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