$$$\frac{\pi \cos{\left(x \right)}}{2 \sqrt{\sin{\left(x \right)}}}$$$ 的積分

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{\pi}{2}$$$ 與 $$$f{\left(x \right)} = \frac{\cos{\left(x \right)}}{\sqrt{\sin{\left(x \right)}}}$$$：

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

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 $$$\cos{\left(x \right)} dx = du$$$。

該積分可改寫為

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

套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=- \frac{1}{2}$$$：

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

回顧一下 $$$u=\sin{\left(x \right)}$$$：

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

因此，

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

加上積分常數：

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

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