$$$\left(1 - \sin{\left(\frac{t}{2} \right)}\right)^{2} \cos{\left(\frac{t}{2} \right)}$$$ 的積分

求$$$\int \left(1 - \sin{\left(\frac{t}{2} \right)}\right)^{2} \cos{\left(\frac{t}{2} \right)}\, dt$$$。

令 $$$u=1 - \sin{\left(\frac{t}{2} \right)}$$$。

則 $$$du=\left(1 - \sin{\left(\frac{t}{2} \right)}\right)^{\prime }dt = - \frac{\cos{\left(\frac{t}{2} \right)}}{2} dt$$$ (步驟見»)，並可得 $$$\cos{\left(\frac{t}{2} \right)} dt = - 2 du$$$。

因此，

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

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=-2$$$ 與 $$$f{\left(u \right)} = u^{2}$$$：

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

套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=2$$$：

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

回顧一下 $$$u=1 - \sin{\left(\frac{t}{2} \right)}$$$：

$$- \frac{2 {\color{red}{u}}^{3}}{3} = - \frac{2 {\color{red}{\left(1 - \sin{\left(\frac{t}{2} \right)}\right)}}^{3}}{3}$$

因此，

$$\int{\left(1 - \sin{\left(\frac{t}{2} \right)}\right)^{2} \cos{\left(\frac{t}{2} \right)} d t} = - \frac{2 \left(1 - \sin{\left(\frac{t}{2} \right)}\right)^{3}}{3}$$

化簡：

$$\int{\left(1 - \sin{\left(\frac{t}{2} \right)}\right)^{2} \cos{\left(\frac{t}{2} \right)} d t} = \frac{2 \left(\sin{\left(\frac{t}{2} \right)} - 1\right)^{3}}{3}$$

加上積分常數：

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

$$$\int \left(1 - \sin{\left(\frac{t}{2} \right)}\right)^{2} \cos{\left(\frac{t}{2} \right)}\, dt = \frac{2 \left(\sin{\left(\frac{t}{2} \right)} - 1\right)^{3}}{3} + C$$$A