$$$4 \tan{\left(3 x \right)}$$$ 的积分
您的输入
求$$$\int 4 \tan{\left(3 x \right)}\, dx$$$。
解答
对 $$$c=4$$$ 和 $$$f{\left(x \right)} = \tan{\left(3 x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$${\color{red}{\int{4 \tan{\left(3 x \right)} d x}}} = {\color{red}{\left(4 \int{\tan{\left(3 x \right)} d x}\right)}}$$
设$$$u=3 x$$$。
则$$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (步骤见»),并有$$$dx = \frac{du}{3}$$$。
积分变为
$$4 {\color{red}{\int{\tan{\left(3 x \right)} d x}}} = 4 {\color{red}{\int{\frac{\tan{\left(u \right)}}{3} d u}}}$$
对 $$$c=\frac{1}{3}$$$ 和 $$$f{\left(u \right)} = \tan{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$$4 {\color{red}{\int{\frac{\tan{\left(u \right)}}{3} d u}}} = 4 {\color{red}{\left(\frac{\int{\tan{\left(u \right)} d u}}{3}\right)}}$$
将正切表示为 $$$\tan\left( u \right)=\frac{\sin\left( u \right)}{\cos\left( u \right)}$$$:
$$\frac{4 {\color{red}{\int{\tan{\left(u \right)} d u}}}}{3} = \frac{4 {\color{red}{\int{\frac{\sin{\left(u \right)}}{\cos{\left(u \right)}} d u}}}}{3}$$
设$$$v=\cos{\left(u \right)}$$$。
则$$$dv=\left(\cos{\left(u \right)}\right)^{\prime }du = - \sin{\left(u \right)} du$$$ (步骤见»),并有$$$\sin{\left(u \right)} du = - dv$$$。
因此,
$$\frac{4 {\color{red}{\int{\frac{\sin{\left(u \right)}}{\cos{\left(u \right)}} d u}}}}{3} = \frac{4 {\color{red}{\int{\left(- \frac{1}{v}\right)d v}}}}{3}$$
对 $$$c=-1$$$ 和 $$$f{\left(v \right)} = \frac{1}{v}$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$:
$$\frac{4 {\color{red}{\int{\left(- \frac{1}{v}\right)d v}}}}{3} = \frac{4 {\color{red}{\left(- \int{\frac{1}{v} d v}\right)}}}{3}$$
$$$\frac{1}{v}$$$ 的积分为 $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:
$$- \frac{4 {\color{red}{\int{\frac{1}{v} d v}}}}{3} = - \frac{4 {\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{3}$$
回忆一下 $$$v=\cos{\left(u \right)}$$$:
$$- \frac{4 \ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{3} = - \frac{4 \ln{\left(\left|{{\color{red}{\cos{\left(u \right)}}}}\right| \right)}}{3}$$
回忆一下 $$$u=3 x$$$:
$$- \frac{4 \ln{\left(\left|{\cos{\left({\color{red}{u}} \right)}}\right| \right)}}{3} = - \frac{4 \ln{\left(\left|{\cos{\left({\color{red}{\left(3 x\right)}} \right)}}\right| \right)}}{3}$$
因此,
$$\int{4 \tan{\left(3 x \right)} d x} = - \frac{4 \ln{\left(\left|{\cos{\left(3 x \right)}}\right| \right)}}{3}$$
加上积分常数:
$$\int{4 \tan{\left(3 x \right)} d x} = - \frac{4 \ln{\left(\left|{\cos{\left(3 x \right)}}\right| \right)}}{3}+C$$
答案
$$$\int 4 \tan{\left(3 x \right)}\, dx = - \frac{4 \ln\left(\left|{\cos{\left(3 x \right)}}\right|\right)}{3} + C$$$A