$$$\sec^{3}{\left(x \right)}$$$ 的積分

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

對於積分 $$$\int{\sec^{3}{\left(x \right)} d x}$$$，使用分部積分法 $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。

令 $$$\operatorname{u}=\sec{\left(x \right)}$$$ 與 $$$\operatorname{dv}=\sec^{2}{\left(x \right)} dx$$$。

則 $$$\operatorname{du}=\left(\sec{\left(x \right)}\right)^{\prime }dx=\tan{\left(x \right)} \sec{\left(x \right)} dx$$$（步驟見 »），且 $$$\operatorname{v}=\int{\sec^{2}{\left(x \right)} d x}=\tan{\left(x \right)}$$$（步驟見 »）。

該積分可改寫為

$$\int{\sec^{3}{\left(x \right)} d x}=\sec{\left(x \right)} \cdot \tan{\left(x \right)}-\int{\tan{\left(x \right)} \cdot \tan{\left(x \right)} \sec{\left(x \right)} d x}=\tan{\left(x \right)} \sec{\left(x \right)} - \int{\tan^{2}{\left(x \right)} \sec{\left(x \right)} d x}$$

套用公式 $$$\tan^{2}{\left(x \right)} = \sec^{2}{\left(x \right)} - 1$$$:

$$\tan{\left(x \right)} \sec{\left(x \right)} - \int{\tan^{2}{\left(x \right)} \sec{\left(x \right)} d x}=\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{2}{\left(x \right)} - 1\right) \sec{\left(x \right)} d x}$$

展開：

$$\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{2}{\left(x \right)} - 1\right) \sec{\left(x \right)} d x}=\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{3}{\left(x \right)} - \sec{\left(x \right)}\right)d x}$$

差的積分等於積分之差：

$$\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{3}{\left(x \right)} - \sec{\left(x \right)}\right)d x}=\tan{\left(x \right)} \sec{\left(x \right)} + \int{\sec{\left(x \right)} d x} - \int{\sec^{3}{\left(x \right)} d x}$$

因此，我們得到關於該積分的下列簡單線性方程：

$${\color{red}{\int{\sec^{3}{\left(x \right)} d x}}}=\tan{\left(x \right)} \sec{\left(x \right)} + \int{\sec{\left(x \right)} d x} - {\color{red}{\int{\sec^{3}{\left(x \right)} d x}}}$$

解得

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

將正割改寫為 $$$\sec\left(x\right)=\frac{1}{\cos\left(x\right)}$$$:

$$\frac{\tan{\left(x \right)} \sec{\left(x \right)}}{2} + \frac{{\color{red}{\int{\sec{\left(x \right)} d x}}}}{2} = \frac{\tan{\left(x \right)} \sec{\left(x \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}}}{2}$$

使用公式 $$$\cos\left(x\right)=\sin\left(x + \frac{\pi}{2}\right)$$$ 將餘弦用正弦表示，然後使用二倍角公式 $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$ 將正弦改寫。:

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

將分子與分母同時乘以 $$$\sec^2\left(\frac{x}{2} + \frac{\pi}{4} \right)$$$:

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

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

則 $$$du=\left(\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2} dx$$$ (步驟見»)，並可得 $$$\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} dx = 2 du$$$。

該積分變為

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

$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：

$$\frac{\tan{\left(x \right)} \sec{\left(x \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = \frac{\tan{\left(x \right)} \sec{\left(x \right)}}{2} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

回顧一下 $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$：

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

因此，

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

加上積分常數：

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

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