$$$\sin^{2}{\left(x \right)} \tan{\left(x \right)} \sec^{2}{\left(x \right)}$$$ 的積分

求$$$\int \sin^{2}{\left(x \right)} \tan{\left(x \right)} \sec^{2}{\left(x \right)}\, dx$$$。

重寫被積函數:

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

提出一個正弦因子，將其餘部分用餘弦表示，使用公式 $$$\sin^2\left(\alpha \right)=-\cos^2\left(\alpha \right)+1$$$，其中 $$$\alpha=x$$$:

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

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

因此，

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

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 $$$f{\left(u \right)} = \frac{1 - u^{2}}{u^{3}}$$$：

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

Expand the expression:

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

逐項積分：

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

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

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

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

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

回顧一下 $$$u=\cos{\left(x \right)}$$$：

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

因此，

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

加上積分常數：

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

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