To solve this we must recall that int sin (u)du = -cos u +c, where u is some function of x. In this case our u is x^3, so du = 3x^2. Thus our first part of the integral will yield -cos (x^3)/3 +c; double checking this by differentiation, we do indeed receive 3x^2sinx^3/3 = x^2sinx^3.
For our second part, recall that cot u = cos u/sin u. Recall also that d/(du)sin u = (cos u) du. If our u is 2x, then du is 2. Recall further that d/(du) ln u = (du)/u. Using these facts...
int -cot2xdx = int -(cos2x)/(sin2x) = int -(du)/(2u) = -1/2 int (du)/u = -1/2 lnu + c = -(ln sin 2x)/2 +c
Thus our general f (x) is:
f (x) = -cosx^3/3 - ln (sin 2x)/2 + c
If f (pi/8) = -1, then...
f (pi/8) = -cos (pi^3/512)/3 -ln sin (pi/4)/2 +c = -1-> c = -1 + cos (pi^3/512)/3 + ln sin (pi/4)/2
Thus...
f (x)=-cosx^3/3 -ln sin (2x)/2 -1 + cos (pi^3/512)/3 + ln sin (pi/4)/2
This answer is somewhat lengthy yes, but note that -1 + cos (pi^3/512)/3 + ln sin (pi/4)/2 is in fact constant and can be calculated; this calculation is left to the student as many professors would simply accept the current form.