$$$\sec{\left(\pi x \right)}$$$ 的積分
您的輸入
求$$$\int \sec{\left(\pi x \right)}\, dx$$$。
解答
令 $$$u=\pi x$$$。
則 $$$du=\left(\pi x\right)^{\prime }dx = \pi dx$$$ (步驟見»),並可得 $$$dx = \frac{du}{\pi}$$$。
該積分可改寫為
$${\color{red}{\int{\sec{\left(\pi x \right)} d x}}} = {\color{red}{\int{\frac{\sec{\left(u \right)}}{\pi} d u}}}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=\frac{1}{\pi}$$$ 與 $$$f{\left(u \right)} = \sec{\left(u \right)}$$$:
$${\color{red}{\int{\frac{\sec{\left(u \right)}}{\pi} d u}}} = {\color{red}{\frac{\int{\sec{\left(u \right)} d u}}{\pi}}}$$
將正割改寫為 $$$\sec\left( u \right)=\frac{1}{\cos\left( u \right)}$$$:
$$\frac{{\color{red}{\int{\sec{\left(u \right)} d u}}}}{\pi} = \frac{{\color{red}{\int{\frac{1}{\cos{\left(u \right)}} d u}}}}{\pi}$$
使用公式 $$$\cos\left( u \right)=\sin\left( u + \frac{\pi}{2}\right)$$$ 將餘弦用正弦表示,然後使用二倍角公式 $$$\sin\left( u \right)=2\sin\left(\frac{ u }{2}\right)\cos\left(\frac{ u }{2}\right)$$$ 將正弦改寫。:
$$\frac{{\color{red}{\int{\frac{1}{\cos{\left(u \right)}} d u}}}}{\pi} = \frac{{\color{red}{\int{\frac{1}{2 \sin{\left(\frac{u}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{\pi}$$
將分子與分母同時乘以 $$$\sec^2\left(\frac{ u }{2} + \frac{\pi}{4} \right)$$$:
$$\frac{{\color{red}{\int{\frac{1}{2 \sin{\left(\frac{u}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{\pi} = \frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{\pi}$$
令 $$$v=\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$。
則 $$$dv=\left(\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}\right)^{\prime }du = \frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2} du$$$ (步驟見»),並可得 $$$\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} du = 2 dv$$$。
所以,
$$\frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{\pi} = \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{\pi}$$
$$$\frac{1}{v}$$$ 的積分是 $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:
$$\frac{{\color{red}{\int{\frac{1}{v} d v}}}}{\pi} = \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{\pi}$$
回顧一下 $$$v=\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$:
$$\frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{\pi} = \frac{\ln{\left(\left|{{\color{red}{\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}}}\right| \right)}}{\pi}$$
回顧一下 $$$u=\pi x$$$:
$$\frac{\ln{\left(\left|{\tan{\left(\frac{\pi}{4} + \frac{{\color{red}{u}}}{2} \right)}}\right| \right)}}{\pi} = \frac{\ln{\left(\left|{\tan{\left(\frac{\pi}{4} + \frac{{\color{red}{\pi x}}}{2} \right)}}\right| \right)}}{\pi}$$
因此,
$$\int{\sec{\left(\pi x \right)} d x} = \frac{\ln{\left(\left|{\tan{\left(\frac{\pi x}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{\pi}$$
化簡:
$$\int{\sec{\left(\pi x \right)} d x} = \frac{\ln{\left(\left|{\tan{\left(\pi \left(\frac{x}{2} + \frac{1}{4}\right) \right)}}\right| \right)}}{\pi}$$
加上積分常數:
$$\int{\sec{\left(\pi x \right)} d x} = \frac{\ln{\left(\left|{\tan{\left(\pi \left(\frac{x}{2} + \frac{1}{4}\right) \right)}}\right| \right)}}{\pi}+C$$
答案
$$$\int \sec{\left(\pi x \right)}\, dx = \frac{\ln\left(\left|{\tan{\left(\pi \left(\frac{x}{2} + \frac{1}{4}\right) \right)}}\right|\right)}{\pi} + C$$$A