$$$\csc^{3}{\left(x \right)} \sec{\left(x \right)}$$$ 的積分
您的輸入
求$$$\int \csc^{3}{\left(x \right)} \sec{\left(x \right)}\, dx$$$。
解答
重寫被積函數:
$${\color{red}{\int{\csc^{3}{\left(x \right)} \sec{\left(x \right)} d x}}} = {\color{red}{\int{\frac{1}{\sin^{3}{\left(x \right)} \cos{\left(x \right)}} d x}}}$$
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:
$${\color{red}{\int{\frac{1}{\sin^{3}{\left(x \right)} \cos{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\cos{\left(x \right)}}{\left(1 - \sin^{2}{\left(x \right)}\right) \sin^{3}{\left(x \right)}} d x}}}$$
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 $$$\cos{\left(x \right)} dx = du$$$。
因此，
$${\color{red}{\int{\frac{\cos{\left(x \right)}}{\left(1 - \sin^{2}{\left(x \right)}\right) \sin^{3}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{u^{3} \left(1 - u^{2}\right)} d u}}}$$
進行部分分式分解（步驟可見 »）:
$${\color{red}{\int{\frac{1}{u^{3} \left(1 - u^{2}\right)} d u}}} = {\color{red}{\int{\left(- \frac{1}{2 \left(u + 1\right)} - \frac{1}{2 \left(u - 1\right)} + \frac{1}{u} + \frac{1}{u^{3}}\right)d u}}}$$
逐項積分：
$${\color{red}{\int{\left(- \frac{1}{2 \left(u + 1\right)} - \frac{1}{2 \left(u - 1\right)} + \frac{1}{u} + \frac{1}{u^{3}}\right)d u}}} = {\color{red}{\left(\int{\frac{1}{u^{3}} d u} + \int{\frac{1}{u} d u} - \int{\frac{1}{2 \left(u - 1\right)} d u} - \int{\frac{1}{2 \left(u + 1\right)} d u}\right)}}$$
$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：
$$\int{\frac{1}{u^{3}} d u} - \int{\frac{1}{2 \left(u - 1\right)} d u} - \int{\frac{1}{2 \left(u + 1\right)} d u} + {\color{red}{\int{\frac{1}{u} d u}}} = \int{\frac{1}{u^{3}} d u} - \int{\frac{1}{2 \left(u - 1\right)} d u} - \int{\frac{1}{2 \left(u + 1\right)} d u} + {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$
套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-3$$$：
$$\ln{\left(\left|{u}\right| \right)} - \int{\frac{1}{2 \left(u - 1\right)} d u} - \int{\frac{1}{2 \left(u + 1\right)} d u} + {\color{red}{\int{\frac{1}{u^{3}} d u}}}=\ln{\left(\left|{u}\right| \right)} - \int{\frac{1}{2 \left(u - 1\right)} d u} - \int{\frac{1}{2 \left(u + 1\right)} d u} + {\color{red}{\int{u^{-3} d u}}}=\ln{\left(\left|{u}\right| \right)} - \int{\frac{1}{2 \left(u - 1\right)} d u} - \int{\frac{1}{2 \left(u + 1\right)} d u} + {\color{red}{\frac{u^{-3 + 1}}{-3 + 1}}}=\ln{\left(\left|{u}\right| \right)} - \int{\frac{1}{2 \left(u - 1\right)} d u} - \int{\frac{1}{2 \left(u + 1\right)} d u} + {\color{red}{\left(- \frac{u^{-2}}{2}\right)}}=\ln{\left(\left|{u}\right| \right)} - \int{\frac{1}{2 \left(u - 1\right)} d u} - \int{\frac{1}{2 \left(u + 1\right)} d u} + {\color{red}{\left(- \frac{1}{2 u^{2}}\right)}}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(u \right)} = \frac{1}{u + 1}$$$：
$$\ln{\left(\left|{u}\right| \right)} - \int{\frac{1}{2 \left(u - 1\right)} d u} - {\color{red}{\int{\frac{1}{2 \left(u + 1\right)} d u}}} - \frac{1}{2 u^{2}} = \ln{\left(\left|{u}\right| \right)} - \int{\frac{1}{2 \left(u - 1\right)} d u} - {\color{red}{\left(\frac{\int{\frac{1}{u + 1} d u}}{2}\right)}} - \frac{1}{2 u^{2}}$$
令 $$$v=u + 1$$$。
則 $$$dv=\left(u + 1\right)^{\prime }du = 1 du$$$ (步驟見»)，並可得 $$$du = dv$$$。
因此，
$$\ln{\left(\left|{u}\right| \right)} - \int{\frac{1}{2 \left(u - 1\right)} d u} - \frac{{\color{red}{\int{\frac{1}{u + 1} d u}}}}{2} - \frac{1}{2 u^{2}} = \ln{\left(\left|{u}\right| \right)} - \int{\frac{1}{2 \left(u - 1\right)} d u} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2} - \frac{1}{2 u^{2}}$$
$$$\frac{1}{v}$$$ 的積分是 $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$：
