I assume you mean #f(x) = int (3x - 3 secx)dx#.
So, We have,
#f(x) = int(3x - 3secx) dx#
#= 3int xdx - 3intsec xdx#
#= 3/2x^2 - 3ln|sec x + tan x| + C#
Now, According to the Question,
#color(white)(xxx)f((7pi)/4) = 0#
#rArr 3/2((7pi)/4)^2 - 3ln|sec ((7pi)/4) + tan ((7pi)/4)| + C = 0#
#rArr 3/2(22/4)^2 - 3ln|(-1/4) + 0| + C = 0#
#rArr 3/2 * 121/4 - 3(-1.39) + C = 0# [As #ln(1/4) = -1.39# (approx)]
#rArr 49.545 + C = 0# [Using Calculator]
#rArr C = -49.545#
So, #f(x) = 3/2x^2 - 3ln|sec x + tan x| - 49.545#
Hope this helps.
