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

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

重寫被積函數:

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

將分子和分母同時乘以一個 cos，並使用公式 $$$\cos^2\left(\alpha \right)=-\sin^2\left(\alpha \right)+1$$$（其中 $$$\alpha=x$$$），把其餘都用正弦表示:

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

令 $$$u=\sin{\left(x \right)}$$$。

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

所以，

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

由於分子次數不小於分母次數，進行多項式長除法（步驟見»）:

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

逐項積分：

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

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

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

令 $$$v=1 - u^{2}$$$。

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

所以，

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

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

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

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

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

回顧一下 $$$v=1 - u^{2}$$$：

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

回顧一下 $$$u=\sin{\left(x \right)}$$$：

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

因此，

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

化簡：

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

加上積分常數（並從表達式中移除常數項）：

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

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