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what is the integration of integral $$\int\cot(\log(\sin(x))\, dx$$

I have tried:

Let $\log(\sin(x))=z$ or $\dfrac{1}{\sin(x)}\cos(x)\cdot dx=dz$ , that means $\cot(x)\cdot dx=dz$ $\Longrightarrow$ integration of $dz$ is $z$ then $c=\dfrac{\log(\sin(x))^2}{2}c$ is it true?

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\begin{align} \text{Let } u &= \ln(\sin(x))\\ \mathrm du &= \cot(x)\ \mathrm dx\\ \\ \int \cot(x)\ln(\sin(x)) \mathrm dx&= \int u\ \mathrm du\\ &= u^2/2 + C\\ &= \left[\ln(\sin(x))\right]^2/2 + C \\ \end{align}

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Your procedure should be wrong!

You will have $\int\cot z\tan x~dz$ and you still have to change $x$ in terms of $z$ !

Let $u=\sin x$ ,

Then $x=\sin^{-1}u$

$dx=\dfrac{1}{\sqrt{1-u^2}}du$

$\therefore\int\cot\log\sin x~dx=\int\dfrac{\cot\log u}{\sqrt{1-u^2}}du$

Let $v=\log u$ ,

Then $u=e^v$

$du=e^v~dv$

$\therefore\int\dfrac{\cot\log u}{\sqrt{1-u^2}}du=\int\dfrac{e^v\cot v}{\sqrt{1-e^{2v}}}dv$

But still highly have no hope to find the nice series form expression......

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