Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 272: 49

$-\dfrac{\csc \theta}{2}+C$

Given: $\int (\dfrac{1}{2} \csc \theta \cot \theta) d\theta$ Thus, $\int (\dfrac{1}{2} \csc \theta \cot \theta) d\theta= \dfrac{1}{2}\int ( \csc \theta \cot \theta) d\theta=-\dfrac{\csc \theta}{2}+C$