$$$\sin^{2}{\left(t \right)} \cos^{4}{\left(t \right)}$$$ 的积分

求$$$\int \sin^{2}{\left(t \right)} \cos^{4}{\left(t \right)}\, dt$$$。

使用降幂公式 $$$\sin^2\left( \alpha \right)=\frac{1}{2}-\frac{1}{2}\cos\left(2 \alpha \right)-$$$（取 $$$\alpha=t$$$）和 $$$\cos^2\left( \beta \right)=\frac{1}{2}+\frac{1}{2}\cos\left(2 \beta \right)+$$$（取 $$$\beta=t$$$）重写被积函数:

$${\color{red}{\int{\sin^{2}{\left(t \right)} \cos^{4}{\left(t \right)} d t}}} = {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right) \left(\frac{\cos{\left(2 t \right)}}{2} + \frac{1}{2}\right)^{2} d t}}}$$

展开该表达式:

$${\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right) \left(\frac{\cos{\left(2 t \right)}}{2} + \frac{1}{2}\right)^{2} d t}}} = {\color{red}{\int{\left(- \frac{\cos^{3}{\left(2 t \right)}}{8} - \frac{\cos^{2}{\left(2 t \right)}}{8} + \frac{\cos{\left(2 t \right)}}{8} + \frac{1}{8}\right)d t}}}$$

逐项积分：

$${\color{red}{\int{\left(- \frac{\cos^{3}{\left(2 t \right)}}{8} - \frac{\cos^{2}{\left(2 t \right)}}{8} + \frac{\cos{\left(2 t \right)}}{8} + \frac{1}{8}\right)d t}}} = {\color{red}{\left(\int{\frac{1}{8} d t} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{2}{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t}\right)}}$$

应用常数法则 $$$\int c\, dt = c t$$$，使用 $$$c=\frac{1}{8}$$$：

$$\int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{2}{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} + {\color{red}{\int{\frac{1}{8} d t}}} = \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{2}{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} + {\color{red}{\left(\frac{t}{8}\right)}}$$

对 $$$c=\frac{1}{8}$$$ 和 $$$f{\left(t \right)} = \cos^{2}{\left(2 t \right)}$$$ 应用常数倍法则 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$：

$$\frac{t}{8} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - {\color{red}{\int{\frac{\cos^{2}{\left(2 t \right)}}{8} d t}}} = \frac{t}{8} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - {\color{red}{\left(\frac{\int{\cos^{2}{\left(2 t \right)} d t}}{8}\right)}}$$

设$$$u=2 t$$$。

则$$$du=\left(2 t\right)^{\prime }dt = 2 dt$$$ (步骤见»)，并有$$$dt = \frac{du}{2}$$$。

积分变为

$$\frac{t}{8} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\cos^{2}{\left(2 t \right)} d t}}}}{8} = \frac{t}{8} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\frac{\cos^{2}{\left(u \right)}}{2} d u}}}}{8}$$

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

$$\frac{t}{8} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\frac{\cos^{2}{\left(u \right)}}{2} d u}}}}{8} = \frac{t}{8} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\left(\frac{\int{\cos^{2}{\left(u \right)} d u}}{2}\right)}}}{8}$$

应用降幂公式 $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$，并令 $$$\alpha= u $$$:

$$\frac{t}{8} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\cos^{2}{\left(u \right)} d u}}}}{16} = \frac{t}{8} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}}{16}$$

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

$$\frac{t}{8} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}}{16} = \frac{t}{8} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}{2}\right)}}}{16}$$

逐项积分：

$$\frac{t}{8} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}}}{32} = \frac{t}{8} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\left(\int{1 d u} + \int{\cos{\left(2 u \right)} d u}\right)}}}{32}$$

