解决三角函数定积分的组合拳：区间再现与点火公式

一、题目

$${\int }_0^{\frac{\pi }{2}}x{\sin }^{2}x{\cos }^{2}x\mathrm{d}x=?$$

难度评级：

二、解析

首先，利用区间在线公式，令 $t$ $=$ $\frac{\pi }{2}$ $+$ $0$ $-$ $x$, 则：

$$t=\frac{\pi }{2}–x$$

$$x=\frac{\pi }{2}–t$$

$$\mathrm{d}x=–\mathrm{d}t$$

$$x\in (0,\frac{\pi }{2})\Rightarrow t\in (\frac{\pi }{2},0)$$

Next

于是：

$${\int }_0^{\frac{\pi }{2}}x{\sin }^{2}x{\cos }^{2}x\mathrm{d}x=$$

$$(-1){\int }_{\frac{\pi }{2}}^0(\frac{\pi }{2}–t){\sin }^2(\frac{\pi }{2}–t){\cos }^2(\frac{\pi }{2}–t)\mathrm{d}t=$$

$${\int }_0^{\frac{\pi }{2}}(\frac{\pi }{2}–t)\sin (\frac{\pi }{2}–t)\sin (\frac{\pi }{2}–t)\cos (\frac{\pi }{2}–t)\cos (\frac{\pi }{2}–t)\mathrm{d}t\Rightarrow$$

Next

三角函数诱导公式（奇变偶不变，符号看象限） $\Rightarrow$

$${\int }_0^{\frac{\pi }{2}}(\frac{\pi }{2}–t)\sin t\sin t\cos t\cos t\mathrm{d}t=$$

$${\int }_0^{\frac{\pi }{2}}(\frac{\pi }{2}–t){\sin }^{2}t{\cos }^{2}t\mathrm{d}t=$$

$$\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}{\sin }^{2}t{\cos }^{2}t\mathrm{d}t–{\int }_0^{\frac{\pi }{2}}t{\sin }^{2}t{\cos }^{2}t\mathrm{d}t\Rightarrow$$

Next

令 $t$ $=$ $x$ $\Rightarrow$

$$\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}{\sin }^{2}x{\cos }^{2}x\mathrm{d}x–{\int }_0^{\frac{\pi }{2}}x{\sin }^{2}x{\cos }^{2}x\mathrm{d}x={\int }_0^{\frac{\pi }{2}}x{\sin }^{2}x{\cos }^{2}x\mathrm{d}x\Rightarrow$$

$$\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}{\sin }^{2}x{\cos }^{2}x\mathrm{d}x=2{\int }_0^{\frac{\pi }{2}}x{\sin }^{2}x{\cos }^{2}x\mathrm{d}x\Rightarrow$$

$${\int }_0^{\frac{\pi }{2}}x{\sin }^{2}x{\cos }^{2}x\mathrm{d}x=\frac{\pi }{4}{\int }_0^{\frac{\pi }{2}}{\sin }^{2}x{\cos }^{2}x\mathrm{d}x=$$

$$\frac{\pi }{4}{\int }_0^{\frac{\pi }{2}}{\sin }^{2}x(1–{\sin }^{2}x)\mathrm{d}x$$

$$\frac{\pi }{4}[{\int }_0^{\frac{\pi }{2}}{\sin }^{2}x\mathrm{d}x–{\int }_0^{\frac{\pi }{2}}{\sin }^{4}x\mathrm{d}x]=$$

Next

点火公式 $\Rightarrow$

$$\frac{\pi }{4}(\frac{1}{2}\cdot \frac{\pi }{2}–\frac{3}{4}\cdot \frac{1}{2}\cdot \frac{\pi }{2})=$$

$$\frac{\pi }{4}(1\cdot \frac{1}{2}\cdot \frac{\pi }{2}–\frac{3}{4}\cdot \frac{1}{2}\cdot \frac{\pi }{2})=$$

$$\frac{\pi }{4}\cdot \frac{1}{4}\cdot \frac{1}{2}\cdot \frac{\pi }{2}=\frac{{\pi }^2}{64}.$$

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