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 $\mathbb{R}$
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.
I used MMA to calculate this indefinite integral and found that his result contains a hypergeometric function, so the primitive function of this indefinite integral cannot be expressed by elementary functions.