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

求$$$\int \frac{\sin{\left(1 \right)} \cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}}\, dx$$$。

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

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

用餘切表示:

$$\sin{\left(1 \right)} {\color{red}{\int{\frac{\cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x}}} = \sin{\left(1 \right)} {\color{red}{\int{\cot^{2}{\left(x \right)} d x}}}$$

令 $$$u=\cot{\left(x \right)}$$$。

則 $$$du=\left(\cot{\left(x \right)}\right)^{\prime }dx = - \csc^{2}{\left(x \right)} dx$$$ (步驟見»)，並可得 $$$\csc^{2}{\left(x \right)} dx = - du$$$。

所以，

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

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 $$$f{\left(u \right)} = \frac{u^{2}}{u^{2} + 1}$$$：

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

重寫並拆分分式:

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

逐項積分：

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

配合 $$$c=1$$$，應用常數法則 $$$\int c\, du = c u$$$：

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

$$$\frac{1}{u^{2} + 1}$$$ 的積分是 $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$：

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

回顧一下 $$$u=\cot{\left(x \right)}$$$：

$$- \sin{\left(1 \right)} \left(- \operatorname{atan}{\left({\color{red}{u}} \right)} + {\color{red}{u}}\right) = - \sin{\left(1 \right)} \left(- \operatorname{atan}{\left({\color{red}{\cot{\left(x \right)}}} \right)} + {\color{red}{\cot{\left(x \right)}}}\right)$$

因此，

$$\int{\frac{\sin{\left(1 \right)} \cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x} = - \left(\cot{\left(x \right)} - \operatorname{atan}{\left(\cot{\left(x \right)} \right)}\right) \sin{\left(1 \right)}$$

化簡：

$$\int{\frac{\sin{\left(1 \right)} \cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x} = \left(- \cot{\left(x \right)} + \operatorname{atan}{\left(\cot{\left(x \right)} \right)}\right) \sin{\left(1 \right)}$$

加上積分常數：

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

$$$\int \frac{\sin{\left(1 \right)} \cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}}\, dx = \left(- \cot{\left(x \right)} + \operatorname{atan}{\left(\cot{\left(x \right)} \right)}\right) \sin{\left(1 \right)} + C$$$A