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

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

將分子與分母同時乘以 $$$\frac{1}{\cos^{2}{\left(x \right)}}$$$，並將 $$$\frac{\cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$$ 轉換為 $$$\frac{1}{\tan^{2}{\left(x \right)}}$$$:

$${\color{red}{\int{\frac{1}{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{\cos^{4}{\left(x \right)} \tan^{2}{\left(x \right)}} d x}}}$$

將兩個餘弦因式提出，並使用公式 $$$\frac{1}{\cos^{2}{\left(x \right)}}=\sec^{2}{\left(x \right)}$$$ 將它們改寫為以正割表示。:

$${\color{red}{\int{\frac{1}{\cos^{4}{\left(x \right)} \tan^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\sec^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)} \tan^{2}{\left(x \right)}} d x}}}$$

使用公式 $$$\cos^{2}{\left(x \right)}=\frac{1}{\tan^{2}{\left(x \right)} + 1}$$$，以正切表示餘弦:

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

令 $$$u=\tan{\left(x \right)}$$$。

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

因此，

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

Expand the expression:

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

逐項積分：

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

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

$$\int{\frac{1}{u^{2}} d u} + {\color{red}{\int{1 d u}}} = \int{\frac{1}{u^{2}} d u} + {\color{red}{u}}$$

套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-2$$$：

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

回顧一下 $$$u=\tan{\left(x \right)}$$$：

$$- {\color{red}{u}}^{-1} + {\color{red}{u}} = - {\color{red}{\tan{\left(x \right)}}}^{-1} + {\color{red}{\tan{\left(x \right)}}}$$

因此，

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

加上積分常數：

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

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