kaoyan2advanced 高等数学 第35题
📝 题目
### 第35题
$\displaystyle \int \frac{1}{\cos ^{2} x \sin ^{4} x} \mathrm{~d} x=$ $\_\_\_\_$ ．$
建衩谷题时问 $\leqslant 5 \mathrm{~min}
💡 答案解析
**答案**：$\displaystyle \frac{1}{3}\tan^{3}x + \frac{2}{\tan x} - \frac{1}{3\tan^{3}x} + C$ **解析**：步骤1：原积分$\displaystyle \int \frac{1}{\cos^{2}x \sin^{4}x}dx = \int \frac{1}{\cos^{2}x} \cdot \frac{1}{\sin^{4}x}dx = \int \sec^{2}x \csc^{4}x dx$。步骤2：令$u=\tan x$，则$du=\sec^{2}x dx$，$\displaystyle \csc^{2}x=1+\cot^{2}x=1+\frac{1}{u^{2}}$，$\displaystyle \csc^{4}x=(1+\frac{1}{u^{2}})^{2}$，积分化为$\displaystyle \int (1+\frac{1}{u^{2}})^{2} du = \int (1+\frac{2}{u^{2}}+\frac{1}{u^{4}})du = u - \frac{2}{u} - \frac{1}{3u^{3}} + C$。步骤3：回代$u=\tan x$，得$\displaystyle \tan x - 2\cot x - \frac{1}{3}\cot^{3}x + C = \frac{1}{3}\tan^{3}x + \frac{2}{\tan x} - \frac{1}{3\tan^{3}x} + C$。 **难度**：★★☆☆☆