If f ' (x) = 24x^2 + 6sec(x)tan(x) and f(0) = 3 then what is f(pi) = ?

If  f ' (x) = 24x^2 + 6sec(x)tan(x), then

f(x) = 8x^3 + 6sec(x) + C

And 

f(0)= 3   ....so

f(0) = 8(0)^3 + 6sec(0) + C  = 3

6 + C =  3

Then C = -3

So

f(pi)  =

8(pi)^3 + 6sec(pi) - 3 =

8(pi)^3 - 6 - 3

8(pi)^3 - 9  =  about 245.05

Nice work Chris, I had to think for a bit about that first integral, it is not one that I have memorized :/