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

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

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 $$$\alpha=x$$$:

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

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 $$$\sin{\left(x \right)} dx = - du$$$。

該積分變為

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

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

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

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，

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

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 $$$f{\left(v \right)} = \frac{1}{v \left(v - 1\right)}$$$：

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

進行部分分式分解（步驟可見 »）:

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

逐項積分：

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

令 $$$w=v - 1$$$。

則 $$$dw=\left(v - 1\right)^{\prime }dv = 1 dv$$$ (步驟見»)，並可得 $$$dv = dw$$$。

該積分變為

$$\frac{\int{\frac{1}{v} d v}}{2} - \frac{{\color{red}{\int{\frac{1}{v - 1} d v}}}}{2} = \frac{\int{\frac{1}{v} d v}}{2} - \frac{{\color{red}{\int{\frac{1}{w} d w}}}}{2}$$

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

$$\frac{\int{\frac{1}{v} d v}}{2} - \frac{{\color{red}{\int{\frac{1}{w} d w}}}}{2} = \frac{\int{\frac{1}{v} d v}}{2} - \frac{{\color{red}{\ln{\left(\left|{w}\right| \right)}}}}{2}$$

回顧一下 $$$w=v - 1$$$：

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

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

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

回顧一下 $$$v=1 - u^{2}$$$：

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

回顧一下 $$$u=\cos{\left(x \right)}$$$：

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

因此，

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

化簡：

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

加上積分常數：

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

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