# 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

$$\int_{0}^{\frac{\pi}{2}} x \sin^{2} x \cos^{2} x \mathrm{d} x = ?$$

## 二、解析

$$t = \frac{\pi}{2} – x$$

$$x = \frac{\pi}{2} – t$$

$$\mathrm{d} x = – \mathrm{d} t$$

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

Next

$$\int_{0}^{\frac{\pi}{2}} x \sin^{2} x \cos^{2} x \mathrm{d} x =$$

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

$$\int_{0}^{\frac{\pi}{2}} (\frac{\pi}{2} – t) \sin (\frac{\pi}{2} – t) \sin (\frac{\pi}{2} – t) \cos (\frac{\pi}{2} – t) \cos (\frac{\pi}{2} – t) \mathrm{d} t \Rightarrow$$

Next

$$\int_{0}^{\frac{\pi}{2}} (\frac{\pi}{2} – t) \sin t \sin t \cos t \cos t \mathrm{d} t =$$

$$\int_{0}^{\frac{\pi}{2}} (\frac{\pi}{2} – t) \sin^{2} t \cos^{2} t \mathrm{d} t =$$

$$\frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} \sin^{2} t \cos^{2} t \mathrm{d} t – \int_{0}^{\frac{\pi}{2}} t \sin^{2} t \cos^{2} t \mathrm{d} t \Rightarrow$$

Next

$$\frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} \sin^{2} x \cos^{2} x \mathrm{d} x – \int_{0}^{\frac{\pi}{2}} x \sin^{2} x \cos^{2} x \mathrm{d} x = \int_{0}^{\frac{\pi}{2}} x \sin^{2} x \cos^{2} x \mathrm{d} x \Rightarrow$$

$$\frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} \sin^{2} x \cos^{2} x \mathrm{d} x = 2 \int_{0}^{\frac{\pi}{2}} x \sin^{2} x \cos^{2} x \mathrm{d} x \Rightarrow$$

$$\int_{0}^{\frac{\pi}{2}} x \sin^{2} x \cos^{2} x \mathrm{d} x = \frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \sin^{2} x \cos^{2} x \mathrm{d} x =$$

$$\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \sin^{2} x (1 – \sin^{2} x) \mathrm{d} x$$

$$\frac{\pi}{4} \Bigg[ \int_{0}^{\frac{\pi}{2}} \sin^{2} x \mathrm{d} x – \int_{0}^{\frac{\pi}{2}} \sin^{4} x \mathrm{d} x \Bigg] =$$

Next

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

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

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