📝 题目
例 6 求 $I = {\int }_{0}^{\frac{\pi }{2}}\left( {\sqrt{\tan x} + \sqrt{\cot x}}\right) \mathrm{d}x$ .
💡 答案与解析
解法 1 用三角函数的恒等变换.
$$ I = \sqrt{2}{\int }_{0}^{\frac{\pi }{2}}\frac{\sin x + \cos x}{\sqrt{2\sin x\cos x}}\mathrm{\;d}x = 2{\int }_{0}^{\frac{\pi }{2}}\frac{\cos \left( {x - \frac{\pi }{4}}\right) }{\sqrt{\cos 2\left( {x - \frac{\pi }{4}}\right) }}\mathrm{d}x $$
$$ \overset{t = x - \pi /4}{ = }2{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{\cos t}{\sqrt{1 - 2{\sin }^{2}t}}\mathrm{\;d}t = 4{\int }_{0}^{\frac{\pi }{4}}\frac{\cos t}{\sqrt{1 - 2{\sin }^{2}t}}\mathrm{\;d}t $$
$$ \overset{v = \sqrt{2}\sin t}{ = }2\sqrt{2}{\int }_{0}^{1}\frac{1}{\sqrt{1 - {v}^{2}}}\mathrm{\;d}v = {\left. 2\sqrt{2}\arcsin v\right| }_{0}^{1} $$
$$ = \sqrt{2}\pi \text{ . } $$
解法 2 令 $u = \sqrt{\tan x}$ ,则 $\mathrm{d}u = \frac{{\sec }^{2}x\mathrm{\;d}x}{2\sqrt{\tan x}} = \frac{1 + {u}^{4}}{2u}\mathrm{\;d}x$ ,因此
$$ I = {\int }_{0}^{\frac{\pi }{2}}\frac{1 + \tan x}{\sqrt{\tan x}}\mathrm{\;d}x = {\int }_{0}^{+\infty }\frac{1 + {u}^{2}}{u} \cdot \frac{2u}{1 + {u}^{4}}\mathrm{\;d}u $$
$$ = 2{\int }_{0}^{+\infty }\frac{1 + \frac{1}{{u}^{2}}}{{u}^{2} + \frac{1}{{u}^{2}}}\mathrm{\;d}u = 2{\int }_{0}^{+\infty }\frac{\mathrm{d}\left( {u - \frac{1}{u}}\right) }{{\left( u - \frac{1}{u}\right) }^{2} + 2} $$
$$ = \sqrt{2}\arctan {\left. \frac{u - \frac{1}{u}}{\sqrt{2}}\right| }_{0}^{+\infty } = \sqrt{2}\pi . $$
解法 3 注意到
$$ x = \frac{\pi }{2} - u \Rightarrow {\int }_{0}^{\frac{\pi }{2}}\sqrt{\cot x}\mathrm{\;d}x = {\int }_{0}^{\frac{\pi }{2}}\sqrt{\tan u}\mathrm{\;d}u = {\int }_{0}^{\frac{\pi }{2}}\sqrt{\tan x}\mathrm{\;d}x. $$
令 $t = \tan x$ ,则有
$$ I = 2{\int }_{0}^{+\infty }\frac{\sqrt{t}}{1 + {t}^{2}}\mathrm{\;d}t\overset{t = {v}^{2}}{ = }{\int }_{0}^{+\infty }\frac{4{v}^{2}}{1 + {v}^{4}}\mathrm{\;d}v $$
$$ = 2{\int }_{0}^{+\infty }\left\{ {\frac{1 + {v}^{2}}{1 + {v}^{4}} - \frac{1 - {v}^{2}}{1 + {v}^{4}}}\right\} \mathrm{d}v $$
$$ = 2{\int }_{0}^{+\infty }\frac{\mathrm{d}\left( {v - \frac{1}{v}}\right) }{{\left( v - \frac{1}{v}\right) }^{2} + 2} - 2{\int }_{0}^{+\infty }\frac{1}{1 + {v}^{4}}\mathrm{\;d}v $$
$$ + 2{\int }_{0}^{+\infty }\frac{{v}^{2}}{1 + {v}^{4}}\mathrm{\;d}v \tag{5.2} $$
又
$$ {\int }_{0}^{+\infty }\frac{{v}^{2}}{1 + {v}^{4}}\mathrm{\;d}v\overset{t = 1/v}{ = }{\int }_{0}^{+\infty }\frac{1}{1 + {t}^{4}}\mathrm{\;d}t = {\int }_{0}^{+\infty }\frac{1}{1 + {v}^{4}}\mathrm{\;d}v, $$
$$ {\int }_{0}^{+\infty }\frac{\mathrm{d}\left( {v - \frac{1}{v}}\right) }{{\left( v - \frac{1}{v}\right) }^{2} + 2}\frac{t = v - 1/v}{}{\int }_{-\infty }^{+\infty }\frac{1}{{t}^{2} + 2}\mathrm{\;d}t = \frac{\pi }{\sqrt{2}}. \tag{5.3} $$
联合 (5.2) 与 (5.3) 式,即得 $\displaystyle{I = 2{\int }_{-\infty }^{+\infty }\frac{\mathrm{d}u}{{u}^{2} + 2} = \sqrt{2}\pi}$ .