$$$\frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}}$$$ 的積分
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您的輸入
求$$$\int \frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}}\, dx$$$。
解答
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(x \right)} = \frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}}$$$:
$${\color{red}{\int{\frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}} d x}}} = {\color{red}{\left(\frac{\int{\frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} d x}}{2}\right)}}$$
令 $$$u=\sin{\left(2 x \right)}$$$。
則 $$$du=\left(\sin{\left(2 x \right)}\right)^{\prime }dx = 2 \cos{\left(2 x \right)} dx$$$ (步驟見»),並可得 $$$\cos{\left(2 x \right)} dx = \frac{du}{2}$$$。
該積分變為
$$\frac{{\color{red}{\int{\frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} d x}}}}{2} = \frac{{\color{red}{\int{\frac{1}{2 u} d u}}}}{2}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(u \right)} = \frac{1}{u}$$$:
$$\frac{{\color{red}{\int{\frac{1}{2 u} d u}}}}{2} = \frac{{\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{2}\right)}}}{2}$$
$$$\frac{1}{u}$$$ 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 $$$u=\sin{\left(2 x \right)}$$$:
$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{4} = \frac{\ln{\left(\left|{{\color{red}{\sin{\left(2 x \right)}}}}\right| \right)}}{4}$$
因此,
$$\int{\frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}} d x} = \frac{\ln{\left(\left|{\sin{\left(2 x \right)}}\right| \right)}}{4}$$
加上積分常數:
$$\int{\frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}} d x} = \frac{\ln{\left(\left|{\sin{\left(2 x \right)}}\right| \right)}}{4}+C$$
答案
$$$\int \frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}}\, dx = \frac{\ln\left(\left|{\sin{\left(2 x \right)}}\right|\right)}{4} + C$$$A