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 $\mathbb{R}$

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.
