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