应用常数法则 $$$\int c\, du = c u$$$，使用 $$$c=1$$$：

$$\frac{t}{8} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{\int{\cos{\left(2 u \right)} d u}}{32} - \frac{{\color{red}{\int{1 d u}}}}{32} = \frac{t}{8} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{\int{\cos{\left(2 u \right)} d u}}{32} - \frac{{\color{red}{u}}}{32}$$

设$$$v=2 u$$$。

则$$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (步骤见»)，并有$$$du = \frac{dv}{2}$$$。

因此，

$$\frac{t}{8} - \frac{u}{32} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{32} = \frac{t}{8} - \frac{u}{32} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{32}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(v \right)} = \cos{\left(v \right)}$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$：

$$\frac{t}{8} - \frac{u}{32} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{32} = \frac{t}{8} - \frac{u}{32} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{32}$$

余弦函数的积分为 $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$：

$$\frac{t}{8} - \frac{u}{32} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{64} = \frac{t}{8} - \frac{u}{32} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\sin{\left(v \right)}}}}{64}$$

回忆一下 $$$v=2 u$$$:

$$\frac{t}{8} - \frac{u}{32} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{\sin{\left({\color{red}{v}} \right)}}{64} = \frac{t}{8} - \frac{u}{32} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{64}$$

回忆一下 $$$u=2 t$$$:

$$\frac{t}{8} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{\sin{\left(2 {\color{red}{u}} \right)}}{64} - \frac{{\color{red}{u}}}{32} = \frac{t}{8} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t} - \frac{\sin{\left(2 {\color{red}{\left(2 t\right)}} \right)}}{64} - \frac{{\color{red}{\left(2 t\right)}}}{32}$$

对 $$$c=\frac{1}{8}$$$ 和 $$$f{\left(t \right)} = \cos^{3}{\left(2 t \right)}$$$ 应用常数倍法则 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$：

$$\frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - {\color{red}{\int{\frac{\cos^{3}{\left(2 t \right)}}{8} d t}}} = \frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - {\color{red}{\left(\frac{\int{\cos^{3}{\left(2 t \right)} d t}}{8}\right)}}$$

设$$$u=2 t$$$。

则$$$du=\left(2 t\right)^{\prime }dt = 2 dt$$$ (步骤见»)，并有$$$dt = \frac{du}{2}$$$。

该积分可以改写为

$$\frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\cos^{3}{\left(2 t \right)} d t}}}}{8} = \frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\frac{\cos^{3}{\left(u \right)}}{2} d u}}}}{8}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \cos^{3}{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$\frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\frac{\cos^{3}{\left(u \right)}}{2} d u}}}}{8} = \frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\left(\frac{\int{\cos^{3}{\left(u \right)} d u}}{2}\right)}}}{8}$$

提出一个余弦，并使用公式 $$$\cos^2\left(\alpha \right)=-\sin^2\left(\alpha \right)+1$$$（令 $$$\alpha= u $$$）将其余部分用正弦表示:

$$\frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\cos^{3}{\left(u \right)} d u}}}}{16} = \frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\left(1 - \sin^{2}{\left(u \right)}\right) \cos{\left(u \right)} d u}}}}{16}$$

设$$$v=\sin{\left(u \right)}$$$。

则$$$dv=\left(\sin{\left(u \right)}\right)^{\prime }du = \cos{\left(u \right)} du$$$ (步骤见»)，并有$$$\cos{\left(u \right)} du = dv$$$。

因此，

$$\frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\left(1 - \sin^{2}{\left(u \right)}\right) \cos{\left(u \right)} d u}}}}{16} = \frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\left(1 - v^{2}\right)d v}}}}{16}$$

逐项积分：

$$\frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\int{\left(1 - v^{2}\right)d v}}}}{16} = \frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\left(\int{1 d v} - \int{v^{2} d v}\right)}}}{16}$$

应用常数法则 $$$\int c\, dv = c v$$$，使用 $$$c=1$$$：

$$\frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} + \frac{\int{v^{2} d v}}{16} - \frac{{\color{red}{\int{1 d v}}}}{16} = \frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} + \frac{\int{v^{2} d v}}{16} - \frac{{\color{red}{v}}}{16}$$

