Answer
$-\dfrac{\csc \theta}{2}+C$
Work Step by Step
As we are given that $\int (\dfrac{1}{2} \csc \theta \cot \theta) d\theta$
Thus,we have $\int (\dfrac{1}{2} \csc \theta \cot \theta) d\theta= \dfrac{1}{2}\int ( \csc \theta \cot \theta) d\theta$
or, $=-\dfrac{\csc \theta}{2}+C$