How to find the antiderivative of f(x)=8sin(2x)

Sorry to ask again, and this must be such a simple problem but I haven't done this for a while. I think I'm over analyzing it. I was going to take the u as 2x and not sure what the du would be..sinx?but then there's no x to put there. I'm confusing myself, ahh!

And it gives you an initial value of F'(0)=5 when F'(x)=f(x). Please help!thank you!

f(x)=8sin(2x)
integrate
= 8 {-cos (2x)}/ 2+C
F(x) = -4 cos(2x) +C................(i)
when x=0
F(0) =5 = - 4cos (0) +C
5 = -4+C
C=9..........................(ii)
F(x) =-4 cos(2x) +9....................Ans

MY MUFUGGIN DIKK !!

GNOMSAYIN BIATTCH !!