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

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

设$$$u=\cos{\left(5 x \right)}$$$。

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

积分变为

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

对 $$$c=- \frac{1}{5}$$$ 和 $$$f{\left(u \right)} = u^{2}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=2$$$：

$$- \frac{{\color{red}{\int{u^{2} d u}}}}{5}=- \frac{{\color{red}{\frac{u^{1 + 2}}{1 + 2}}}}{5}=- \frac{{\color{red}{\left(\frac{u^{3}}{3}\right)}}}{5}$$

回忆一下 $$$u=\cos{\left(5 x \right)}$$$:

$$- \frac{{\color{red}{u}}^{3}}{15} = - \frac{{\color{red}{\cos{\left(5 x \right)}}}^{3}}{15}$$

因此，

$$\int{\sin{\left(5 x \right)} \cos^{2}{\left(5 x \right)} d x} = - \frac{\cos^{3}{\left(5 x \right)}}{15}$$

加上积分常数：

$$\int{\sin{\left(5 x \right)} \cos^{2}{\left(5 x \right)} d x} = - \frac{\cos^{3}{\left(5 x \right)}}{15}+C$$

$$$\int \sin{\left(5 x \right)} \cos^{2}{\left(5 x \right)}\, dx = - \frac{\cos^{3}{\left(5 x \right)}}{15} + C$$$A