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.
Best Answer
$$\begin{align}\int_0^{\pi/3}\ln(1+\sin x)\ln(1-\sin x)\,dx=&\frac{29\pi^3}{216}+\frac{5\pi}6\ln^2\left(2+\sqrt3\right)+\frac\pi3\ln^22+\frac{\pi^2}{3\sqrt3}\ln2\\+&\frac{8G}3\ln\left(2+\sqrt3\right)-4 \operatorname{Ti}_3\left(2+\sqrt3\right)-\frac{\psi^{(1)}\!\left(\tfrac13\right)}{2 \sqrt{3}}\ln2,\end{align}$$ 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.