Yet another log-sin integral $\int\limits_0^{\pi/3}\log(1+\sin x)\log(1-\sin x)\,dx$ There has been much interest to various log-trig integrals on this site (e.g. see [1][2][3][4][5][6][7][8][9]).
Here is another one I'm trying to solve:
$$\int\limits_0^{\pi/3}\log(1+\sin x)\log(1-\sin x)\,dx\approx-0.41142425522824105371...$$
I tried to feed it to Maple and Mathematica, but they are unable to evaluate in this form. After changing the variable $x=2\arctan z,$ and factoring rational functions under logarithms, the integrand takes the form
$$\frac{2 \log ^2\left(z^2+1\right)}{z^2+1}-\frac{4 \log (1-z) \log \left(z^2+1\right)}{z^2+1}\\-\frac{4 \log (z+1) \log
   \left(z^2+1\right)}{z^2+1}+\frac{8 \log (1-z) \log (z+1)}{z^2+1}$$
in which it can be evaluated by Mathematica. It spits out a huge ugly expression with complex numbers, polylogarithms, polygammas and generalized hypergeometric functions (that indeed matches numerical estimates of the integral). It takes a long time to simplify and with only little improvement (see here if you are curious). 
I'm looking for a better approach to this integral that can produce the answer in a simpler form.
 A: $$\begin{align}\int_0^{\pi/3}\ln(1+\sin x)\ln(1-\sin x)\,dx=&\frac{29\pi^3}{216}+\frac{5\pi}6\ln^2\left(2+\sqrt3\right)+\frac\pi3\ln^22+\frac{\pi^2}{3\sqrt3}\ln2\\+&\frac{8G}3\ln\left(2+\sqrt3\right)-4 \operatorname{Ti}_3\left(2+\sqrt3\right)-\frac{\psi^{(1)}\!\left(\tfrac13\right)}{2 \sqrt{3}}\ln2,\end{align}$$
where $G$ is the Catalan constant, $\operatorname{Ti}_3(z)=\Im\operatorname{Li}_3(iz)$ is the generalized inverse tangent integral, and $\psi^{(1)}(z)$ is the trigamma function.
A: Another way is to use Maclaurin series:
$$\log(1\pm \sin x) = \pm\sin x +\dfrac12 \sin^2x\pm \dfrac13 \sin^3x +\dfrac14\sin^4x\pm \dfrac15\sin^5x+\dfrac16\sin^6x+\dots,$$where $\sin^2 x<\dfrac 34.$  Expression
$$ \int\limits_0^{\dfrac{\pi}3}\left(\dfrac14\sin^4x\left(1+ \dfrac12\sin^2x+\dfrac13\sin^4x+\dfrac14\sin^6x+\dots\right)^2 - {\sin^2x\left(1 + \dfrac13 \sin^2x + \dfrac15 \sin^4x + \dfrac17\sin^4x+\dots\right)^2}\right)dx$$
looks convenient to approximate calculations.
A: Too long for a comment : Using the fact that $\sin t=\cos\bigg(\dfrac\pi2-t\bigg),$ together with the 
well-known formulas for $1\pm\cos(2u),$ and the properties of the natural logarithm, we have 

$$\begin{align}I(a)&=\int_0^a\ln(1-\sin x)\ln(1+\sin x)~dx=\\&=(2\pi-a)\ln^22-\dfrac{\pi^3}{12}+2\ln2\displaystyle\int_0^a\ln\cos x~dx-8\int_0^b\ln\sin x\ln\cos x~dx,\end{align}$$ 

where $a\in\bigg(0,~\dfrac\pi2\bigg)$ and $b=\dfrac\pi4-\dfrac a2.~$ Even in the absence of any particularly bright ideas, the 
last two integrals are still expressible in terms of the derivatives of the $($ incomplete $)$ beta  function. See Wallis' integrals for more information. In this specific case, $a=\dfrac\pi3$ and $b=\dfrac\pi{12}.$
