#### Answer

$\dfrac{1}{2} t +\dfrac{1}{12} \sin 6t+C$

#### Work Step by Step

Solve. $ \dfrac{1}{2} \int 1 dt -\dfrac{1}{2} \int (\cos 6t) dt$
Thus, we have $(\dfrac{1}{2}) \int 1 dt -\dfrac{1}{2} \int (\cos 6t) dt= \dfrac{1}{2} t -(\dfrac{1}{2})(\dfrac{1}{6}) (\sin 6t) +C$
or, $=(\dfrac{1}{2}) t +\dfrac{1}{12} \sin 6t+C$