$$$\left(\cos{\left(t \right)} + 1\right) \sin{\left(t \right)}$$$ 的積分

求$$$\int \left(\cos{\left(t \right)} + 1\right) \sin{\left(t \right)}\, dt$$$。

令 $$$u=\cos{\left(t \right)} + 1$$$。

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

因此，

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

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=-1$$$ 與 $$$f{\left(u \right)} = u$$$：

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

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

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

回顧一下 $$$u=\cos{\left(t \right)} + 1$$$：

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

因此，

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

加上積分常數：

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

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