Answer
$\dfrac{1}{2} t +\dfrac{1}{8} \sin 4t+C$
Work Step by Step
Solve. $ \dfrac{1}{2} \int 1 dt +\dfrac{1}{2} \int (\cos 4t) dt$
Thus, we have $ \dfrac{1}{2} \int 1 dt +\dfrac{1}{2} \int (\cos 4t) dt= \dfrac{1}{2} t +(\dfrac{1}{2})(\dfrac{1}{4}) (\sin 4t) +C$
or, $=(\dfrac{1}{2}) t +(\dfrac{1}{8}) \sin 4t+C$
