\, dx = \frac{\ln\left(\left|{\tan{\left(\frac{x^{3}}{2} + \frac{\pi}{4} \right)}}\right|\right)}{3} + C$$$A