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

$$\frac{t}{16} - \frac{v}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} + \frac{{\color{red}{\int{v^{2} d v}}}}{16}=\frac{t}{16} - \frac{v}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} + \frac{{\color{red}{\frac{v^{1 + 2}}{1 + 2}}}}{16}=\frac{t}{16} - \frac{v}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} + \frac{{\color{red}{\left(\frac{v^{3}}{3}\right)}}}{16}$$

回忆一下 $$$v=\sin{\left(u \right)}$$$:

$$\frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{v}}}{16} + \frac{{\color{red}{v}}^{3}}{48} = \frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \frac{{\color{red}{\sin{\left(u \right)}}}}{16} + \frac{{\color{red}{\sin{\left(u \right)}}}^{3}}{48}$$

回忆一下 $$$u=2 t$$$:

$$\frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \frac{\sin{\left({\color{red}{u}} \right)}}{16} + \frac{\sin^{3}{\left({\color{red}{u}} \right)}}{48} = \frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64} + \int{\frac{\cos{\left(2 t \right)}}{8} d t} - \frac{\sin{\left({\color{red}{\left(2 t\right)}} \right)}}{16} + \frac{\sin^{3}{\left({\color{red}{\left(2 t\right)}} \right)}}{48}$$

对 $$$c=\frac{1}{8}$$$ 和 $$$f{\left(t \right)} = \cos{\left(2 t \right)}$$$ 应用常数倍法则 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$：

$$\frac{t}{16} + \frac{\sin^{3}{\left(2 t \right)}}{48} - \frac{\sin{\left(2 t \right)}}{16} - \frac{\sin{\left(4 t \right)}}{64} + {\color{red}{\int{\frac{\cos{\left(2 t \right)}}{8} d t}}} = \frac{t}{16} + \frac{\sin^{3}{\left(2 t \right)}}{48} - \frac{\sin{\left(2 t \right)}}{16} - \frac{\sin{\left(4 t \right)}}{64} + {\color{red}{\left(\frac{\int{\cos{\left(2 t \right)} d t}}{8}\right)}}$$

积分 $$$\int{\cos{\left(2 t \right)} d t}$$$ 已经计算过：

$$\int{\cos{\left(2 t \right)} d t} = \frac{\sin{\left(2 t \right)}}{2}$$

因此，

$$\frac{t}{16} + \frac{\sin^{3}{\left(2 t \right)}}{48} - \frac{\sin{\left(2 t \right)}}{16} - \frac{\sin{\left(4 t \right)}}{64} + \frac{{\color{red}{\int{\cos{\left(2 t \right)} d t}}}}{8} = \frac{t}{16} + \frac{\sin^{3}{\left(2 t \right)}}{48} - \frac{\sin{\left(2 t \right)}}{16} - \frac{\sin{\left(4 t \right)}}{64} + \frac{{\color{red}{\left(\frac{\sin{\left(2 t \right)}}{2}\right)}}}{8}$$

因此，

$$\int{\sin^{2}{\left(t \right)} \cos^{4}{\left(t \right)} d t} = \frac{t}{16} + \frac{\sin^{3}{\left(2 t \right)}}{48} - \frac{\sin{\left(4 t \right)}}{64}$$

加上积分常数：

$$\int{\sin^{2}{\left(t \right)} \cos^{4}{\left(t \right)} d t} = \frac{t}{16} + \frac{\sin^{3}{\left(2 t \right)}}{48} - \frac{\sin{\left(4 t \right)}}{64}+C$$

$$$\int \sin^{2}{\left(t \right)} \cos^{4}{\left(t \right)}\, dt = \left(\frac{t}{16} + \frac{\sin^{3}{\left(2 t \right)}}{48} - \frac{\sin{\left(4 t \right)}}{64}\right) + C$$$A