By Parts Is Not Recommended

Relevant wiki: Integration Tricks

$\text{Using the Identity : } \displaystyle \int_{a}^{b} f(x)\cdot dx = \displaystyle \int_{a}^{b} f(a+b-x)\cdot dx \\ I = \displaystyle \int_{0}^{\pi} \sin{x} \cdot \ln{\left( \cot{\frac{x}{2}}\right)}\cdot dx \\ I = \displaystyle \int_{0}^{\pi} \sin{(\pi - x)} \cdot \ln{\left( \cot{\frac{\pi-x}{2}}\right)}\cdot dx \\ \text{Add the two integrals } \\ 2I= \displaystyle \int_{0}^{\pi} \sin{x} \cdot \ln{1} \cdot dx \\ \boxed{I = 0}$

the d theta is meant to be a du, and du is meant to be d theta for 1st piece of the problem as we're solving for d theta.

First, make that arguments of the trigonometric functions identical to $\frac{x}{2}=y$ , replace $\cot y = \frac{\cos y}{\sin y}$ , and distribute the logarithm: $\sin x \cdot \ln\left(\cot\frac{x}{2}\right) = 2 \sin y \cos y \ln \cos y - 2 \sin y \cos y \ln \sin y$ Using that $\sin y = \cos(\pi/2-y)$ , and similarly $\cos y = \sin(\pi/2-y)$ , this rewrites into $= 2 \sin y \cos y \ln \cos y - 2 \sin(\pi/2-y) \cos(\pi/2-y) \ln \cos(\pi/2-y)$ Now, observe that $y$ ranges from $0$ to $\pi/2$ , and that $\int_0^{\frac{\pi}{2}} f(y)\ dy = \int_0^{\frac{\pi}{2}} f(\pi/2 -y)\ dy$ . Thus, we find $\int_0^{\frac{\pi}{2}} 2 \sin y \cos y \ln \cos y\ dy = \int_0^{\frac{\pi}{2}} 2 \sin(\pi/2-y) \cos(\pi/2-y) \ln \cos(\pi/2-y)\ dy$ Hence, we have $\int_0^{\pi} \sin x\cdot\ln\left(\cot\frac{x}{2}\right) dx = 0$

The integrand is symmetrical respect to $x=\frac{\pi}2$ then the integral is 0.