$$$4 x^{3} - \frac{1}{\cos{\left(2 x \right)}}$$$ 的积分

求$$$\int \left(4 x^{3} - \frac{1}{\cos{\left(2 x \right)}}\right)\, dx$$$。

逐项积分：

$${\color{red}{\int{\left(4 x^{3} - \frac{1}{\cos{\left(2 x \right)}}\right)d x}}} = {\color{red}{\left(\int{4 x^{3} d x} - \int{\frac{1}{\cos{\left(2 x \right)}} d x}\right)}}$$

设$$$u=2 x$$$。

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

因此，

$$\int{4 x^{3} d x} - {\color{red}{\int{\frac{1}{\cos{\left(2 x \right)}} d x}}} = \int{4 x^{3} d x} - {\color{red}{\int{\frac{1}{2 \cos{\left(u \right)}} d u}}}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \frac{1}{\cos{\left(u \right)}}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$\int{4 x^{3} d x} - {\color{red}{\int{\frac{1}{2 \cos{\left(u \right)}} d u}}} = \int{4 x^{3} d x} - {\color{red}{\left(\frac{\int{\frac{1}{\cos{\left(u \right)}} d u}}{2}\right)}}$$

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。:

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

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

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

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$$$\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} du = 2 dv$$$。

所以，

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

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

$$\int{4 x^{3} d x} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2} = \int{4 x^{3} d x} - \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{2}$$

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

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

回忆一下 $$$u=2 x$$$:

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

对 $$$c=4$$$ 和 $$$f{\left(x \right)} = x^{3}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

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

应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=3$$$：

$$- \frac{\ln{\left(\left|{\tan{\left(x + \frac{\pi}{4} \right)}}\right| \right)}}{2} + 4 {\color{red}{\int{x^{3} d x}}}=- \frac{\ln{\left(\left|{\tan{\left(x + \frac{\pi}{4} \right)}}\right| \right)}}{2} + 4 {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}=- \frac{\ln{\left(\left|{\tan{\left(x + \frac{\pi}{4} \right)}}\right| \right)}}{2} + 4 {\color{red}{\left(\frac{x^{4}}{4}\right)}}$$

因此，

$$\int{\left(4 x^{3} - \frac{1}{\cos{\left(2 x \right)}}\right)d x} = x^{4} - \frac{\ln{\left(\left|{\tan{\left(x + \frac{\pi}{4} \right)}}\right| \right)}}{2}$$

加上积分常数：

$$\int{\left(4 x^{3} - \frac{1}{\cos{\left(2 x \right)}}\right)d x} = x^{4} - \frac{\ln{\left(\left|{\tan{\left(x + \frac{\pi}{4} \right)}}\right| \right)}}{2}+C$$

$$$\int \left(4 x^{3} - \frac{1}{\cos{\left(2 x \right)}}\right)\, dx = \left(x^{4} - \frac{\ln\left(\left|{\tan{\left(x + \frac{\pi}{4} \right)}}\right|\right)}{2}\right) + C$$$A