📝 题目
例 6 计算定积分 $I = {\int }_{0}^{\frac{\pi }{2}}\frac{\cos x\mathrm{\;d}x}{{a}^{2}{\sin }^{2}x + {b}^{2}{\cos }^{2}x}\left( {a,b > 0}\right)$ .
💡 答案与解析
解 当 $a = b$ 时, $\displaystyle{I = \frac{1}{{a}^{2}}{\int }_{0}^{\frac{\pi }{2}}\cos x\mathrm{\;d}x = \frac{1}{{a}^{2}}}$ ;
当 $a > b$ 时,
$$ I = {\int }_{0}^{\frac{\pi }{2}}\frac{d\sin x}{{a}^{2}{\sin }^{2}x + {b}^{2}\left( {1 - {\sin }^{2}x}\right) } $$
$$ = \frac{1}{\sqrt{{a}^{2} - {b}^{2}}}{\int }_{0}^{\frac{\pi }{2}}\frac{d\sqrt{{a}^{2} - {b}^{2}}\sin x}{{b}^{2} + \left( {{a}^{2} - {b}^{2}}\right) {\sin }^{2}x} $$
$$ = \frac{1}{\sqrt{{a}^{2} - {b}^{2}}} \cdot {\left. \frac{1}{b}\arctan \frac{\sqrt{{a}^{2} - {b}^{2}}\sin x}{b}\right| }_{0}^{\frac{\pi }{2}} $$
$$ = \frac{1}{b\sqrt{{a}^{2} - {b}^{2}}}\arctan \frac{\sqrt{{a}^{2} - {b}^{2}}}{b}; $$
当 $a < b$ 时,
$$ I = \frac{1}{\sqrt{{b}^{2} - {a}^{2}}}{\int }_{0}^{\frac{\pi }{2}}\frac{d\sqrt{{b}^{2} - {a}^{2}}\sin x}{{b}^{2} - \left( {{b}^{2} - {a}^{2}}\right) {\sin }^{2}x} $$
$$ = {\left. \frac{1}{{2b}\sqrt{{b}^{2} - {a}^{2}}}\ln \frac{b + \sqrt{{b}^{2} - {a}^{2}}\sin x}{b - \sqrt{{b}^{2} - {a}^{2}}\sin x}\right| }_{0}^{\frac{\pi }{2}} $$
$$ = \frac{1}{b\sqrt{{b}^{2} - {a}^{2}}}\ln \frac{b + \sqrt{{b}^{2} - {a}^{2}}}{a}. $$