$$\int \frac{\ln(1-\sin(2x))}{\sin(2x)}\,\mathrm{d}x$$
I tried to find the solution through WolframAlpha and I got the result, but I do not know how the solution is: $$1/2 (2 \operatorname{Li}_2(1 - \tan(x)) + \operatorname{Li}_2(-i \tan(x)) + \operatorname{Li}_2(i \tan(x)) + \log(1 - i \tan(x)) \log(\tan(x)) + \log(1 + i \tan(x)) \log(\tan(x)) + \log(1 - \sin(2 x)) \log(\tan(x))) + \operatorname{const.}$$
We want to prove that $$I := {\displaystyle\int}\dfrac{\ln\left(1-2\sin\left(x\right)\right)}{\sin\left(2x\right)}\,\mathrm{d}x = -\dfrac{\operatorname{Li}_2\left(2\sin\left(x\right)-1\right)}{4}-\dfrac{\operatorname{Li}_2\left(2\sin\left(x\right)\right)}{2}-\dfrac{\ln\left(1-2\sin\left(x\right)\right)\ln\left(2-2\sin\left(x\right)\right)}{4}-\dfrac{\ln\left(3\right)\ln\left(-2\sin\left(x\right)-2\right)}{4}+\dfrac{\operatorname{Li}_2\left(-\frac{-2\sin\left(x\right)-2}{3}\right)}{4}.$$
Rewrite
$$={\displaystyle\int}-\class{steps-node}{\cssId{steps-node-1}{2\cos\left(x\right)}}\cdot\class{steps-node}{\cssId{steps-node-2}{\dfrac{\ln\left(1-2\sin\left(x\right)\right)}{\left(-2\sin\left(x\right)-2\right)\left(2-2\sin\left(x\right)\right)\sin\left(x\right)}}}\,\mathrm{d}x$$ With the substitution $u=1-2\sin(x)$ we get to $$-\class{steps-node}{\cssId{steps-node-3}{2}}{\displaystyle\int}\dfrac{\ln\left(u\right)}{\left(u-3\right)\left(u-1\right)\left(u+1\right)}\,\mathrm{d}u.$$ By using partial fraction decomposition $$\class{steps-node}{\cssId{steps-node-4}{\dfrac{1}{8}}}\underbrace{\int\dfrac{\ln\left(u\right)}{u+1}\,\mathrm{d}u}_{I_1}-\class{steps-node}{\cssId{steps-node-5}{\dfrac{1}{4}}}\underbrace{{\int}\dfrac{\ln\left(u\right)}{u-1}\,\mathrm{d}u}_{I_2}+\class{steps-node}{\cssId{steps-node-6}{\dfrac{1}{8}}}\underbrace{{\int}\dfrac{\ln\left(u\right)}{u-3}\,\mathrm{d}u}_{I_3}.$$
$I_1$
Partial integration yields $$I_1=\ln\left(u\right)\ln\left(u+1\right)-{\displaystyle\int}\dfrac{\ln\left(u+1\right)}{u}\,\mathrm{d}u$$ If you substitute $v=-u$ in the integral, this is equal to $\operatorname{Li_2}(v)$. Substituting back yields $$I_1=\ln\left(u\right)\ln\left(u+1\right)+\operatorname{Li_2}(-u).$$
$I_2$
Substitute $v=u-1$. This will get us to a very similar integral we calculated in $I_1$. In conclusion $$\operatorname{Li_2}(1-u)$$
$I_3$
After substituting $v=u-3$: $$I_3={\displaystyle\int}\left(\dfrac{\ln\left(\frac{v}{3}+1\right)}{v}+\dfrac{\ln\left(3\right)}{v}\right)\mathrm{d}v$$ We can split the integral up. The first integral can be evaluated with the same methods as earlier. This will yields $-\operatorname{Li}_2\left(-\dfrac{v}{3}\right)$. The second integral easily evaluates to $\ln 3\ln v$. $$I_3=\ln\left(3\right)\ln\left(u-3\right)-\operatorname{Li}_2\left(-\dfrac{u-3}{3}\right).$$
$I$
Putting all of the above together: $$I=-2\left(\dfrac{\ln\left(u\right)\ln\left(u+1\right)}{8}+\dfrac{\operatorname{Li}_2\left(-u\right)}{8}+\dfrac{\ln\left(3\right)\ln\left(u-3\right)}{8}-\dfrac{\operatorname{Li}_2\left(-\frac{u-3}{3}\right)}{8}+\dfrac{\operatorname{Li}_2\left(1-u\right)}{4}\right).$$ Now remember $u=1-2\sin x$.
$$I=-\dfrac{\operatorname{Li}_2\left(2\sin\left(x\right)-1\right)}{4}-\dfrac{\operatorname{Li}_2\left(2\sin\left(x\right)\right)}{2}-\dfrac{\ln\left(1-2\sin\left(x\right)\right)\ln\left(2-2\sin\left(x\right)\right)}{4}-\dfrac{\ln\left(3\right)\ln\left(-2\sin\left(x\right)-2\right)}{4}+\dfrac{\operatorname{Li}_2\left(-\frac{-2\sin\left(x\right)-2}{3}\right)}{4}.$$