📝 题目
例 1 求积分 $\displaystyle{\int }_{0}^{a}\arctan \sqrt{\frac{a - x}{a + x}}\mathrm{\;d}x\left( {a > 0}\right)$ .
💡 答案与解析
解法 1 用分部积分法. 记 $w\left( t\right) = \sqrt{\frac{a - x}{a + x}}$ ,则有 $w\left( a\right) = 0$ ,
原式 $= {\left. x\arctan w\left( x\right) \right| }_{0}^{a} - {\int }_{0}^{a}x \cdot \frac{1}{1 + {w}^{2}} \cdot \frac{1}{2w} \cdot \frac{-{2a}}{{\left( a + x\right) }^{2}}\mathrm{\;d}x$
$$ = {\int }_{0}^{a}\frac{x}{2\sqrt{{a}^{2} - {x}^{2}}}\mathrm{\;d}x = - \frac{1}{4}{\int }_{0}^{a}\frac{\mathrm{d}\left( {{a}^{2} - {x}^{2}}\right) }{\sqrt{{a}^{2} - {x}^{2}}} $$
$$ = - {\left. \frac{1}{2}\sqrt{{a}^{2} - {x}^{2}}\right| }_{0}^{a} = \frac{a}{2}. $$
解法 2 令 $x = a\cos t\left( {0 \leq t \leq \frac{\pi }{2}}\right)$ ,则
$$ \text{ 原式 } = a{\int }_{\frac{\pi }{2}}^{0}\frac{t}{2}\mathrm{\;d}\cos t = {\left. a \cdot \frac{t}{2}\cos t\right| }_{\frac{\pi }{2}}^{0} - {\int }_{\frac{\pi }{2}}^{0}\frac{a}{2}\cos t\mathrm{\;d}t $$
$$ = - {\left. \frac{a}{2}\sin t\right| }_{\frac{\pi }{2}}^{0} = \frac{a}{2}. $$
解法 3 令 $t = \arctan \sqrt{\frac{a - x}{a + x}}$ ,则
$$ \cos {2t} = \frac{1 - {\tan }^{2}t}{1 + {\tan }^{2}t} = \frac{1 - \frac{a - x}{a + x}}{1 + \frac{a - x}{a + x}} = \frac{x}{a}, $$
$$ \text{ 原式 } = {\int }_{\frac{\pi }{2}}^{0}t\mathrm{\;d}\left( {a\cos {2t}}\right) = {\left. at\cos 2t\right| }_{\frac{\pi }{4}}^{0} + a{\int }_{0}^{\frac{\pi }{4}}\cos {2t}\mathrm{\;d}t $$
$$ = {\left. \frac{a}{2}\sin 2t\right| }_{0}^{\frac{\pi }{4}} = \frac{a}{2}. $$