2020-07-03 积分题

题目

计算不定积分 $I=\int \frac{1}{{\sin }^{6}x+{\cos }^{6}x}\mathrm{d}x$ .

注意到恒等式

$$\left(a^6+b^6\right)=\left(a^2+b^2\right)\left(a^4-a^2{b}^2+b^4\right)$$

因此

$$\begin{array}{rl}{\sin }^{6}x+{\cos }^{6}x& =\left({\sin }^{2}x+{\cos }^{2}x\right)\left({\sin }^{4}x-{\sin }^{2}x{\cos }^{2}x+{\cos }^{4}x\right)\\ & ={\cos }^{4}x\left({\tan }^{4}x-{\tan }^{2}x+1\right)\end{array}$$

从而

$$\begin{array}{rl}I& =\int \frac{{\sec }^{4}x}{\left({\sin }^{6}x+{\cos }^{6}x\right){\sec }^{4}x}\mathrm{d}x=\int \frac{\left(1+{\tan }^2\right){\sec }^{2}x}{1-{\tan }^{2}x+{\tan }^{4}x}\mathrm{d}x\\ & \stackrel{u=\tan x}{=}\int \frac{1+u^2}{1-u^2+u^4}\mathrm{d}u=\int \frac{\frac{1}{u^2}+1}{\frac{1}{u^2}-1+u^2}\mathrm{d}u\\ & =\int \frac{1}{{\left(\frac{1}{u}-u\right)}^2+1}\mathrm{d}\left(u-\frac{1}{u}\right)=\arctan \left(u-\frac{1}{u}\right)+C\\ & =\arctan \left(\tan x-\cot x\right)+C\end{array}$$