$$$\frac{\cos^{4}{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$$ 的積分

求$$$\int \frac{\cos^{4}{\left(x \right)}}{\sin^{2}{\left(x \right)}}\, dx$$$。

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 $$$\frac{1}{\tan^{2}{\left(x \right)}}$$$:

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

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

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

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

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

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 $$$\sec^{2}{\left(x \right)} dx = du$$$。

該積分變為

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

進行部分分式分解（步驟可見 »）:

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

逐項積分：

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

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

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

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

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

為了計算積分 $$$\int{\frac{1}{\left(u^{2} + 1\right)^{2}} d u}$$$，對積分 $$$\int{\frac{1}{u^{2} + 1} d u}$$$ 使用分部積分法 $$$\int \operatorname{c} \operatorname{dv} = \operatorname{c}\operatorname{v} - \int \operatorname{v} \operatorname{dc}$$$。

令 $$$\operatorname{c}=\frac{1}{u^{2} + 1}$$$ 與 $$$\operatorname{dv}=du$$$。

則 $$$\operatorname{dc}=\left(\frac{1}{u^{2} + 1}\right)^{\prime }du=- \frac{2 u}{\left(u^{2} + 1\right)^{2}} du$$$（步驟見 »），且 $$$\operatorname{v}=\int{1 d u}=u$$$（步驟見 »）。

因此，

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

將常數提出：

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

將被積分函數的分子改寫為 $$$u^{2}=u^{2}{\color{red}{+1}}{\color{red}{-1}}$$$，並拆分：

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

將積分拆分：

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

因此，我們得到如下關於該積分的簡單線性方程：

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

解得

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

因此，

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

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

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

回顧一下 $$$u=\tan{\left(x \right)}$$$：

$$- \frac{3 \operatorname{atan}{\left({\color{red}{u}} \right)}}{2} - {\color{red}{u}}^{-1} - \frac{{\color{red}{u}} \left(1 + {\color{red}{u}}^{2}\right)^{-1}}{2} = - \frac{3 \operatorname{atan}{\left({\color{red}{\tan{\left(x \right)}}} \right)}}{2} - {\color{red}{\tan{\left(x \right)}}}^{-1} - \frac{{\color{red}{\tan{\left(x \right)}}} \left(1 + {\color{red}{\tan{\left(x \right)}}}^{2}\right)^{-1}}{2}$$

因此，

$$\int{\frac{\cos^{4}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x} = - \frac{3 \operatorname{atan}{\left(\tan{\left(x \right)} \right)}}{2} - \frac{1}{\tan{\left(x \right)}} - \frac{\tan{\left(x \right)}}{2 \left(\tan^{2}{\left(x \right)} + 1\right)}$$

化簡：

$$\int{\frac{\cos^{4}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x} = - \frac{3 x}{2} - \frac{\sin{\left(2 x \right)}}{4} - \frac{1}{\tan{\left(x \right)}}$$

加上積分常數：

$$\int{\frac{\cos^{4}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x} = - \frac{3 x}{2} - \frac{\sin{\left(2 x \right)}}{4} - \frac{1}{\tan{\left(x \right)}}+C$$

$$$\int \frac{\cos^{4}{\left(x \right)}}{\sin^{2}{\left(x \right)}}\, dx = \left(- \frac{3 x}{2} - \frac{\sin{\left(2 x \right)}}{4} - \frac{1}{\tan{\left(x \right)}}\right) + C$$$A