$$$\frac{\pi \cos{\left(x \right)}}{2 \sqrt{\sin{\left(x \right)}}}$$$ 的积分

求$$$\int \frac{\pi \cos{\left(x \right)}}{2 \sqrt{\sin{\left(x \right)}}}\, dx$$$。

对 $$$c=\frac{\pi}{2}$$$ 和 $$$f{\left(x \right)} = \frac{\cos{\left(x \right)}}{\sqrt{\sin{\left(x \right)}}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{\frac{\pi \cos{\left(x \right)}}{2 \sqrt{\sin{\left(x \right)}}} d x}}} = {\color{red}{\left(\frac{\pi \int{\frac{\cos{\left(x \right)}}{\sqrt{\sin{\left(x \right)}}} d x}}{2}\right)}}$$

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$$$\cos{\left(x \right)} dx = du$$$。

因此，

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

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=- \frac{1}{2}$$$：

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

回忆一下 $$$u=\sin{\left(x \right)}$$$:

$$\pi \sqrt{{\color{red}{u}}} = \pi \sqrt{{\color{red}{\sin{\left(x \right)}}}}$$

因此，

$$\int{\frac{\pi \cos{\left(x \right)}}{2 \sqrt{\sin{\left(x \right)}}} d x} = \pi \sqrt{\sin{\left(x \right)}}$$

加上积分常数：

$$\int{\frac{\pi \cos{\left(x \right)}}{2 \sqrt{\sin{\left(x \right)}}} d x} = \pi \sqrt{\sin{\left(x \right)}}+C$$

$$$\int \frac{\pi \cos{\left(x \right)}}{2 \sqrt{\sin{\left(x \right)}}}\, dx = \pi \sqrt{\sin{\left(x \right)}} + C$$$A