$$$- \frac{1}{\sin{\left(a - x \right)} \cos{\left(b - x \right)}}$$$ 关于$$$x$$$的积分
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您的输入
求$$$\int \left(- \frac{1}{\sin{\left(a - x \right)} \cos{\left(b - x \right)}}\right)\, dx$$$。
解答
改写被积函数:
$${\color{red}{\int{\left(- \frac{1}{\sin{\left(a - x \right)} \cos{\left(b - x \right)}}\right)d x}}} = {\color{red}{\int{\left(- \frac{\sin{\left(b - x \right)}}{\cos{\left(a - b \right)} \cos{\left(b - x \right)}} - \frac{\cos{\left(a - x \right)}}{\sin{\left(a - x \right)} \cos{\left(a - b \right)}}\right)d x}}}$$
逐项积分:
$${\color{red}{\int{\left(- \frac{\sin{\left(b - x \right)}}{\cos{\left(a - b \right)} \cos{\left(b - x \right)}} - \frac{\cos{\left(a - x \right)}}{\sin{\left(a - x \right)} \cos{\left(a - b \right)}}\right)d x}}} = {\color{red}{\left(- \int{\frac{\cos{\left(a - x \right)}}{\sin{\left(a - x \right)} \cos{\left(a - b \right)}} d x} - \int{\frac{\sin{\left(b - x \right)}}{\cos{\left(a - b \right)} \cos{\left(b - x \right)}} d x}\right)}}$$
对 $$$c=\frac{1}{\cos{\left(a - b \right)}}$$$ 和 $$$f{\left(x \right)} = \frac{\sin{\left(b - x \right)}}{\cos{\left(b - x \right)}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$$- \int{\frac{\cos{\left(a - x \right)}}{\sin{\left(a - x \right)} \cos{\left(a - b \right)}} d x} - {\color{red}{\int{\frac{\sin{\left(b - x \right)}}{\cos{\left(a - b \right)} \cos{\left(b - x \right)}} d x}}} = - \int{\frac{\cos{\left(a - x \right)}}{\sin{\left(a - x \right)} \cos{\left(a - b \right)}} d x} - {\color{red}{\frac{\int{\frac{\sin{\left(b - x \right)}}{\cos{\left(b - x \right)}} d x}}{\cos{\left(a - b \right)}}}}$$
设$$$u=\cos{\left(b - x \right)}$$$。
则$$$du=\left(\cos{\left(b - x \right)}\right)^{\prime }dx = \sin{\left(b - x \right)} dx$$$ (步骤见»),并有$$$\sin{\left(b - x \right)} dx = du$$$。
因此,
$$- \int{\frac{\cos{\left(a - x \right)}}{\sin{\left(a - x \right)} \cos{\left(a - b \right)}} d x} - \frac{{\color{red}{\int{\frac{\sin{\left(b - x \right)}}{\cos{\left(b - x \right)}} d x}}}}{\cos{\left(a - b \right)}} = - \int{\frac{\cos{\left(a - x \right)}}{\sin{\left(a - x \right)} \cos{\left(a - b \right)}} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{\cos{\left(a - b \right)}}$$
$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$- \int{\frac{\cos{\left(a - x \right)}}{\sin{\left(a - x \right)} \cos{\left(a - b \right)}} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{\cos{\left(a - b \right)}} = - \int{\frac{\cos{\left(a - x \right)}}{\sin{\left(a - x \right)} \cos{\left(a - b \right)}} d x} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{\cos{\left(a - b \right)}}$$
回忆一下 $$$u=\cos{\left(b - x \right)}$$$:
$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{\cos{\left(a - b \right)}} - \int{\frac{\cos{\left(a - x \right)}}{\sin{\left(a - x \right)} \cos{\left(a - b \right)}} d x} = - \frac{\ln{\left(\left|{{\color{red}{\cos{\left(b - x \right)}}}}\right| \right)}}{\cos{\left(a - b \right)}} - \int{\frac{\cos{\left(a - x \right)}}{\sin{\left(a - x \right)} \cos{\left(a - b \right)}} d x}$$
