$$$\frac{\cos{\left(x \right)}}{\sin^{3}{\left(x \right)}}$$$ 的积分
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您的输入
求$$$\int \frac{\cos{\left(x \right)}}{\sin^{3}{\left(x \right)}}\, dx$$$。
解答
设$$$u=\sin{\left(x \right)}$$$。
则$$$du=\left(\sin{\left(x \right)}\right)^{\prime }dx = \cos{\left(x \right)} dx$$$ (步骤见»),并有$$$\cos{\left(x \right)} dx = du$$$。
该积分可以改写为
$${\color{red}{\int{\frac{\cos{\left(x \right)}}{\sin^{3}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{u^{3}} d u}}}$$
应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=-3$$$:
$${\color{red}{\int{\frac{1}{u^{3}} d u}}}={\color{red}{\int{u^{-3} d u}}}={\color{red}{\frac{u^{-3 + 1}}{-3 + 1}}}={\color{red}{\left(- \frac{u^{-2}}{2}\right)}}={\color{red}{\left(- \frac{1}{2 u^{2}}\right)}}$$
回忆一下 $$$u=\sin{\left(x \right)}$$$:
$$- \frac{{\color{red}{u}}^{-2}}{2} = - \frac{{\color{red}{\sin{\left(x \right)}}}^{-2}}{2}$$
因此,
$$\int{\frac{\cos{\left(x \right)}}{\sin^{3}{\left(x \right)}} d x} = - \frac{1}{2 \sin^{2}{\left(x \right)}}$$
加上积分常数:
$$\int{\frac{\cos{\left(x \right)}}{\sin^{3}{\left(x \right)}} d x} = - \frac{1}{2 \sin^{2}{\left(x \right)}}+C$$
答案
$$$\int \frac{\cos{\left(x \right)}}{\sin^{3}{\left(x \right)}}\, dx = - \frac{1}{2 \sin^{2}{\left(x \right)}} + C$$$A