又一道MMA出虚数的三角函数不定积分

青青子衿

\begin{align*}
&\int \dfrac{1}{\cos\left(3x\right)\sin\left(4x\right)}\mathrm{d}x\\
=\,&2\int \dfrac{1}{\sin x+\sin7x}\mathrm{d}x=2\int \dfrac{\sin x}{\sin x\left(\sin x+\sin7x\right)}\mathrm{d}x\\
=\,&2\int \dfrac{\sin x}{\sin x\left(\sin x+\sin7x\right)}\mathrm{d}x=2\int \dfrac{\sin x}{\sin^2x+\sin\left(x\right)\sin\left(7x\right)}\mathrm{d}x\\
=\,&2\int \dfrac{\sin x}{\frac{1-\,\cos2x}{2}+\frac{\cos6x-\,\cos8x}{2}}\mathrm{d}x=4\int \dfrac{\sin x}{1-\cos2x+\cos6x-\cos8x}\mathrm{d}x\\
=\,&4\int \dfrac{\mathrm{d}\left(\cos x\right)}{128\cos^8x-288\cos^6x+208\cos^4x-48\cos^2x}\\
=\,&-\frac{1}{4}\int \dfrac{\mathrm{d}\left(\cos x\right)}{\cos^2x\left(1-\cos x\right)\left(1+\cos x\right)\left(2\cos^2x-1\right)\left(4\cos^2x-3\right)}\\
=\,&-\frac{1}{4}\int \dfrac{\mathrm{d}t}{t^2\left(1-t\right)\left(1+t\right)\left(2t^2-1\right)\left(4t^2-3\right)}\\
\end{align*}

1. \frac{1}{2}\ln\left(\frac{x}{a}+\sqrt[4]{1+\frac{x^4}{a^4}}\right)+\frac{1}{4}\ln\left(\frac{x^2}{a^2}+\sqrt{1+\frac{x^4}{a^4}}\right)+\frac{1}{2}\arctan\left(\frac{x}{\sqrt[4]{a^4+x^4}}\right)

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青青子衿

回复 1# 青青子衿
\begin{align*}
&-\frac{1}{4}\int \dfrac{\mathrm{d}t}{t^2\left(1-t\right)\left(1+t\right)\left(2t^2-1\right)\left(4t^2-3\right)}\\  
=\,&-\frac{1}{12}\int\frac{1}{t^2}\mathrm{d}t-\frac{1}{4}\int\frac{1}{1-t^2}\mathrm{d}t-\int\frac{1}{1-2t^2}\mathrm{d}t+\frac{8}{3}\int\frac{1}{3-4t^2}\mathrm{d}t\\
=\,&\frac{1}{12t}-\frac{1}{8}\ln\left|\frac{1+t}{1-t}\right|-\frac{1}{2\sqrt{\,2\,}}\ln\left|\frac{1+\sqrt{\,2\,}\,t}{1-\sqrt{\,2\,}\,t}\right|+\frac{2}{3\sqrt{\,3\,}}\ln\left|\frac{\sqrt{\,3\,}+2t}{\sqrt{\,3\,}-2t}\right|+C\\
\end{align*}

\begin{align*}
\frac{1}{12t^2}+\frac{1}{4\left(1-t^2\right)}-\frac{1}{2t^2-1}+\frac{8}{3\left(4t^2-3\right)}\\
\frac{1}{4t^2\left(1-t^2\right)\left(2t^2-1\right)\left(4t^2-3\right)}
\end{align*}

\begin{align*}  
&\int \dfrac{1}{\cos\left(3x\right)\sin\left(4x\right)}\mathrm{d}x\\  

=\,&4\int \dfrac{\sin x}{1-\cos2x+\cos6x-\cos8x}\mathrm{d}x\\  
=\,&4\int \dfrac{\mathrm{d}\left(\cos x\right)}{128\cos^8x-288\cos^6x+208\cos^4x-48\cos^2x}\\  
=\,&-\frac{1}{4}\int \dfrac{\mathrm{d}\left(\cos x\right)}{\cos^2x\left(1-\cos x\right)\left(1+\cos x\right)\left(2\cos^2x-1\right)\left(4\cos^2x-3\right)}\\  
=\,&-\frac{1}{12}\int\frac{\mathrm{d}\left(\cos x\right)}{\cos^2x}-\frac{1}{4}\int\frac{\mathrm{d}\left(\cos x\right)}{1-\cos^2x}-\int\frac{\mathrm{d}\left(\cos x\right)}{1-2\cos^2x}+\frac{8}{3}\int\frac{\mathrm{d}\left(\cos x\right)}{3-4\cos^2x}\\  

=\,&\,\frac{1}{12\cos x}-\frac{1}{8}\ln\left(\frac{1+\cos x}{1-\cos x}\right)-\frac{1}{2\sqrt{\,2\,}}\ln\left|\frac{1+\sqrt{\,2\,}\cos x}{1-\sqrt{\,2\,}\cos x}\right|+\frac{2}{3\sqrt{\,3\,}}\ln\left|\frac{\sqrt{\,3\,}+2\cos x}{\sqrt{\,3\,}-2\cos x}\right|+C\\

\end{align*}

\begin{align*}
\operatorname{artanh}\left(\cos x\right)&=\frac{1}{2}\ln\left(\frac{1+\cos x}{1-\cos x}\right)\\
&=-\ln\left|\tan\frac{x}{2}\right|
\end{align*}

kuing

我倒是想知道软件是用什么方法弄的，结果有几个 tan(x/2)，难道用了万能公式？

青青子衿

回复 3# kuing
手机上下载一个：Wolfram Alpha
输入：∫1/(cos3xsin4x)dx
点击"Step-by-step solution"
有的不定积分会给出求解过程

kuing

回复 4# 青青子衿
我用网页版的 wolframalpha 试了下：wolframalpha.com/input/?i=%E2%88%AB1%2F(cos3xsin4x)dx 点那 step-by-step 要 Get pro 才能看

业余的业余

$\cos 3x=\cos 2x \cos x-\sin 2x\sin x=\cos x(\cos^2 x - \sin^2 x-2\sin^2 x)=\cos x(4 \cos^2 x-3)$

$\sin 4x=2 \sin 2x \cos 2x=4 \sin x \cos x (2\cos^2 x-1)$

$\begin{align*}I&=\int \cfrac {\sin x}{4 \cos^2 x (1-\cos^2 x)(2\cos^2 x-1)(4 \cos^2 x-3)} \mathrm{d}x\\
&=-\cfrac 14 \int \cfrac{dt}{t^2(1-t^2)(2t^2-1)(4t^2-3)}\hspace{1cm}\text{令} t=\cos x\end{align*}$

与一楼结果同。

hbghlyj

kuing 发表于 2019-2-14 13:22
回复 4# 青青子衿
点那 step-by-step 要 Get pro 才能看

可以使用WolfreeAlpha wolfreealpha.on.fleek.co/:

Wolfree bypasses paywalls and shares step-by-step solutions. (It's like Sci-Hub, but Sci-Hub shares scholarly literature.) Wolfree is free and open-source.