Computing the value of $\operatorname{Li}_{3}\left(\frac{1}{2} \right) $ 
How to prove the following identity
  $$ \operatorname{Li}_{3}\left(\frac{1}{2} \right) = \sum_{n=1}^{\infty}\frac{1}{2^n n^3}= \frac{1}{24} \left( 21\zeta(3)+4\ln^3 (2)-2\pi^2 \ln2\right)\,?$$

Where $\operatorname{Li}_3 (x)$ is the trilogarithm also the result from above can be found here in $(2)$. 
In particular for $x=\frac12$ we have:$$\operatorname{Li}_3\left(\frac12\right)=\sum_{n=1}^\infty \frac{1}{n^3}\frac{1}{2^n}=\frac14 \sum_{n=1}^\infty \frac{1}{2^{n-1}} \int_0^\frac12 x^{n-1}\ln^2 xdx=\frac14 \int_0^1 \frac{\ln^2 x}{1-\frac{x}{2}}dx$$

Here is a partial solution:
Substitute $x=2\sin^2 \theta$ to obtain:
$$\int_0^{\frac{\pi}{4}} \left(\ln^2 2 + 4 \ln 2 \ln (\sin \theta) + 4 \ln^2 (\sin \theta)\right) \tan \theta \, \text{d}\theta$$
The $\ln^2 2$ part is trivial and evaluates to $\frac{\ln^3 2}{2}$.
The $\ln 2 \ln (\sin \theta)$ part is relatively straightforward.
Let $\displaystyle I = 4\ln 2 \int_0^\frac{\pi}{4} \ln (\sin \theta) \tan{\theta} \, \text{d}\theta$.
Integrate by parts to obtain $\displaystyle I = -\ln^3 2 + 4\ln 2 \int_0^\frac{\pi}{4} \ln (\cos \theta) \cot{\theta} \, \text{d}\theta$
Substitute $\theta \mapsto \frac{\pi}{2} - \theta$ to obtain $\displaystyle I = -\ln^3 2 + 4\ln 2 \int_\frac{\pi}{4}^\frac{\pi}{2} \ln (\sin \theta) \tan{\theta} \, \text{d}\theta$
which can then be averaged to obtain $\displaystyle I = -\frac{\ln^3 2}{2} + 2\ln 2 \int_0^\frac{\pi}{2} \ln (\sin \theta) \tan{\theta} \, \text{d}\theta$
Then use the substitution $u = \cos^2 \theta$ or $u = \sin^2 \theta$ to obtain the classic dilogarithm integral for $\zeta(2)$.
This simplifies down to $\displaystyle I = -\frac{\ln^3 2}{2} - \frac{\pi^2 \ln 2}{12}$
It then remains to show that $\displaystyle 4 \int_0^\frac{\pi}{4} \ln^2 (\sin \theta) \tan \theta = \frac{\ln^3 2}{6} + \frac{7}{8} \zeta(3)$.
I will continue attempting this integral and come back when/if I progress.