$$\ln{\left(\left|{u}\right| \right)} - \int{\frac{1}{2 \left(u - 1\right)} d u} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2} - \frac{1}{2 u^{2}} = \ln{\left(\left|{u}\right| \right)} - \int{\frac{1}{2 \left(u - 1\right)} d u} - \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{2} - \frac{1}{2 u^{2}}$$
回顧一下 $$$v=u + 1$$$：
$$\ln{\left(\left|{u}\right| \right)} - \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} - \int{\frac{1}{2 \left(u - 1\right)} d u} - \frac{1}{2 u^{2}} = \ln{\left(\left|{u}\right| \right)} - \frac{\ln{\left(\left|{{\color{red}{\left(u + 1\right)}}}\right| \right)}}{2} - \int{\frac{1}{2 \left(u - 1\right)} d u} - \frac{1}{2 u^{2}}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(u \right)} = \frac{1}{u - 1}$$$：
$$\ln{\left(\left|{u}\right| \right)} - \frac{\ln{\left(\left|{u + 1}\right| \right)}}{2} - {\color{red}{\int{\frac{1}{2 \left(u - 1\right)} d u}}} - \frac{1}{2 u^{2}} = \ln{\left(\left|{u}\right| \right)} - \frac{\ln{\left(\left|{u + 1}\right| \right)}}{2} - {\color{red}{\left(\frac{\int{\frac{1}{u - 1} d u}}{2}\right)}} - \frac{1}{2 u^{2}}$$
令 $$$v=u - 1$$$。
則 $$$dv=\left(u - 1\right)^{\prime }du = 1 du$$$ (步驟見»)，並可得 $$$du = dv$$$。
該積分可改寫為
$$\ln{\left(\left|{u}\right| \right)} - \frac{\ln{\left(\left|{u + 1}\right| \right)}}{2} - \frac{{\color{red}{\int{\frac{1}{u - 1} d u}}}}{2} - \frac{1}{2 u^{2}} = \ln{\left(\left|{u}\right| \right)} - \frac{\ln{\left(\left|{u + 1}\right| \right)}}{2} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2} - \frac{1}{2 u^{2}}$$
$$$\frac{1}{v}$$$ 的積分是 $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$：
$$\ln{\left(\left|{u}\right| \right)} - \frac{\ln{\left(\left|{u + 1}\right| \right)}}{2} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2} - \frac{1}{2 u^{2}} = \ln{\left(\left|{u}\right| \right)} - \frac{\ln{\left(\left|{u + 1}\right| \right)}}{2} - \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{2} - \frac{1}{2 u^{2}}$$
回顧一下 $$$v=u - 1$$$：
$$\ln{\left(\left|{u}\right| \right)} - \frac{\ln{\left(\left|{u + 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} - \frac{1}{2 u^{2}} = \ln{\left(\left|{u}\right| \right)} - \frac{\ln{\left(\left|{u + 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{\left(u - 1\right)}}}\right| \right)}}{2} - \frac{1}{2 u^{2}}$$
回顧一下 $$$u=\sin{\left(x \right)}$$$：
$$- \frac{\ln{\left(\left|{-1 + {\color{red}{u}}}\right| \right)}}{2} - \frac{\ln{\left(\left|{1 + {\color{red}{u}}}\right| \right)}}{2} + \ln{\left(\left|{{\color{red}{u}}}\right| \right)} - \frac{{\color{red}{u}}^{-2}}{2} = - \frac{\ln{\left(\left|{-1 + {\color{red}{\sin{\left(x \right)}}}}\right| \right)}}{2} - \frac{\ln{\left(\left|{1 + {\color{red}{\sin{\left(x \right)}}}}\right| \right)}}{2} + \ln{\left(\left|{{\color{red}{\sin{\left(x \right)}}}}\right| \right)} - \frac{{\color{red}{\sin{\left(x \right)}}}^{-2}}{2}$$
因此，
$$\int{\csc^{3}{\left(x \right)} \sec{\left(x \right)} d x} = - \frac{\ln{\left(\left|{\sin{\left(x \right)} - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{\sin{\left(x \right)} + 1}\right| \right)}}{2} + \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \frac{1}{2 \sin^{2}{\left(x \right)}}$$
加上積分常數：
$$\int{\csc^{3}{\left(x \right)} \sec{\left(x \right)} d x} = - \frac{\ln{\left(\left|{\sin{\left(x \right)} - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{\sin{\left(x \right)} + 1}\right| \right)}}{2} + \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \frac{1}{2 \sin^{2}{\left(x \right)}}+C$$
答案
$$$\int \csc^{3}{\left(x \right)} \sec{\left(x \right)}\, dx = \left(- \frac{\ln\left(\left|{\sin{\left(x \right)} - 1}\right|\right)}{2} - \frac{\ln\left(\left|{\sin{\left(x \right)} + 1}\right|\right)}{2} + \ln\left(\left|{\sin{\left(x \right)}}\right|\right) - \frac{1}{2 \sin^{2}{\left(x \right)}}\right) + C$$$A