How to evaluate $\int \frac{dx}{\sin(\ln(x))}$? I am wondering how to evaluate the indefinite integral
$$\int \frac{dx}{\sin(\ln(x))} \quad (1)$$

Attempt 1
I tried using Weierstrass substitution.

The Weierstrass substitution, (named after K.Weierstrass (1815)), is a substitution used in order to convert trigonometric functions rational expressions to polynomial rational expressions. Integrals of this type are usually easier to evaluate.

This substitution is constructed by letting:
$$t = \tan\left(\frac{x}{2}\right) \iff x = 2\arctan(t)  \iff dx = \frac{2}{t^2+1}$$
Using basic trigonometric identities it is easy to prove that:
$$\cos x = \dfrac{1 - t^2}{1 +  t^2}$$
$$\sin x = \dfrac{2t}{1 + t^2}$$
But I couldn't express $\ln(x)$ in terms of $t$.

Attempt 2
I tried using integration by parts but I couldn't find a workaround, it gets more complicated, really fast.
$$ \int \frac{dx}{\sin(\ln(x))} \ = x \sin(\ln(x)) - \int \frac{\cot \left(\ln \left(x\right)\right)}{x\sin \left(\ln \left(x\right)\right)} $$


Attempt 3
The most logical substitution I could think of. It doesn't seem to lead anywhere though.
Let, $\ln(x) = u \iff dx = \, e^u du$
$$ (1) \iff \int \frac{dx}{\sin(\ln(x))} = \int \frac{e^u}{\sin(u)} du = \int \frac{(e^u)'}{\sin(u)} du = $$
$$ \frac{(e^u)'}{\sin(u)} - \int e^u \left(\frac{1}{\sin(u)}\right)' = \frac{(e^u)'}{\sin(u)} - \int e^u \frac{\cos(u)}{\sin^2(u)} =  ?$$

Attempt 4
A combination of attempts 1,2, 3.
Let $\ln(x) = t$ then $dx = e^t dt$, therefore,
$$\int \frac{dx}{\sin(\ln(x))} dx = \int \frac{e^t }{\sin(t)}dt \quad (1)$$
Let's first evaluate $$ \int \frac{1\:}{\sin\left(t\right)}dt \quad (2)$$
Using the Weierstrass substitution $$ t = \arctan(\frac{x}{2})$$
it is easy to prove that
$$ (2) = \int \frac{1\:}{\sin\left(t\right)}dt= \ln \left|\tan \left(\frac{t}{2}\right)\right|+C$$
Therefore,
$$ (1) \iff I = \int e^x\left(\ln \:\left|\tan \:\left(\frac{t}{2}\right)\right|\right)'dt = e^x \ln \:\left|\tan \:\left(\frac{t}{2}\right)\right| - \int (e^x)' \ln \:\left|\tan \:\left(\frac{t}{2}\right)\right|dt = $$
$$  e^x \ln \:\left|\tan \:\left(\frac{t}{2}\right)\right| - \left( e^x \ln \:\left|\tan \:\left(\frac{t}{2}\right)\right|  - \int e^x \left(\ln \:\left|\tan \:\left(\frac{t}{2}\right)\right|\right)'dt \right) $$
$$ I = 0 + I \iff 0=0$$
Tautology. No answer here.

Attempt 5
Ask a question on MathExchange: Any ideas?
Note: A complex-plane solution was proposed in the comments, but I am evaluating this on $\mathbb{R}$
 A: To avoid hypergeometric functions use  the partial fraction decomposition:
$$\frac{1}{\sin x} = \frac{1}{x} + \sum\limits _{n=1} ^{\infty}(-1)^n \left (\frac{1}{x-n\pi}-\frac{1}{x+n\pi}\right ),
$$
replace $x$ with $\ln x$ and use
$$\int\frac{dx}{\ln x \pm n\pi}=e^{\mp n\pi}\text{Ei}(x\pm n\pi)+\text{const}$$
where $\text{Ei}(x)=-\int_{-x}^{\infty}\frac{e^{-t}dt}{t}$ is the so called exponential integral
A: Change variables $y=\sin x$ to convert this to the indefinite integral
$$
\int\frac{e^y}{\sin y}\;dy .
$$
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.
A: 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, $\ln(x) = u \iff dx = e^u \, du$
$$ \int \frac{dx}{\sin(\ln(x))} = I=  \int \frac{e^u}{\sin(u)} du = \int \frac{(e^u)'}{\sin(u)} du = $$
$$ \frac{e^u}{\sin(u)} - \int e^u \left(\frac{1}{\sin(u)}\right)' = \frac{(e^u)'}{\sin(u)} - \int e^u \frac{\cos(u)}{\sin^2(u)} =  $$
$$ \frac{e^u}{\sin(u)} - \int e^u \frac{\cos(u)}{2cos(u)sin(u)} =  $$
$$ \frac{e^u}{\sin(u)} - \frac{1}{2}\int \frac{e^u}{sin(u)} = $$
$$ \frac{e^u}{\sin(u)} - I \iff $$
$$ 2I = \frac{(e^u)'}{\sin(u)} $$
$$ I = \frac{e^u}{2\sin(u)} $$
Therefore, substituting back the initial variables:
$$ I = \frac{e^{\ln(x)}}{2\sin(\ln(x))} $$
A: By Euler's formula,
$$\sin(\ln(x))=\frac{e^{i\ln(x)}-e^{-i\ln(x)}}{2i}=\frac{x^i-x^{-i}}{2i}$$
In the integral, this works out to give us
$$\int\frac{\mathrm dx}{\sin(\ln(x))}=\int\frac{2i~\mathrm dx}{x^i-x^{-i}}=2i\int\frac{x^i~\mathrm dx}{x^{2i}-1}=-2i\int\frac{x^i~\mathrm dx}{1-x^{2i}}$$
By expanding with geometric series, this then becomes
$$\int\frac{x^i~\mathrm dx}{1-x^{2i}}=\sum_{k=0}^\infty\int x^{(2k+1)i}~\mathrm dx=\sum_{k=0}^\infty\frac{x^{1+(2k+1)i}}{1+(2k+1)i}$$
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:
$$\sum_{k=0}^\infty\frac{x^{1+(2k+1)i}}{1+(2k+1)i}=x^{1+i}{}_2F_1\left(\color{#3377cc}{\frac{1+i}2},\color{#3377cc}1;\color{#339999}{\frac{1+3i}2};\color{#dd3333}{x^{2i}}\right)$$
and altogether,
$$\int\frac{\mathrm dx}{\sin(\ln(x))}=-2ix^{1+i}{}_2F_1\left(\frac{1+i}2,1;\frac{1+3i}2;x^{2i}\right)\color{#999999}{{}+C}$$
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.
A: Let's prove the theorem, stated by @PleasecorrectGrammarMistakes, that$$\tfrac{d}{dx}\left[(-1-i)x^{1+i}F\left(\tfrac{1-i}{2},\,1;\,\tfrac{3-i}{2};\,x^{2i}\right)\right]=\csc\ln x$$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$$\sum_{n\ge0}(-1-i)\frac{1-i}{2n+1-i}(1+(2n+1)i)x^{(2n+1)i}=-2i\sum_{n\ge0}x^{(2n+1)i}=\frac{-2ix^i}{1-x^{2i}}.$$We can rewrite this as$$\frac{2i}{x^i-x^{-i}}=\frac{2i}{e^{i\ln x}-e^{-i\ln x}}=\csc\ln x,$$as required.
A: 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.
