Who's up to the challenge? 45

Using Integration by Parts, $\int \ln(1 - \cos(x)) dx = x \ln(1 - \cos(x)) - \int \frac{x \sin(x)}{1 - \cos(x)} dx$ $= x \ln(1 - \cos(x)) - \int x \cot\left(\frac{x}{2}\right)dx$ $= x \ln(1 - \cos(x)) - \int ix \coth\left(\frac{ix}{2}\right)dx$ $= x \ln(1 - \cos(x)) + \int ix \left( \frac{1 + e^{ix}}{1 - e^{ix}} \right) dx$ $= x \ln(1 - \cos(x)) + \int ix \cdot dx -2 \int x \left( \frac{-i e^{ix}}{1 - e^{ix}} \right) dx$ $= x \ln(1 - \cos(x)) + \frac{i x^2}{2} - 2 \int x \cdot \frac{\delta}{\delta x} \left\{\ln(1 - e^{ix})\right\} dx$ Let's use Integration by Parts again: $= x \ln(1 - \cos(x)) + \frac{i x^2}{2} - 2\left[x\ln(1 - e^{ix}) - \int \ln(1 - e^{ix}) dx \right]$ $= x \ln(1 - \cos(x)) + \frac{i x^2}{2} - 2x\ln(1 - e^{ix}) -2 \int \sum_{k = 1}^{\infty} \frac{e^{kix}}{k}dx$ $= x \ln(1 - \cos(x)) + \frac{i x^2}{2} - 2x\ln(1 - e^{ix}) -2 \sum_{k = 1}^{\infty} \frac{e^{kix}}{k(ki)} + C$ $= x \ln(1 - \cos(x)) + \frac{i x^2}{2} - 2x\ln(1 - e^{ix}) + 2i \sum_{k = 1}^{\infty} \frac{(e^{ix})^k}{k^2} + C$ $= x \ln(1 - \cos(x)) + \frac{i x^2}{2} - 2x\ln(1 - e^{ix}) + 2i \cdot \text{Li}_2(e^{ix}) + C$ Therefore, $\int_{0}^{1} \ln(1 - \cos(x)) dx$ $= \ln(1 - \cos(1)) + \frac{i}{2} - 2\ln(1 - e^{i}) + 2i \cdot \text{Li}_2(e^{i}) - \left[ \lim_{x\to 0} [x \ln(1 - \cos(x)) - 2x\ln(1 - e^{ix})] + 2i \cdot \text{Li}_2(1) \right]$ $= \ln(1 - \cos(1)) + \frac{i}{2} - 2\ln(1 - e^{i}) + 2i \cdot \text{Li}_2(e^{i}) - \frac{\pi^2}{3}i$ $= \boxed{\dfrac{i}{2} - \dfrac{i \pi^2}{3} + \ln\left(\dfrac{1}{(e^i-1)^2}\right) + \ln(1 - \cos(1)) + 2i \cdot \text{Li}_2(e^{i})}$