积分运算去根号的方式之一：三角函数凑平方去根号

一、题目

$$I={\int }_0^{\pi }\sqrt{1-\sin x}\mathrm{d}x=?$$

难度评级：

二、解析

由于：

$$\sin 2\alpha =2\sin \alpha \cos \alpha \Rightarrow$$

$$\begin{array}{rl}(\sin \alpha -\cos \alpha {)}^2& ={\sin }^2\alpha +{\cos }^2\alpha -2\sin \alpha \cos \alpha \\ & =1–\sin 2\alpha \end{array}$$

于是：

$$\begin{array}{rl}I& ={\int }_0^{\pi }\sqrt{1-\sin x}\mathrm{d}x\\ & ={\int }_0^{\pi }\sqrt{{(\sin \frac{x}{2}-\cos \frac{x}{2})}^2}\mathrm{d}x\\ & ={\int }_0^{\pi }|\sin \frac{x}{2}-\cos \frac{x}{2}|\mathrm{d}x\end{array}$$

又：

$$\begin{array}{rl}& \{\begin{array}{ll}\cos x>\sin x,& x\in (0,\frac{\pi }{4})\\ \sin x>\cos x,& x\in (\frac{\pi }{4},\frac{\pi }{2})\end{array}\Rightarrow \\ \\ & \{\begin{array}{ll}\cos \frac{x}{2}>\sin \frac{x}{2},& \frac{x}{2}\in (0,\frac{\pi }{4})\\ \sin \frac{x}{2}>\cos \frac{x}{2},& \frac{x}{2}\in (\frac{\pi }{4},\frac{\pi }{2})\end{array}\Rightarrow \\ \\ & \{\begin{array}{ll}\cos \frac{x}{2}>\sin \frac{x}{2},& x\in (0,\frac{\pi }{2})\\ \sin \frac{x}{2}>\cos \frac{x}{2},& x\in (\frac{\pi }{2},\pi )\end{array}\end{array}$$

于是：

$$\begin{array}{rl}I& ={\int }_0^{\frac{\pi }{2}}(\cos \frac{x}{2}-\sin \frac{x}{2})\mathrm{d}x+{\int }_{\frac{\pi }{2}}^{\pi }(\sin \frac{x}{2}-\cos \frac{x}{2})\mathrm{d}x\\ & =2{\int }_0^{\frac{\pi }{2}}\cos \frac{x}{2}\mathrm{d}\left(\frac{x}{2}\right)–2{\int }_0^{\frac{\pi }{2}}\sin \frac{x}{2}\mathrm{d}\left(\frac{x}{2}\right)\\ & +2{\int }_{\frac{\pi }{2}}^{\pi }\sin \frac{x}{2}\mathrm{d}\left(\frac{x}{2}\right)–2{\int }_{\frac{\pi }{2}}^{\pi }\cos \frac{x}{2}\mathrm{d}\left(\frac{x}{2}\right)\\ & =2[{\sin \frac{x}{2}|}_0^{\frac{\pi }{2}}+{\cos \frac{x}{2}|}_0^{\frac{\pi }{2}}-{\cos \frac{\pi }{2}|}_{\frac{\pi }{2}}^{\pi }-{\sin \frac{x}{2}|}_{\frac{\pi }{2}}^{\pi }]\\ & =2[\frac{\sqrt{2}}{2}-0+\frac{\sqrt{2}}{2}-1+\frac{\sqrt{2}}{2}-1+\frac{\sqrt{2}}{2}]\\ & =4\sqrt{2}-4\end{array}$$

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