$$$\sin^{2}{\left(t \right)} \cos^{2}{\left(t \right)}$$$ 的積分

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

使用倍角公式 $$$\sin\left(t \right)\cos\left(t \right)=\frac{1}{2}\sin\left( 2 t \right)$$$ 改寫被積函數:

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

套用常數倍法則 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$，使用 $$$c=\frac{1}{4}$$$ 與 $$$f{\left(t \right)} = \sin^{2}{\left(2 t \right)}$$$：

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

套用降冪公式 $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$，令 $$$\alpha=2 t$$$:

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

套用常數倍法則 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(t \right)} = 1 - \cos{\left(4 t \right)}$$$：

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

逐項積分：

$$\frac{{\color{red}{\int{\left(1 - \cos{\left(4 t \right)}\right)d t}}}}{8} = \frac{{\color{red}{\left(\int{1 d t} - \int{\cos{\left(4 t \right)} d t}\right)}}}{8}$$

配合 $$$c=1$$$，應用常數法則 $$$\int c\, dt = c t$$$：

$$- \frac{\int{\cos{\left(4 t \right)} d t}}{8} + \frac{{\color{red}{\int{1 d t}}}}{8} = - \frac{\int{\cos{\left(4 t \right)} d t}}{8} + \frac{{\color{red}{t}}}{8}$$

令 $$$u=4 t$$$。

則 $$$du=\left(4 t\right)^{\prime }dt = 4 dt$$$ (步驟見»)，並可得 $$$dt = \frac{du}{4}$$$。

所以，

$$\frac{t}{8} - \frac{{\color{red}{\int{\cos{\left(4 t \right)} d t}}}}{8} = \frac{t}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8}$$

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 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$：

$$\frac{t}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8} = \frac{t}{8} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{8}$$

餘弦函數的積分為 $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$：

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

回顧一下 $$$u=4 t$$$：

$$\frac{t}{8} - \frac{\sin{\left({\color{red}{u}} \right)}}{32} = \frac{t}{8} - \frac{\sin{\left({\color{red}{\left(4 t\right)}} \right)}}{32}$$

因此，

$$\int{\sin^{2}{\left(t \right)} \cos^{2}{\left(t \right)} d t} = \frac{t}{8} - \frac{\sin{\left(4 t \right)}}{32}$$

加上積分常數：

$$\int{\sin^{2}{\left(t \right)} \cos^{2}{\left(t \right)} d t} = \frac{t}{8} - \frac{\sin{\left(4 t \right)}}{32}+C$$

$$$\int \sin^{2}{\left(t \right)} \cos^{2}{\left(t \right)}\, dt = \left(\frac{t}{8} - \frac{\sin{\left(4 t \right)}}{32}\right) + C$$$A