📝 题目
例 3 求下列不定积分:
(1) $\displaystyle{\int \frac{\mathrm{d}x}{{\tan }^{2}x}}$ ; (2) $\displaystyle{\int \frac{\cos {2x}}{{\sin }^{2}{2x}}\mathrm{\;d}x}$ ； (3) $\displaystyle{\int \frac{x\mathrm{\;d}x}{\sqrt{1 + x}}}$ .
💡 答案与解析
解 (1) 原式 $\displaystyle{= \int \frac{{\cos }^{2}x}{{\sin }^{2}x}\mathrm{\;d}x = \int \frac{\mathrm{d}x}{{\sin }^{2}x} - \int \mathrm{d}x = - \cot x - x + C}$ .
(2)原式 $\displaystyle{= \int \frac{{\cos }^{2}x - {\sin }^{2}x}{4{\sin }^{2}x{\cos }^{2}x}\mathrm{\;d}x = \frac{1}{4}\left\lbrack {\int \frac{\mathrm{d}x}{{\sin }^{2}x}-\int \frac{\mathrm{d}x}{{\cos }^{2}x}}\right\rbrack}$
$$ = \frac{1}{4}\left\lbrack {\cot x + \tan x}\right\rbrack + C, $$
或
$$ \text{ 原式 } = \frac{1}{2}\int \frac{\mathrm{d}\sin {2x}}{{\sin }^{2}{2x}}\mathrm{\;d}x = - \frac{1}{2\sin {2x}} + C\text{ . } $$
(3) 原式 $\displaystyle{= \int \frac{x + 1 - 1}{\sqrt{1 + x}}\mathrm{\;d}x = \int \sqrt{1 + x}\mathrm{\;d}x - \int \frac{1}{\sqrt{1 + x}}\mathrm{\;d}x}$
$$ = \frac{2}{3}x\sqrt{x + 1} - \frac{4}{3}\sqrt{x + 1} + C. $$
\subsubsection{二、换元法}