Answer
$\dfrac {\cos ^{3}x}{\sin x}$
or
$\cot{x}-\sin{x}\cos{x}$
Work Step by Step
$\dfrac {d}{dx}\left( \ln \sin x-\dfrac {1}{2}\sin ^{2}x\right) =\dfrac {1}{\sin x}\times \left( \dfrac {d}{dx}\sin x\right) -\sin x\times \left( \dfrac {d}{dx}\sin x\right) =\cos x\left( \dfrac {1}{\sin x}-\sin x\right) =\dfrac {\cos ^{3}x}{\sin x}$
