Integral of $$$\ln\left(\sin{\left(x \right)}\right)$$$

This integral does not have a closed form:

$${\color{red}{\int{\ln{\left(\sin{\left(x \right)} \right)} d x}}} = {\color{red}{\left(\frac{i x^{2}}{2} - x \ln{\left(1 - e^{2 i x} \right)} + x \ln{\left(\sin{\left(x \right)} \right)} + \frac{i \operatorname{Li}_{2}\left(e^{2 i x}\right)}{2}\right)}}$$

Therefore,

$$\int{\ln{\left(\sin{\left(x \right)} \right)} d x} = \frac{i x^{2}}{2} - x \ln{\left(1 - e^{2 i x} \right)} + x \ln{\left(\sin{\left(x \right)} \right)} + \frac{i \operatorname{Li}_{2}\left(e^{2 i x}\right)}{2}$$

Add the constant of integration:

$$\int{\ln{\left(\sin{\left(x \right)} \right)} d x} = \frac{i x^{2}}{2} - x \ln{\left(1 - e^{2 i x} \right)} + x \ln{\left(\sin{\left(x \right)} \right)} + \frac{i \operatorname{Li}_{2}\left(e^{2 i x}\right)}{2}+C$$

$$$\int \ln\left(\sin{\left(x \right)}\right)\, dx = \left(\frac{i x^{2}}{2} - x \ln\left(1 - e^{2 i x}\right) + x \ln\left(\sin{\left(x \right)}\right) + \frac{i \operatorname{Li}_{2}\left(e^{2 i x}\right)}{2}\right) + C$$$A