$$$\cos^{2}{\left(5 x \right)} \tan{\left(5 x \right)}$$$ 的积分

求$$$\int \cos^{2}{\left(5 x \right)} \tan{\left(5 x \right)}\, dx$$$。

改写被积函数:

$${\color{red}{\int{\cos^{2}{\left(5 x \right)} \tan{\left(5 x \right)} d x}}} = {\color{red}{\int{\sin{\left(5 x \right)} \cos{\left(5 x \right)} d x}}}$$

设$$$u=\sin{\left(5 x \right)}$$$。

则$$$du=\left(\sin{\left(5 x \right)}\right)^{\prime }dx = 5 \cos{\left(5 x \right)} dx$$$ (步骤见»)，并有$$$\cos{\left(5 x \right)} dx = \frac{du}{5}$$$。

该积分可以改写为

$${\color{red}{\int{\sin{\left(5 x \right)} \cos{\left(5 x \right)} d x}}} = {\color{red}{\int{\frac{u}{5} d u}}}$$

对 $$$c=\frac{1}{5}$$$ 和 $$$f{\left(u \right)} = u$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{\frac{u}{5} d u}}} = {\color{red}{\left(\frac{\int{u d u}}{5}\right)}}$$

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=1$$$：

$$\frac{{\color{red}{\int{u d u}}}}{5}=\frac{{\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{5}=\frac{{\color{red}{\left(\frac{u^{2}}{2}\right)}}}{5}$$

回忆一下 $$$u=\sin{\left(5 x \right)}$$$:

$$\frac{{\color{red}{u}}^{2}}{10} = \frac{{\color{red}{\sin{\left(5 x \right)}}}^{2}}{10}$$

因此，

$$\int{\cos^{2}{\left(5 x \right)} \tan{\left(5 x \right)} d x} = \frac{\sin^{2}{\left(5 x \right)}}{10}$$

加上积分常数：

$$\int{\cos^{2}{\left(5 x \right)} \tan{\left(5 x \right)} d x} = \frac{\sin^{2}{\left(5 x \right)}}{10}+C$$

$$$\int \cos^{2}{\left(5 x \right)} \tan{\left(5 x \right)}\, dx = \frac{\sin^{2}{\left(5 x \right)}}{10} + C$$$A