方企勤 第三章 一元函数积分学 第5题
📝 题目
例 5 求不定积分 $\displaystyle{\int \frac{\mathrm{d}x}{2 + {\tan }^{2}x}}$ .
💡 答案解析
解法 1
$$ \text{ 原式 } = \int \frac{{\cos }^{2}x}{1 + {\cos }^{2}x}\mathrm{\;d}x = \int \frac{{\cos }^{2}x + 1 - 1}{1 + {\cos }^{2}x}\mathrm{\;d}x $$
$$ = x - \int \frac{1}{1 + {\cos }^{2}x}\mathrm{\;d}x = x - \int \frac{{\sec }^{2}x}{2 + {\tan }^{2}x}\mathrm{\;d}x $$
$$ = x - \int \frac{\mathrm{d}\tan x}{2 + {\tan }^{2}x} = x - \frac{1}{\sqrt{2}}\arctan \frac{\tan x}{\sqrt{2}} + C. $$
解法 2
$$ \text{ 原式 } = \int \frac{{\sec }^{2}x}{\left( {2 + {\tan }^{2}x}\right) \left( {1 + {\tan }^{2}x}\right) }\mathrm{d}x $$
$$ \overset{u = \tan x}{ = }\int \frac{1}{\left( {{u}^{2} + 1}\right) \left( {{u}^{2} + 2}\right) }\mathrm{d}u $$
$$ = \int \left( {\frac{1}{{u}^{2} + 1} - \frac{1}{{u}^{2} + 2}}\right) \mathrm{d}u $$
$$ = \arctan u - \frac{1}{2}\sqrt{2}\arctan \frac{1}{2}u\sqrt{2} + C $$
$$ = x - \frac{1}{\sqrt{2}}\arctan \frac{\tan x}{\sqrt{2}} + C. $$