$$$\frac{\cos^{3}{\left(x \right)}}{\sin{\left(x \right)}}$$$ 的積分

求$$$\int \frac{\cos^{3}{\left(x \right)}}{\sin{\left(x \right)}}\, dx$$$。

把分子與分母同乘以一個正弦，並將其餘全部用餘弦表示，使用公式 $$$\sin^2\left(\alpha \right)=-\cos^2\left(\alpha \right)+1$$$，其中 $$$\alpha=x$$$:

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

令 $$$u=\cos{\left(x \right)}$$$。

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

該積分可改寫為

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

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=-1$$$ 與 $$$f{\left(u \right)} = \frac{u^{3}}{1 - u^{2}}$$$：

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

由於分子次數不小於分母次數，進行多項式長除法（步驟見»）:

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

逐項積分：

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

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

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

令 $$$v=1 - u^{2}$$$。

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

因此，

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

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

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

$$$\frac{1}{v}$$$ 的積分是 $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$：

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

回顧一下 $$$v=1 - u^{2}$$$：

$$\frac{u^{2}}{2} + \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} = \frac{u^{2}}{2} + \frac{\ln{\left(\left|{{\color{red}{\left(1 - u^{2}\right)}}}\right| \right)}}{2}$$

回顧一下 $$$u=\cos{\left(x \right)}$$$：

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

因此，

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

加上積分常數：

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

$$$\int \frac{\cos^{3}{\left(x \right)}}{\sin{\left(x \right)}}\, dx = \left(\frac{\ln\left(\left|{\cos^{2}{\left(x \right)} - 1}\right|\right)}{2} + \frac{\cos^{2}{\left(x \right)}}{2}\right) + C$$$A