对 $$$c=\frac{1}{\cos{\left(a - b \right)}}$$$ 和 $$$f{\left(x \right)} = \frac{\cos{\left(a - x \right)}}{\sin{\left(a - x \right)}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$$- \frac{\ln{\left(\left|{\cos{\left(b - x \right)}}\right| \right)}}{\cos{\left(a - b \right)}} - {\color{red}{\int{\frac{\cos{\left(a - x \right)}}{\sin{\left(a - x \right)} \cos{\left(a - b \right)}} d x}}} = - \frac{\ln{\left(\left|{\cos{\left(b - x \right)}}\right| \right)}}{\cos{\left(a - b \right)}} - {\color{red}{\frac{\int{\frac{\cos{\left(a - x \right)}}{\sin{\left(a - x \right)}} d x}}{\cos{\left(a - b \right)}}}}$$
设$$$u=\sin{\left(a - x \right)}$$$。
则$$$du=\left(\sin{\left(a - x \right)}\right)^{\prime }dx = - \cos{\left(a - x \right)} dx$$$ (步骤见»),并有$$$\cos{\left(a - x \right)} dx = - du$$$。
因此,
$$- \frac{\ln{\left(\left|{\cos{\left(b - x \right)}}\right| \right)}}{\cos{\left(a - b \right)}} - \frac{{\color{red}{\int{\frac{\cos{\left(a - x \right)}}{\sin{\left(a - x \right)}} d x}}}}{\cos{\left(a - b \right)}} = - \frac{\ln{\left(\left|{\cos{\left(b - x \right)}}\right| \right)}}{\cos{\left(a - b \right)}} - \frac{{\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}}{\cos{\left(a - b \right)}}$$
对 $$$c=-1$$$ 和 $$$f{\left(u \right)} = \frac{1}{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$$- \frac{\ln{\left(\left|{\cos{\left(b - x \right)}}\right| \right)}}{\cos{\left(a - b \right)}} - \frac{{\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}}{\cos{\left(a - b \right)}} = - \frac{\ln{\left(\left|{\cos{\left(b - x \right)}}\right| \right)}}{\cos{\left(a - b \right)}} - \frac{{\color{red}{\left(- \int{\frac{1}{u} d u}\right)}}}{\cos{\left(a - b \right)}}$$
$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$- \frac{\ln{\left(\left|{\cos{\left(b - x \right)}}\right| \right)}}{\cos{\left(a - b \right)}} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{\cos{\left(a - b \right)}} = - \frac{\ln{\left(\left|{\cos{\left(b - x \right)}}\right| \right)}}{\cos{\left(a - b \right)}} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{\cos{\left(a - b \right)}}$$
回忆一下 $$$u=\sin{\left(a - x \right)}$$$:
$$- \frac{\ln{\left(\left|{\cos{\left(b - x \right)}}\right| \right)}}{\cos{\left(a - b \right)}} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{\cos{\left(a - b \right)}} = - \frac{\ln{\left(\left|{\cos{\left(b - x \right)}}\right| \right)}}{\cos{\left(a - b \right)}} + \frac{\ln{\left(\left|{{\color{red}{\sin{\left(a - x \right)}}}}\right| \right)}}{\cos{\left(a - b \right)}}$$
因此,
$$\int{\left(- \frac{1}{\sin{\left(a - x \right)} \cos{\left(b - x \right)}}\right)d x} = \frac{\ln{\left(\left|{\sin{\left(a - x \right)}}\right| \right)}}{\cos{\left(a - b \right)}} - \frac{\ln{\left(\left|{\cos{\left(b - x \right)}}\right| \right)}}{\cos{\left(a - b \right)}}$$
化简:
$$\int{\left(- \frac{1}{\sin{\left(a - x \right)} \cos{\left(b - x \right)}}\right)d x} = \frac{\ln{\left(\left|{\sin{\left(a - x \right)}}\right| \right)} - \ln{\left(\left|{\cos{\left(b - x \right)}}\right| \right)}}{\cos{\left(a - b \right)}}$$
加上积分常数:
$$\int{\left(- \frac{1}{\sin{\left(a - x \right)} \cos{\left(b - x \right)}}\right)d x} = \frac{\ln{\left(\left|{\sin{\left(a - x \right)}}\right| \right)} - \ln{\left(\left|{\cos{\left(b - x \right)}}\right| \right)}}{\cos{\left(a - b \right)}}+C$$
答案
$$$\int \left(- \frac{1}{\sin{\left(a - x \right)} \cos{\left(b - x \right)}}\right)\, dx = \frac{\ln\left(\left|{\sin{\left(a - x \right)}}\right|\right) - \ln\left(\left|{\cos{\left(b - x \right)}}\right|\right)}{\cos{\left(a - b \right)}} + C$$$A