$$$\tan{\left(4 x \right)}$$$ 的積分
您的輸入
求$$$\int \tan{\left(4 x \right)}\, dx$$$。
解答
令 $$$u=4 x$$$。
則 $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (步驟見»),並可得 $$$dx = \frac{du}{4}$$$。
該積分可改寫為
$${\color{red}{\int{\tan{\left(4 x \right)} d x}}} = {\color{red}{\int{\frac{\tan{\left(u \right)}}{4} d u}}}$$
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 $$$f{\left(u \right)} = \tan{\left(u \right)}$$$:
$${\color{red}{\int{\frac{\tan{\left(u \right)}}{4} d u}}} = {\color{red}{\left(\frac{\int{\tan{\left(u \right)} d u}}{4}\right)}}$$
將切線改寫為 $$$\tan\left( u \right)=\frac{\sin\left( u \right)}{\cos\left( u \right)}$$$:
$$\frac{{\color{red}{\int{\tan{\left(u \right)} d u}}}}{4} = \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{\cos{\left(u \right)}} d u}}}}{4}$$
令 $$$v=\cos{\left(u \right)}$$$。
則 $$$dv=\left(\cos{\left(u \right)}\right)^{\prime }du = - \sin{\left(u \right)} du$$$ (步驟見»),並可得 $$$\sin{\left(u \right)} du = - dv$$$。
該積分變為
$$\frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{\cos{\left(u \right)}} d u}}}}{4} = \frac{{\color{red}{\int{\left(- \frac{1}{v}\right)d v}}}}{4}$$
套用常數倍法則 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$,使用 $$$c=-1$$$ 與 $$$f{\left(v \right)} = \frac{1}{v}$$$:
$$\frac{{\color{red}{\int{\left(- \frac{1}{v}\right)d v}}}}{4} = \frac{{\color{red}{\left(- \int{\frac{1}{v} d v}\right)}}}{4}$$
$$$\frac{1}{v}$$$ 的積分是 $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:
$$- \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{4} = - \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{4}$$
回顧一下 $$$v=\cos{\left(u \right)}$$$:
$$- \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{4} = - \frac{\ln{\left(\left|{{\color{red}{\cos{\left(u \right)}}}}\right| \right)}}{4}$$
回顧一下 $$$u=4 x$$$:
$$- \frac{\ln{\left(\left|{\cos{\left({\color{red}{u}} \right)}}\right| \right)}}{4} = - \frac{\ln{\left(\left|{\cos{\left({\color{red}{\left(4 x\right)}} \right)}}\right| \right)}}{4}$$
因此,
$$\int{\tan{\left(4 x \right)} d x} = - \frac{\ln{\left(\left|{\cos{\left(4 x \right)}}\right| \right)}}{4}$$
加上積分常數:
$$\int{\tan{\left(4 x \right)} d x} = - \frac{\ln{\left(\left|{\cos{\left(4 x \right)}}\right| \right)}}{4}+C$$
答案
$$$\int \tan{\left(4 x \right)}\, dx = - \frac{\ln\left(\left|{\cos{\left(4 x \right)}}\right|\right)}{4} + C$$$A