How to evaluate $int frac{dx}{sin(ln(x))}$?

Based on the hypergeometric answers of J.G. and Simply Beautiful Art.

Taking another branch of the solution of the hypergeometric differential equation from those answers, we can get solutions like this: $$ f(x) = mathrm{Re}left[ {frac { left( 1+i right) {x}^{1+i}}{{x}^{2,i}-1} ;{mbox{$_2$F$_1$}left(1,1;frac{3-i}{2};,{frac {{x}^{2,i}}{{x}^{2,i}-1}}right)} } right] $$ which satisfies $$ f'(x) = frac{1}{sin(log x)} $$ in the interval $(0.21 , 0.55)$. Here we are inside the radius of convergence of the hypergeometric function.

Change variables $y=sin x$ to convert this to the indefinite integral $$ intfrac{e^y}{sin y};dy . $$

This is not an elementary function.
Reference: 2.665 in

Gradshteyn, I. S.; Ryzhik, I. M.; Zwillinger, Daniel (ed.); Moll, Victor (ed.), Table of integrals, series, and products. Translated from the Russian. Translation edited and with a preface by Victor Moll and Daniel Zwillinger, Amsterdam: Elsevier/Academic Press (ISBN 978-0-12-384933-5/hbk; 978-0-12-384934-2/ebook). xlv, 1133 p. (2015). ZBL1300.65001.

To avoid hypergeometric functions use the partial fraction decomposition:

$$frac{1}{sin x} = frac{1}{x} + sumlimits _{n=1} ^{infty}(-1)^n left (frac{1}{x-npi}-frac{1}{x+npi}right ), $$

replace $x$ with $ln x$ and use

$$intfrac{dx}{ln x pm npi}=e^{mp npi}text{Ei}(xpm npi)+text{const}$$

where $text{Ei}(x)=-int_{-x}^{infty}frac{e^{-t}dt}{t}$ is the so called exponential integral

By Euler's formula,

$$sin(ln(x))=frac{e^{iln(x)}-e^{-iln(x)}}{2i}=frac{x^i-x^{-i}}{2i}$$

In the integral, this works out to give us

$$intfrac{mathrm dx}{sin(ln(x))}=intfrac{2i~mathrm dx}{x^i-x^{-i}}=2iintfrac{x^i~mathrm dx}{x^{2i}-1}=-2iintfrac{x^i~mathrm dx}{1-x^{2i}}$$

By expanding with geometric series, this then becomes

$$intfrac{x^i~mathrm dx}{1-x^{2i}}=sum_{k=0}^inftyint x^{(2k+1)i}~mathrm dx=sum_{k=0}^inftyfrac{x^{1+(2k+1)i}}{1+(2k+1)i}$$

Observe that the ratio of consecutive terms in this series is given by

$$frac{x^{1+(2k+3)i}/(1+(2k+3)i)}{x^{1+(2k+1)i}/(1+(2k+1)i)}=frac{(2k+1)i+1}{(2k+3)i+1}x^{2i}=frac{(k+color{#3377cc}{frac{1+i}2})(k+color{#3377cc}1)}{k+color{#339999}{frac{1+3i}2}}frac{color{#dd3333}{x^{2i}}}{k+1}$$

which implies the series is a hypergeometric function:

$$sum_{k=0}^inftyfrac{x^{1+(2k+1)i}}{1+(2k+1)i}=x^{1+i}{}_2F_1left(color{#3377cc}{frac{1+i}2},color{#3377cc}1;color{#339999}{frac{1+3i}2};color{#dd3333}{x^{2i}}right)$$

and altogether,

$$intfrac{mathrm dx}{sin(ln(x))}=-2ix^{1+i}{}_2F_1left(frac{1+i}2,1;frac{1+3i}2;x^{2i}right)color{#999999}{{}+C}$$

which likely cannot be simplified further, though can be rewritten using various hypergeometric identities.

Note: The above manipulations require that the series converges, but the end results in terms of hypergeometric functions hold everywhere they both exist, as they are defined through the use of analytic continuation.

Let's prove the theorem, stated by @PleasecorrectGrammarMistakes, that$$tfrac{d}{dx}left[(-1-i)x^{1+i}Fleft(tfrac{1-i}{2},,1;,tfrac{3-i}{2};,x^{2i}right)right]=cscln x$$provided $|x^{2i}<1$. We'll begin by noting$$[y^n]F(a,,1;,a+1;,y)=frac{Gamma(a+n)Gamma(n+1)Gamma(a+1)}{Gamma(a)Gamma(1)Gamma(a+n+1)n!}=frac{a}{a+n},$$so the left-hand side is$$sum_{nge0}(-1-i)frac{1-i}{2n+1-i}(1+(2n+1)i)x^{(2n+1)i}=-2isum_{nge0}x^{(2n+1)i}=frac{-2ix^i}{1-x^{2i}}.$$We can rewrite this as$$frac{2i}{x^i-x^{-i}}=frac{2i}{e^{iln x}-e^{-iln x}}=cscln x,$$as required.

This answer is wrong, but I am leaving it here for the archive (comments discussion), until we find a proper solution.

Thanks to your comments, here is the solution. (Apparently approach 3 was on the right path)

Let, $ln(x) = u iff dx = e^u , du$

$$ int frac{dx}{sin(ln(x))} = I= int frac{e^u}{sin(u)} du = int frac{(e^u)'}{sin(u)} du = $$

$$ frac{e^u}{sin(u)} - int e^u left(frac{1}{sin(u)}right)' = frac{(e^u)'}{sin(u)} - int e^u frac{cos(u)}{sin^2(u)} = $$

$$ frac{e^u}{sin(u)} - int e^u frac{cos(u)}{2cos(u)sin(u)} = $$

$$ frac{e^u}{sin(u)} - frac{1}{2}int frac{e^u}{sin(u)} = $$

$$ frac{e^u}{sin(u)} - I iff $$

$$ 2I = frac{(e^u)'}{sin(u)} $$

$$ I = frac{e^u}{2sin(u)} $$

Therefore, substituting back the initial variables:

$$ I = frac{e^{ln(x)}}{2sin(ln(x))} $$