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Find $\displaystyle \int \cos(x) \frac {\ln(\sin(x))} {\ln(\tan(x))}dx.$

A friend and I found this problem statement in our high school mathematics textbook under "hard integrals", and after hours trying every trick we've been taught, we can't solve it.

How do you solve this?

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  • $\begingroup$ Please refrain from using "displaystyle" in titles. Better yet, repeat the specific question in the body of your post. $\endgroup$
    – amWhy
    Feb 11 '20 at 17:15
  • $\begingroup$ Wolfram Alpha seems to say there is no easy way to do this $\endgroup$
    – gt6989b
    Feb 11 '20 at 17:15
  • $\begingroup$ If you got this problem from a high school textbook, could you double-check it? Or maybe it's a misprint. $\endgroup$
    – J.G.
    Feb 11 '20 at 17:17
  • $\begingroup$ It turns out my friend got the problem from a friend, who said it was from a textbook, but it was actually from a website. The answer is apparently "impossible to integrate", which is deeply disappointing. Can't believe I spend so long on a problem to get that answer. $\endgroup$ Feb 12 '20 at 9:12
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According to Maple this antiderivative is not elementary. Since this involves the "purely transcendental" case of the Risch-Norman algorithm, which I believe Maple has implemented completely, I'm confident that this is correct.

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$$I=\int\cos(x)\frac{\ln(\sin x)}{\ln(\tan x)}dx$$ we could try and use that: $$\frac{\ln(\sin x)}{\ln(\tan x)}=\frac{\ln(\sin x)}{\ln(\sin x)-\ln(\cos x)}$$ or that $\ln(\tan x)=-2\ln(\csc^2x-1)$ and maybe use $u=\sin(x)$ although with the two logs present this doesn't look easy either.

EDIT: Another thought would be use feynmans rule: $$I(t)=\int\cos(x)\frac{\ln(\sin xt)}{\ln(\tan x)}dx$$ or something similar then differentiate wrt t.

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    $\begingroup$ We tried all of those things and they didn't lead to an answer. We also tried substituting $e^v = sin(x)$ $\endgroup$ Feb 11 '20 at 17:45
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For the fun of it, I tried series expansion of the integrand. We obtain things like $$\cos (x)\frac{ \log (\sin (x))}{\log (\tan (x))}=1-\frac{ (t+1)}{2 t}x^2+\frac{ (t+2)^2}{24 t^2}x^4-\frac{ \left(t^3+t^2+12t+40\right)}{720 t^3}x^6+\frac{\left(9 t^4+576 t^3-568 t^2-2688 t+6720\right)}{362880 t^4}x^8+\cdots$$ where $t=\log(x)$. So, the general form is $$\cos (x)\frac{ \log (\sin (x))}{\log (\tan (x))}=\sum_{n=0}^\infty \frac{P_n(t)}{t^n} x^{2n}$$

This means that we face a bunch of integrals $$I_{m,n}=\int\frac {x^m}{\log^n(x)}\,dx=-\frac{E_n(-(m+1) \log (x)) } {\log^{n-1}(x)}$$

Limited to the truncated expansion given at the beginning, for a limited range, the results do not look too bad for the integral between $0$ and $k$ $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 0.05 & 0.049986 & 0.049986 \\ 0.10 & 0.099897 & 0.099897 \\ 0.15 & 0.149694 & 0.149694 \\ 0.20 & 0.199369 & 0.199369 \\ 0.25 & 0.248956 & 0.248956 \\ 0.30 & 0.298540 & 0.298540 \\ 0.35 & 0.348275 & 0.348275 \\ 0.40 & 0.398411 & 0.398411 \\ 0.45 & 0.449341 & 0.449341 \\ 0.50 & 0.501690 & 0.501692 \\ 0.55 & 0.556477 & 0.556491 \\ 0.60 & 0.615470 & 0.615546 \\ 0.65 & 0.682009 & 0.682434 \\ 0.70 & 0.763211 & 0.765857 \\ 0.75 & 0.876866 & 0.899287 \end{array} \right)$$

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