一个有趣的不定积分

一个有趣的积分:$\displaystyle \int \cfrac{d\theta}{\textrm{sin}^3 \theta+\textrm{cos}^3 \theta}$
已邀请:
在开始之前,先介绍几个恒等式:
⑴ $\textrm{sin}^3\theta+\textrm{cos}^3\theta$
$\equiv\left( \textrm{sin}\theta+\textrm{cos}\theta\right)\left(\textrm{sin}^2\theta-\textrm{sin}\theta\textrm{cos}\theta+\textrm{cos}^2\theta \right)$
$\equiv\left( \textrm{sin}\theta+\textrm{cos}\theta\right)\left(1-\textrm{sin}\theta\textrm{cos}\theta \right).$

⑵ $\textrm{sin}\theta+\textrm{cos}\theta$
$\equiv\sqrt{2}\textrm{cos}\left(\frac{\pi}{4}\right)\textrm{sin}\left( \theta\right)+\sqrt{2}\textrm{sin}\left(\frac{\pi}{4}\right)\textrm{cos}\left( \theta\right)$
$\equiv\sqrt{2}\textrm{sin}\left( \theta+\frac{\pi}{4}\right).$

⑶ $1+2\textrm{sin}\theta\textrm{cos}\theta$
$\equiv\textrm{sin}^2\theta+2\textrm{sin}\theta\textrm{cos}\theta+\textrm{cos}^2\theta $
$\equiv\left( \textrm{sin}\theta+\textrm{cos}\theta\right)^2.$

⑷ $\textrm{sin}\theta\textrm{cos}\theta\equiv\frac{1}{2}\textrm{sin}2\theta.$

⑸ $\textrm{sin}\left(2\phi - \frac{\pi}{2}\right)$
$\equiv-\textrm{cos}2\phi$
$\equiv1-2\textrm{cos}^2\phi.$

⑹ $\sqrt{2}\textrm{cos}\left(\theta+\frac{\pi}{4} \right)\equiv\textrm{cos}\theta-\textrm{sin}\theta.$

⑺ $\left[ \textrm{csc}\left( \theta+\frac{\pi}{4}\right)+\textrm{cot}\left( \theta+\frac{\pi}{4}\right)\right]^{-1}\equiv\textrm{tan}\left( \frac{\theta}{2}+\frac{\pi}{8}\right).$
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设$\displaystyle I= \int \cfrac{d\theta}{\textrm{sin}^3\theta+\textrm{cos}^3\theta}.$
我们引入一个辅助计算的积分,设为$J$:
$\displaystyle J= \int \cfrac{\textrm{sin}\theta\ \textrm{cos}\theta\ d\theta}{\textrm{sin}^3\theta+\textrm{cos}^3\theta}.$


$\displaystyle I-J= \int \cfrac{\left(1-\textrm{sin}\theta\textrm{cos}\theta \right)d\theta}{\left( \textrm{sin}\theta+\textrm{cos}\theta\right)\left(1-\textrm{sin}\theta\textrm{cos}\theta \right)}=\int\cfrac{d\theta}{\textrm{sin}\theta+\textrm{cos}\theta}$
$\displaystyle =\cfrac{\sqrt{2}}{2} \int \textrm{csc}\left( \theta+\frac{\pi}{4}\right)d\left( \theta+\frac{\pi}{4}\right)$
$\displaystyle =-\cfrac{\sqrt{2}}{2} \textrm{ln}\left[ \textrm{csc}\left( \theta+\frac{\pi}{4}\right)+\textrm{cot}\left( \theta+\frac{\pi}{4}\right)\right]+C_1\ \cdot\cdot\cdot\cdot\cdot\cdot\ \textbf{①}$


$\displaystyle I+2J= \int \cfrac{\left(1+2\textrm{sin}\theta\textrm{cos}\theta \right)d\theta}{\left( \textrm{sin}\theta+\textrm{cos}\theta\right)\left(1-\textrm{sin}\theta\textrm{cos}\theta \right)}=\int \cfrac{\left(\textrm{sin}\theta+\textrm{cos}\theta \right)d\theta}{1-\textrm{sin}\theta\textrm{cos}\theta}$
$\displaystyle =2\sqrt{2} \int \cfrac{\left[\textrm{sin}\left(\theta +\frac{\pi}{4}\right)\right]d\theta}{2-\textrm{sin} 2\theta}$ (设$\phi=\theta+\frac{\pi}{4}$)
$\displaystyle =2\sqrt{2} \int \cfrac{\textrm{sin}\phi \ d\phi}{2-(1-2\textrm{cos}^2\phi)}$
$\displaystyle =-\sqrt{2} \int \cfrac{\ d(\textrm{cos}\phi)}{\left(\frac{1}{\sqrt{2}}\right)^2+\textrm{cos}^2\phi}$
$\displaystyle =-\sqrt{2} \left\{ \sqrt{2}\textrm{arctan}\left( \sqrt{2}\textrm{cos}\phi\right)\right\}+C_2$
$\displaystyle =-2\textrm{arctan}\left( \textrm{cos}\theta-\textrm{sin}\theta\right)+C_2$
$\displaystyle =2\textrm{arctan}\left( \textrm{sin}\theta-\textrm{cos}\theta\right)+C_2\ \cdot\cdot\cdot\cdot\cdot\cdot\ \textbf{②}$


$2\textbf{①}+\textbf{②}=3I$
$=-\sqrt{2}\textrm{ln}\left[ \textrm{csc}\left( \theta+\frac{\pi}{4}\right)+\textrm{cot}\left( \theta+\frac{\pi}{4}\right)\right]+2C_1+2\textrm{arctan}\left( \textrm{sin}\theta-\textrm{cos}\theta\right)+C_2$

$I=-\cfrac{\sqrt{2}}{3}\textrm{ln}\left[ \textrm{csc}\left( \theta+\frac{\pi}{4}\right)+\textrm{cot}\left( \theta+\frac{\pi}{4}\right)\right]+\cfrac{2}{3}\textrm{arctan}\left( \textrm{sin}\theta-\textrm{cos}\theta\right)+C$
$\underline {\underline{ =\cfrac{\sqrt{2}}{3}\textrm{ln}\ \textrm{tan}\left( \cfrac{\theta}{2}+\cfrac{\pi}{8}\right)+\cfrac{2}{3}\textrm{arctan}\left( \textrm{sin}\theta-\textrm{cos}\theta\right)+C}}.$

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