Answer
$\dfrac{1}{2} t -\dfrac{1}{12} \sin 6t+C$
Work Step by Step
Given: $ \dfrac{1}{2} \int 1 dt -\dfrac{1}{2} \int (\cos 6t) dt$
Thus, $\dfrac{1}{2} \int 1 dt -\dfrac{1}{2} \int (\cos 6t) dt= \dfrac{1}{2} t -(\dfrac{1}{2})(\dfrac{1}{6}) (\sin 6t) +C=\dfrac{1}{2} t -\dfrac{1}{12} \sin 6t+C$