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


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

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

海马非马

赞同来自:

用了★永ぁ恒ギ★帖中消去$\cfrac{\pi}{4}$的技巧。

$\displaystyle f(\theta)=\int \cfrac{d\theta}{\textrm{sin}^3\theta+\textrm{cos}^3\theta}$
$\displaystyle =2\int{\cfrac{d\theta}{(\textrm{sin}\theta+\cos\theta)(2-\sin 2\theta)}}$
$\displaystyle =\sqrt{2}\int\cfrac{d\theta}{\sin(\theta+\pi/4)(2-\sin 2\theta)}$
令$\alpha=\theta+\pi/4$, 则
$\displaystyle f(\theta)=f(\alpha-\pi/4)=\sqrt{2}\int\cfrac{d\alpha}{\sin\alpha(2-\cos 2\alpha)}$
$\displaystyle =\sqrt{2}\int\cfrac{\sin\alpha d\alpha}{\sin^2\alpha(2-\cos 2\alpha)}$
$\displaystyle =-\sqrt{2}\int\cfrac{d\cos\alpha}{(1-\cos^2\alpha)(3-2\cos^2\alpha)}$
$\displaystyle =-\sqrt{2}\int\cfrac{d\cos\alpha}{1-\cos^2\alpha}+2\sqrt{2}\int\cfrac{d\cos\alpha}{3-2\cos^2\alpha}$
$\displaystyle =-\sqrt{2}\ln\sqrt{\cfrac{1+\cos\alpha}{1-\cos\alpha}}+\cfrac{\sqrt{6}}{3}\ln\cfrac{\sqrt{3}+\sqrt{2}\cos\alpha}{\sqrt{3}-\sqrt{2}\cos\alpha}+C$
$\displaystyle =-\sqrt{2}\ln\sqrt{\cfrac{1+\cos(\theta+\pi/4)}{1-\cos(\theta+\pi/4)}}+\cfrac{\sqrt{6}}{3}\ln\cfrac{\sqrt{3}+\sqrt{2}\cos(\theta+\pi/4)}{\sqrt{3}-\sqrt{2}\cos(\theta+\pi/4)}+C$

又,$\displaystyle \int \cfrac {a dx} {a^2-x^2}=\tanh^{-1}\cfrac x a +C$
$\displaystyle \int \cfrac {dx} {a^2-b^2 x^2}=\cfrac{1}{ab}\tanh^{-1}\cfrac x {ab} +C$
$\displaystyle f(\theta)=-\sqrt{2}\tanh^{-1}(\cos\theta+\pi/4)+\cfrac{2\sqrt{3}}{3}\tanh^{-1}\cfrac{\sqrt{6}}{3}(\cos\theta+\pi/4)+C$

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