Answer
$\dfrac{1}{2} t +\dfrac{1}{8} \sin 4t+C$
Work Step by Step
Given: $ \dfrac{1}{2} \int 1 dt +\dfrac{1}{2} \int (\cos 4t) dt$
Thus, $ \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=\dfrac{1}{2} t +\dfrac{1}{8} \sin 4t+C$
