📝 题目
例 6 求不定积分 $\displaystyle \int \frac{\mathrm{d}x}{1 + a\cos x}\left( {a > 0}\right)$ .
💡 答案与解析
解 当 $a = 1$ 时,
$$ \text{ 原式 } = {\int }_{\overline{{\cos }^{2}\frac{x}{2}}}^{\mathrm{d}\left( \frac{x}{2}\right) } = \tan \frac{x}{2} + C. $$
当 $a \neq 1$ . 时,
$$ \text{ 原式 } = \int \frac{\mathrm{d}x}{1 + a\left( {{\cos }^{2}\frac{x}{2} - {\sin }^{2}\frac{x}{2}}\right) } $$
$$ = \int \frac{\mathrm{d}\left( {\tan \frac{x}{2}}\right) }{\left( {1 + a}\right) + \left( {1 - a}\right) {\tan }^{2}\frac{x}{2}} $$
$$ = \left\{ \begin{array}{ll} \frac{2}{\sqrt{1 - {a}^{2}}}\arctan \sqrt{\frac{1 - a}{1 + a}}\tan \frac{x}{2} + C, & 0 < a < 1, \\ \frac{2}{\sqrt{{a}^{2} - 1}}\ln \left| \frac{\sqrt{a + 1} + \sqrt{a - 1}\tan \frac{x}{2}}{\sqrt{a + 1} - \sqrt{a - 1}\tan \frac{x}{2}}\right| + C, & a > 1. \end{array}\right. $$