📝 题目
例 7 设 $f\left( x\right)$ 在 $\left\lbrack {a,b}\right\rbrack$ 上连续,求证:
$$ {\int }_{a}^{b}f\left( x\right) \mathrm{d}x = {\int }_{a}^{b}f\left( {a + b - x}\right) \mathrm{d}x, $$
并由此计算
$$ {\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{{\cos }^{2}x}{x\left( {\pi - {2x}}\right) }\mathrm{d}x. $$
💡 答案与解析
解 令 $t = a + b - x$ ,则
$$ {\int }_{a}^{b}f\left( x\right) \mathrm{d}x = {\int }_{a}^{b}f\left( {a + b - t}\right) \mathrm{d}t = {\int }_{a}^{b}f\left( {a + b - x}\right) \mathrm{d}x. $$
对 $a = \frac{\pi }{6},b = \frac{\pi }{3}$ 用前一部分结果,有
原式 $= {\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{{\sin }^{2}x}{x\left( {\pi - {2x}}\right) }\mathrm{d}x = \frac{1}{2}{\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{1}{x\left( {\pi - {2x}}\right) }\mathrm{d}x$
$$ = \frac{1}{2\pi }{\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\left\lbrack {\frac{1}{x} + \frac{2}{\pi - {2x}}}\right\rbrack \mathrm{d}x = {\left. \frac{1}{2\pi }\ln \frac{x}{\pi - {2x}}\right| }_{\frac{\pi }{6}}^{\frac{\pi }{3}} = \frac{1}{\pi }\ln 2. $$