$$$\tan^{5}{\left(x \right)}$$$ 的積分

求$$$\int \tan^{5}{\left(x \right)}\, dx$$$。

令 $$$u=\tan{\left(x \right)}$$$。

則 $$$x=\operatorname{atan}{\left(u \right)}$$$ 與 $$$dx=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$（步驟見»）。

因此，

$${\color{red}{\int{\tan^{5}{\left(x \right)} d x}}} = {\color{red}{\int{\frac{u^{5}}{u^{2} + 1} d u}}}$$

由於分子次數不小於分母次數，進行多項式長除法（步驟見»）:

$${\color{red}{\int{\frac{u^{5}}{u^{2} + 1} d u}}} = {\color{red}{\int{\left(u^{3} - u + \frac{u}{u^{2} + 1}\right)d u}}}$$

逐項積分：

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

套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=3$$$：

$$- \int{u d u} + \int{\frac{u}{u^{2} + 1} d u} + {\color{red}{\int{u^{3} d u}}}=- \int{u d u} + \int{\frac{u}{u^{2} + 1} d u} + {\color{red}{\frac{u^{1 + 3}}{1 + 3}}}=- \int{u d u} + \int{\frac{u}{u^{2} + 1} d u} + {\color{red}{\left(\frac{u^{4}}{4}\right)}}$$

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

$$\frac{u^{4}}{4} + \int{\frac{u}{u^{2} + 1} d u} - {\color{red}{\int{u d u}}}=\frac{u^{4}}{4} + \int{\frac{u}{u^{2} + 1} d u} - {\color{red}{\frac{u^{1 + 1}}{1 + 1}}}=\frac{u^{4}}{4} + \int{\frac{u}{u^{2} + 1} d u} - {\color{red}{\left(\frac{u^{2}}{2}\right)}}$$

令 $$$v=u^{2} + 1$$$。

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

因此，

$$\frac{u^{4}}{4} - \frac{u^{2}}{2} + {\color{red}{\int{\frac{u}{u^{2} + 1} d u}}} = \frac{u^{4}}{4} - \frac{u^{2}}{2} + {\color{red}{\int{\frac{1}{2 v} d v}}}$$

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

$$\frac{u^{4}}{4} - \frac{u^{2}}{2} + {\color{red}{\int{\frac{1}{2 v} d v}}} = \frac{u^{4}}{4} - \frac{u^{2}}{2} + {\color{red}{\left(\frac{\int{\frac{1}{v} d v}}{2}\right)}}$$

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

$$\frac{u^{4}}{4} - \frac{u^{2}}{2} + \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2} = \frac{u^{4}}{4} - \frac{u^{2}}{2} + \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{2}$$

回顧一下 $$$v=u^{2} + 1$$$：

$$\frac{u^{4}}{4} - \frac{u^{2}}{2} + \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} = \frac{u^{4}}{4} - \frac{u^{2}}{2} + \frac{\ln{\left(\left|{{\color{red}{\left(u^{2} + 1\right)}}}\right| \right)}}{2}$$

回顧一下 $$$u=\tan{\left(x \right)}$$$：

$$\frac{\ln{\left(1 + {\color{red}{u}}^{2} \right)}}{2} - \frac{{\color{red}{u}}^{2}}{2} + \frac{{\color{red}{u}}^{4}}{4} = \frac{\ln{\left(1 + {\color{red}{\tan{\left(x \right)}}}^{2} \right)}}{2} - \frac{{\color{red}{\tan{\left(x \right)}}}^{2}}{2} + \frac{{\color{red}{\tan{\left(x \right)}}}^{4}}{4}$$

因此，

$$\int{\tan^{5}{\left(x \right)} d x} = \frac{\ln{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{2} + \frac{\tan^{4}{\left(x \right)}}{4} - \frac{\tan^{2}{\left(x \right)}}{2}$$

加上積分常數：

$$\int{\tan^{5}{\left(x \right)} d x} = \frac{\ln{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{2} + \frac{\tan^{4}{\left(x \right)}}{4} - \frac{\tan^{2}{\left(x \right)}}{2}+C$$

$$$\int \tan^{5}{\left(x \right)}\, dx = \left(\frac{\ln\left(\tan^{2}{\left(x \right)} + 1\right)}{2} + \frac{\tan^{4}{\left(x \right)}}{4} - \frac{\tan^{2}{\left(x \right)}}{2}\right) + C$$$A