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

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\cos{\left(2 \right)}$$$ 與 $$$f{\left(x \right)} = \sin{\left(x \right)} \cos{\left(x \right)}$$$：

$${\color{red}{\int{\sin{\left(x \right)} \cos{\left(2 \right)} \cos{\left(x \right)} d x}}} = {\color{red}{\cos{\left(2 \right)} \int{\sin{\left(x \right)} \cos{\left(x \right)} d x}}}$$

令 $$$u=\sin{\left(x \right)}$$$。

則 $$$du=\left(\sin{\left(x \right)}\right)^{\prime }dx = \cos{\left(x \right)} dx$$$ (步驟見»)，並可得 $$$\cos{\left(x \right)} dx = du$$$。

該積分變為

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

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

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

回顧一下 $$$u=\sin{\left(x \right)}$$$：

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

因此，

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

加上積分常數：

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

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