## Thomas' Calculus 13th Edition

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