Answer
$$ - \frac{1}{4}\cos 4t + 4\cos \frac{t}{4} + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {\sin 4t - \sin \frac{t}{4}} \right)} dt \cr
& {\text{sum Rule}} \cr
& = \int {\sin 4t} dt - \int {\sin \frac{t}{4}} dt \cr
& {\text{use the formula for indefinite integrals of trigonometric functions}} \cr
& \int {\sin ax} dx = - \frac{1}{a}\cos ax + C \cr
& = - \frac{1}{4}\cos 4t - \left( { - \frac{1}{{1/4}}\cos \frac{t}{4}} \right) + C \cr
& {\text{Simplify}} \cr
& = - \frac{1}{4}\cos 4t + 4\cos \frac{t}{4} + C \cr
& {\text{check by differentiation}} \cr
& {\text{ = }}\frac{d}{{dt}}\left( { - \frac{1}{4}\cos 4t + 4\cos \frac{t}{4} + C} \right) \cr
& = - \frac{1}{4}\left( { - 4\sin 4t} \right) + 4\left( { - \frac{1}{4}\sin \frac{t}{4}} \right) + 0 \cr
& = \sin 4t - \sin \frac{t}{4} \cr} $$
