#### Answer

$$f\left( u \right) = 4\sin u + 2\cos 2u - 3$$

#### Work Step by Step

$$\eqalign{
& f'\left( u \right) = 4\left( {\cos u - \sin 2u} \right) \cr
& {\text{multiply}} \cr
& f'\left( u \right) = 4\cos u - 4\sin 2u \cr
& f\left( u \right) = \int {f'\left( u \right)} du \cr
& then \cr
& f\left( u \right) = \int {\left( {4\cos u - 4\sin 2u} \right)} du \cr
& find{\text{ the general solution}} \cr
& f\left( u \right) = 4\sin u + 2\cos 2u + C \cr
& {\text{using the initial condition }}f\left( {\pi /6} \right) = 0 \cr
& 0 = 4\sin \left( {\pi /6} \right) + 2\cos 2\left( {\pi /6} \right) + C \cr
& 0 = 4\left( {1/2} \right) + 2\left( {1/2} \right) + C \cr
& 0 = 2 + 1 + C \cr
& C = - 3 \cr
& {\text{the solution to the initial value problem is}} \cr
& f\left( u \right) = 4\sin u + 2\cos 2u - 3 \cr} $$