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.
We can show fairly straightforwardly that this reduces to an elliptic integral, which cannot be an elementary function: put $ x = \arctan(u^2) $. Then $ dx = 2u/(1+u^4) \, du $, $\tan x = u^2$ and $\sin x = u^2/\sqrt{1+u^4}$, and rationalising implies that the integral becomes
$$ \int \bigg( 2u^4 + \frac{2u^6}{\sqrt{1+u^4}} \bigg) \, du , $$
and we just need to worry about the second term. It so happens that this was one of the earliest integrals Liouville considered when he became interested in when an integral is algebraic (See Lützen's Joseph Liouville 1809–1882 pp. 374ff. for the details). An integration by parts reduces us to $ \int \frac{u^2}{\sqrt{1+u^4}} \, du $, which is known to not be elementary (see either Liouville's work, or Ritt's book Integration in finite terms). Thus the "elementary part" is
$$ \frac{2}{5} ( u^5 + u^3 \sqrt{1+u^4}) , $$
while the non-elementary part is the elliptic integral
$$ - \frac{6}{5} \int \frac{u^2}{\sqrt{1+u^4}} \, du = \frac{6}{5}\sqrt{i} ( F(\arcsin(\sqrt{i}u) \mid -1) - F(\arcsin(\sqrt{i}u) \mid -1) . $$
One could write in terms of $x$ again, but there doesn't seem much point.
Best Answer
$$\int_0^{\pi/2}x\ln(\sin x)dx=\int_0^{\pi/2}x\left(-\ln2-\sum_{n=1}^\infty\frac{\cos(2nx)}{n}\right)dx$$
$$=-\frac{\pi^2}{8}\ln2-\sum_{n=1}^\infty\frac{1}{n}\int_0^{\pi/2}x\cos(2nx)dx$$
$$=-\frac{\pi^2}{8}\ln2-\sum_{n=1}^\infty\frac{1}{n}\left(\frac{\cos(n\pi)}{4n^2}+\frac{\pi\sin(n\pi)}{4n}-\frac{1}{4n^2}\right)$$
$$=-\frac{\pi^2}{8}\ln2-\sum_{n=1}^\infty\frac{1}{n}\left(\frac{(-1)^n}{4n^2}+\frac{0}{4n}-\frac{1}{4n^2}\right)$$
$$=-\frac{\pi^2}{8}\ln2-\frac14\text{Li}_3(-1)+\frac14\zeta(3)$$
$$=-\frac{\pi^2}{8}\ln2+\frac{7}{16}\zeta(3)$$
Bonus: With subbing $x\to \pi/2-x$ we have
$$\int_0^{\pi/2}x\ln(\cos x)dx=\int_0^{\pi/2}(\pi/2-x)\ln(\sin x)dx$$
$$=\frac{\pi}{2}\int_0^{\pi/2}\ln(\sin x)dx-\int_0^{\pi/2}x\ln(\sin x)dx$$
$$=\frac{\pi}{2}\left(-\frac{\pi}{2}\ln2\right)-\left(-\frac{\pi^2}{8}\ln2+\frac{7}{16}\zeta(3)\right)$$ $$=-\frac{\pi^2}{8}\ln(2)-\frac7{16}\zeta(3)$$
Or we can use the Fourier series of $\ \ln(\cos x)=-\ln2-\sum_{n=1}^\infty\frac{(-1)^n\cos(2nx)}{n}$.
Also by subtracting the two integrals gives
$$\int_0^{\pi/2}x\ln(\tan x)dx=\frac78\zeta(3)$$
Or we can use the Fourier series of $\ \ln(\tan x)=-2\sum_{n=1}^\infty\frac{\cos((4n-2)x)}{2n-1}.$